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7.2: Multiplying and Dividing Rational Expressions

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Learning Objectives

• Multiply rational expressions.
• Divide rational expressions.
• Multiply and divide rational functions.

Multiplying Rational Expressions

When multiplying fractions, we can multiply the numerators and denominators together and then reduce, as illustrated:

\(\dfrac{3}{5} \cdot \dfrac{5}{9}=\dfrac{3 \cdot 5}{5 \cdot 9}=\dfrac{\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{3}}}}\color{black}{\cdot}\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{5}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{5}}}{1}}\color{black}{\cdot}\color{Cerulean}{\stackrel{\cancel{\color{black}{9}}}{3}}}\color{black}{=\dfrac{1}{3} }\)

Multiplying rational expressions is performed in a similar manner. For example,

\(\dfrac{y}{x} \cdot \dfrac{x}{y^{2}}=\dfrac{y \cdot x}{x \cdot y^{2}}=\dfrac{\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{y}}}}\color{black}{\cdot}\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{x}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{x}}}{1}}\color{black}{\cdot}\color{Cerulean}{\stackrel{\cancel{\color{black}{y^{2}}}}{3}}}\color{black}{=\dfrac{1}{y} }\)

In general, given polynomials \(P\), \(Q\), \(R\), and \(S\), where \(Q≠0\) and \(S≠0\), we have

\[\dfrac{P}{Q} \cdot \dfrac{R}{S}=\dfrac{P R}{Q S}\]

In this section, assume that all variable expressions in the denominator are nonzero unless otherwise stated.

Example \(\PageIndex{1}\)

\(\dfrac{12 x^{2}}{5 y^{3}} \cdot \dfrac{20 y^{4}}{6 x^{3}}\)

Multiply numerators and denominators and then cancel common factors.

\(\begin{aligned} \dfrac{12 x^{2}}{5 y^{3}} \cdot \dfrac{20 y^{4}}{6 x^{3}}&=\dfrac{240x^{2}y^{4}}{30x^{3}y^{3}}\qquad\quad\:\:\color{Cerulean}{Multiply.} \\&=\dfrac{\color{Cerulean}{\stackrel{8}{\cancel{\color{black}{240}}}}\color{Cerulean}{\stackrel{1}{ \cancel{\color{black}{x^{2}}}}}\color{Cerulean}{\stackrel{y}{ \cancel{\color{black}{y^{4}}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{30}}}{1}}\color{Cerulean}{ \stackrel{\cancel{\color{black}{x^{3}}}}{x}}\color{Cerulean}{ \stackrel{\cancel{\color{black}{y^{3}}}}{1}}} \qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{8 y}{x} \end{aligned}\)

\(\dfrac{8y}{x}\)

Example \(\PageIndex{2}\)

\(\dfrac{x-3}{x+5} \cdot \dfrac{x+5}{x+7}\)

Leave the product in factored form and cancel the common factors.

\(\begin{aligned} \dfrac{x-3}{x+5} \cdot \dfrac{x+5}{x+7} &=\dfrac{(x-3) \cdot\color{Cerulean}{\cancel{\color{black}{(x+5)}}}}{\color{Cerulean}{\cancel{\color{black}{(x+5)}}} \color{black}{\cdot(x+7)}} \\ &=\dfrac{x-3}{x+7} \end{aligned}\)

\(\dfrac{x-3}{x+7}\)

Example \(\PageIndex{3}\)

\(\dfrac{15 x^{2} y^{3}}{(2 x-1)} \cdot \dfrac{x(2 x-1)}{3 x^{2} y(x+3)}\)

Leave the polynomials in the numerator and denominator factored so that we can cancel the factors. In other words, do not apply the distributive property.

\(\begin{aligned} \dfrac{15 x^{2} y^{3}}{(2 x-1)} \cdot \dfrac{x(2 x-1)}{3 x^{2} y(x+3)}&=\dfrac{15 x^{3} y^{3}(2 x-1)}{3 x^{2} y(2 x-1)(x+3)}\qquad\qquad\color{Cerulean}{Multiply.} \\ &=\dfrac{\color{Cerulean}{\stackrel{5}{\cancel{\color{black}{15}}}}\color{Cerulean}{\stackrel{x}{\cancel{\color{black}{x^{3}}}}}\color{Cerulean}{\stackrel{y^{2}}{\cancel{\color{black}{y^{3}}}}}\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{(2x-1)}}}}}{\color{Cerulean}{\cancel{\color{black}{3}}\cancel{\color{black}{x^{2}}}\cancel{\color{black}{y}}\cancel{\color{black}{(2x-1)}}}\color{black}{(x+3)}}\qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{5xy^{2}}{x+3} \end{aligned}\)

\(\dfrac{5 x y^{2}}{x+3}\)

Typically, rational expressions will not be given in factored form. In this case, first factor all numerators and denominators completely. Next, multiply and cancel any common factors, if there are any.

Example \(\PageIndex{4}\)

\(\dfrac{x+5}{x-5} \cdot \dfrac{x-5}{x^{2}-25}\)

Factor the denominator \(x^{2}−25\) as a difference of squares. Then multiply and cancel.

\(\begin{aligned} \dfrac{x+5}{x-5} \cdot \dfrac{x-5}{x^{2}-25} &=\dfrac{x+5}{x-5} \cdot \dfrac{x-5}{(x+5)(x-5)} \qquad\qquad\color{Cerulean}{Factor.}\\ &=\dfrac{\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{(x+5)}}}}\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{(x-5)}}}}}{\color{Cerulean}{\cancel{\color{black}{(x-5)}}}\color{Cerulean}{\cancel{\color{black}{(x+5)}}}\color{black}{(x-5)}}\qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{1}{x-5} \end{aligned}\)

Keep in mind that 1 is always a factor; so when the entire numerator cancels out, make sure to write the factor 1.

\(\dfrac{1}{x-5}\)

Example \(\PageIndex{5}\)

It is a best practice to leave the final answer in factored form.

\(\dfrac{(x+2)(x-4)}{(x-2)(x+7)}\)

Example \(\PageIndex{6}\)

The trinomial \(−2x^{2}+x+3\) in the numerator has a negative leading coefficient. Recall that it is a best practice to first factor out a \(−1\) and then factor the resulting trinomial.

\(-\dfrac{3(2 x-3)}{x(x+4)}\)

Example \(\PageIndex{7}\)

\(\dfrac{7-x}{x^{2}+3 x} \cdot \dfrac{x^{2}+10 x+21}{x^{2}-49}\)

We replace \(7−x\) with \(−1(x−7)\) so that we can cancel this factor.

\(\begin{aligned} \dfrac{7-x}{x^{2}+3 x} \cdot \dfrac{x^{2}+10 x+21}{x^{2}-49} &=\dfrac{-1(x-7)}{x(x+3)} \cdot \dfrac{(x+3)(x+7)}{(x+7)(x-7)} \\ &=\dfrac{-1\color{Cerulean}{\cancel{\color{black}{(x-7)}}\cancel{\color{black}{(x+3)}}\cancel{\color{black}{(x+7)}}}}{x\color{Cerulean}{\cancel{\color{black}{(x+3)}}\cancel{\color{black}{(x+7)}}\cancel{\color{black}{(x-7)}}}} \\ &=\dfrac{-1}{x} \\ &=-\dfrac{1}{x} \end{aligned}\)

\(-\dfrac{1}{x}\)

Exercise \(\PageIndex{1}\)

\(\dfrac{x^{2}-64}{8-x} \cdot \dfrac{x+x^{2}}{x^{2}+9 x+8}\)

Dividing Rational Expressions

To divide two fractions, we multiply by the reciprocal of the divisor, as illustrated:

\(\dfrac{5}{8} \div \color{OliveGreen}{ \dfrac{1}{2}}\color{black}{=}\dfrac{5}{8} \cdot \color{OliveGreen}{\dfrac{2}{1}}\color{black}{=}\dfrac{5 \cdot \color{Cerulean}{\stackrel{1}{\cancel{\color{black}{2}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{8}}}{4}} \color{black}{\cdot 1}}=\dfrac{5}{4}\)

Dividing rational expressions is performed in a similar manner. For example,

\(\dfrac{x}{y^{2}} \div \color{OliveGreen}{ \dfrac{1}{y}}\color{black}{=}\dfrac{x}{y^{2}} \cdot \color{OliveGreen}{\dfrac{y}{1}}\color{black}{=}\dfrac{x \cdot \color{Cerulean}{\stackrel{1}{\cancel{\color{black}{y}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{y^{2}}}}{y}} \color{black}{\cdot 1}}=\dfrac{x}{y}\)

In general, given polynomials P , Q , R , and S , where \(Q≠0\), \(R≠0\), and \(S≠0\), we have

\[\dfrac{P}{Q} \div \dfrac{R}{S}=\dfrac{P}{Q} \cdot \dfrac{S}{R}=\dfrac{P S}{Q R}\]

Example \(\PageIndex{8}\)

\(\dfrac{8 x^{5} y}{25 z^{6}} \div \dfrac{20 x y^{4}}{15 z^{3}}\)

First, multiply by the reciprocal of the divisor and then cancel.

\(\begin{aligned} \dfrac{8 x^{5} y}{25 z^{6}} \div \color{Cerulean}{\dfrac{20 x y^{4}}{15 z^{3}}} &\color{black}{=}\dfrac{8 x^{5} y}{25 z^{6}} \cdot\color{Cerulean}{ \dfrac{15 z^{3}}{20 x y^{4}}}\qquad\color{Cerulean}{}Multiply\:by\:the\:reciprocal\:of\:the\:divisor. \\ &=\dfrac{120 x^{5} y z^{3}}{500 x y^{4} z^{6}}\\ &=\dfrac{\color{Cerulean}{\stackrel{6}{\cancel{\color{black}{120}}}\stackrel{x^{4}}{\cancel{\color{black}{x^{5}}}}\cancel{\color{black}{y}}\cancel{\color{black}{z^{3}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{500}}}{25}\cancel{\color{black}{x}}\stackrel{\cancel{\color{black}{y^{4}}}}{y^{3}}\stackrel{\cancel{\color{black}{z^{6}}}}{z^{3}}}}\qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{6x^{4}}{25y^{3}z^{3}} \end{aligned}\)

\(\dfrac{6 x^{4}}{25 y^{3} z^{3}}\)

Example \(\PageIndex{9}\)

\(\dfrac{x+2}{x^{2}-4} \div \dfrac{x+3}{x-2}\)

After multiplying by the reciprocal of the divisor, factor and cancel.

\(\begin{aligned} \dfrac{x+2}{x^{2}-4} \div \color{Cerulean}{\dfrac{x+3}{x-2}} &=\dfrac{x+2}{x^{2}-4} \cdot \color{Cerulean}{\dfrac{x-2}{x+3}}\qquad\quad\:\qquad\qquad\color{Cerulean}{Multiply\:by\:the\:reciprocal\:of\:the\:divisor.} \\ &=\dfrac{(x+2)}{(x+2)(x-2)} \cdot \dfrac{(x-2)}{(x+3)}\qquad\quad\color{Cerulean}{Factor.} \\ &=\dfrac{\color{Cerulean}{\cancel{\color{black}{(x+2)}}\cancel{\color{black}{(x-2)}}}}{\color{Cerulean}{\cancel{\color{black}{(x+2)}}\cancel{\color{black}{(x-2)}}}\color{black}{(x+3)}}\qquad\quad\color{Cerulean}{Cancel.} \\ &=\dfrac{1}{x+3} \end{aligned}\)

\(\dfrac{1}{x+3}\)

Example \(\PageIndex{10}\)

Begin by multiplying by the reciprocal of the divisor. After doing so, factor and cancel.

\(\dfrac{(x-8)(x-5)}{(x+7)^{2}}\)

Example \(\PageIndex{11}\)

Just as we do with fractions, think of the divisor \((2x−3)\) as an algebraic fraction over 1.

\(-\dfrac{2 x+3}{x+2}\)

Exercise \(\PageIndex{2}\)

\(-4 x^{3}-8 x^{2}\)

Multiplying and Dividing Rational Functions

The product and quotient of two rational functions can be simplified using the techniques described in this section. The restrictions to the domain of a product consist of the restrictions of each function.

Example \(\PageIndex{12}\)

Calculate \((f⋅g)(x)\) and determine the restrictions to the domain.

In this case, the domain of \(f(x)\) consists of all real numbers except 0, and the domain of \(g(x)\) consists of all real numbers except \(\dfrac{1}{4}\).

Therefore, the domain of the product consists of all real numbers except 0 and \(\dfrac{1}{4}\). Multiply the functions and then simplify the result.

\((f \cdot g)(x)=-\dfrac{4 x+1}{5 x}\), where \(x\neq 0, \dfrac{1}{4}\)

The restrictions to the domain of a quotient will consist of the restrictions of each function as well as the restrictions on the reciprocal of the divisor.

Example \(\PageIndex{13}\)

Calculate \((f/g)(x)\) and determine the restrictions.

In this case, the domain of \(f(x)\) consists of all real numbers except 3 and 8, and the domain of \(g(x)\) consists all real numbers except 3. In addition, the reciprocal of \(g(x)\) has a restriction of −8. Therefore, the domain of this quotient consists of all real numbers except 3, 8, and −8.

\((f / g)(x) = 1\), where \(x\neq 3, 8, -8\)

Key Takeaways

• After multiplying rational expressions, factor both the numerator and denominator and then cancel common factors. Make note of the restrictions to the domain. The values that give a value of 0 in the denominator are the restrictions.
• To divide rational expressions, multiply by the reciprocal of the divisor.
• The restrictions to the domain of a product consist of the restrictions to the domain of each factor.
• The restrictions to the domain of a quotient consist of the restrictions to the domain of each rational expression as well as the restrictions on the reciprocal of the divisor.

Exercise \(\PageIndex{2}\) Multiplying Rational Expressions

Multiply. (Assume all denominators are nonzero.)

• \(\dfrac{2 x}{3} \cdot \dfrac{9}{4 x^{2}}\)
• \(-\dfrac{5 x}{3 y} \cdot \dfrac{y^{2}}{25 x}\)
• \(\dfrac{5 x^{2}}{2 y} \cdot \dfrac{4 y^{2}}{15 x^{3}}\)
• \(\dfrac{16 a^{4}}{7 b^{2}} \cdot \dfrac{49 b^{3}}{2 a^{3}}\)
• \(\dfrac{x-6}{12 x^{3}} \cdot \dfrac{24 x^{2}}{x-6}\)
• \(\dfrac{x+10}{2x−1}\cdot\dfrac{x−2}{x+10}\)
• \(\dfrac{(y-1)^{2}}{y+1} \cdot \dfrac{1}{y-1}\)
• \(\dfrac{y^{2}-9}{y+3} \cdot \dfrac{2 y-3}{y-3}\)
• \(\dfrac{2 a-5}{a-5} \cdot \dfrac{2 a+5}{4 a^{2}-25}\)
• \(\dfrac{2 a^{2}-9 a+4}{a^{2}-16} \cdot\left(a^{2}+4 a\right)\)
• \(\dfrac{2 x^{2}+3 x-2}{(2 x-1)^{2}} \cdot \dfrac{2 x}{x+2}\)
• \(\dfrac{9x^{2}+19x+2}{4−x^{2}}\cdot\dfrac{x^{2}−4x+4}{9x^{2}−8x−1}\)
• \(\dfrac{x^{2}+8x+16}{16−x^{2}}\cdot\dfrac{x^{2}−3x−4}{x^{2}+5x+4}\)
• \(\dfrac{x^{2}−x−2}{x^{2}+8x+7}\cdot\dfrac{x^{2}+2x−15}{x^{2}−5x+6}\)
• \(\dfrac{x+1}{x−3}\cdot\dfrac{3−x}{x+5}\)
• \(\dfrac{2x−1}{x−1}\cdot\dfrac{x+6}{1−2x}\)
• \(\dfrac{9+x}{3x+1}\cdot\dfrac{3}{x+9}\)
• \(\dfrac{1}{2+5x}\cdot\dfrac{5x+2}{5x}\)
• \(\dfrac{100-y^{2}}{y-10} \cdot \dfrac{25 y^{2}}{y+10}\)
• \(\dfrac{3 y^{3}}{6 y-5} \cdot \dfrac{36 y^{2}-25}{5+6 y}\)
• \(\dfrac{3 a^{2}+14 a-5}{a^{2}+1} \cdot \dfrac{3 a+1}{1-9 a^{2}}\)
• \(\dfrac{4a^{2}−16a}{4a−1}\cdot\dfrac{1−16a^{2}}{4a^{2}−15a−4}\)
• \(\dfrac{x+9}{-x^{2}+14 x-45} \cdot\left(x^{2}-81\right) \)
• \(\dfrac{1}{2+5 x} \cdot\left(25 x^{2}+20 x+4\right)\)
• \(\dfrac{x^{2}+x−6}{3x^{2}+15x+18}\cdot\dfrac{2x^{2}−8}{x^{2}−4x+4}\)
• \(\dfrac{5x^{2}−4x−1}{5x^{2}−6x+1}\cdot\dfrac{25x^{2}−10x+1}{3−75x^{2}}\)

1. \(\dfrac{3}{2x}\)

3. \(\dfrac{2y}{3x}\)

5. \(\dfrac{2}{x}\)

7. \(\dfrac{y−1}{y+1}\)

9. \(\dfrac{1}{a−5}\)

11. \(\dfrac{2x}{2x−1}\)

13. \(−1\)

15. \(−\dfrac{x+1}{x+5}\)

17. \(\dfrac{3}{3x+1}\)

19. \(−25y^{2}\)

21. \(-\dfrac{a+5}{a^{2}+1}\)

23. \(-\dfrac{(x+9)^{2}}{x-5}\)

25. \(\dfrac{2}{3}\)

Exercise \(\PageIndex{3}\) Dividing Rational Expressions

Divide. (Assume all denominators are nonzero.)

• \(\dfrac{5 x}{8} \div \dfrac{15 x^{2}}{4}\)
• \(\dfrac{3}{8 y} \div \dfrac{15}{2 y^{2}}\)
• \(\dfrac{\dfrac{5 x^{9}}{3 y^{3}}}{\dfrac{25 x^{10}}{9 y^{5}}}\)
• \(\dfrac{\dfrac{12 x^{4} y^{2}}{21 z^{5}}}{\dfrac{6 x^{3} y^{2}}{7 z^{3}}}\)
• \(\dfrac{(x-4)^{2}}{30 x^{4}} \div \dfrac{x-4}{15 x}\)
• \(\dfrac{5 y^{4}}{10(3 y-5)^{2}} \div \dfrac{10 y^{5}}{2(3 y-5)^{3}}\)
• \(\dfrac{x^{2}-9}{5 x} \div(x-3)\)
• \(\dfrac{y^{2}-64}{8 y} \div(8+y)\)
• \(\dfrac{(a-8)^{2}}{2 a^{2}+10 a} \div \dfrac{a-8}{a}\)
• \(\dfrac{2}{4 a^{2} b^{3}(a-2 b)} \div 12 a b(a-2 b)^{5}\)
• \(\dfrac{x^{2}+7 x+10}{x^{2}+4 x+4} \div \dfrac{1}{x^{2}-4}\)
• \(\dfrac{2 x^{2}-x-1}{2 x^{2}-3 x+1} \div \dfrac{1}{4 x^{2}-1}\)
• \(\dfrac{y+1}{y^{2}-3 y} \div \dfrac{y^{2}-1}{y^{2}-6 y+9}\)
• \(\dfrac{9-a^{2}}{a^{2}-8 a+15} \div \dfrac{2 a^{2}-10 a}{a^{2}-10 a+25}\)
• \(\dfrac{a^{2}-3 a-18}{2 a^{2}-11 a-6} \div \dfrac{a^{2}+a-6}{2 a^{2}-a-1}\)
• \(\dfrac{y^{2}-7 y+10}{y^{2}+5 y-14} \div \dfrac{2 y^{2}-9 y-5}{y^{2}+14 y+49}\)
• \(\dfrac{6 y^{2}+y-1}{4 y^{2}+4 y+1} \div \dfrac{3 y^{2}+2 y-1}{2 y^{2}-7 y-4}\)
• \(\dfrac{x^{2}−7x−18}{x^{2}+8x+12}\div\dfrac{x^{2}−81}{x^{2}+12x+36}\)
• \(\dfrac{4a^{2}−b^{2}}{b+2a}\div (b−2a)^{2}\)
• \(\dfrac{x^{2}−y^{2}}{y+x}\div (y−x)^{2}\)
• \(\dfrac{5 y^{2}(y-3)}{4 x^{3}} \div \dfrac{25 y(3-y)}{2 x^{2}}\)
• \(\dfrac{15 x^{3}}{3(y+7)} \div \dfrac{25 x^{6}}{9(7+y)^{2}} \)
• \(\dfrac{3 x+4}{x-8} \div \dfrac{7 x}{8-x}\)
• \(\dfrac{3x−2}{2x+1}\div \dfrac{2−3x}{3x}\)
• \(\dfrac{(7 x-1)^{2}}{4 x+1} \div \dfrac{28 x^{2}-11 x+1}{1-4 x}\)
• \(\dfrac{4 x}{(x+2)^{2}} \div \dfrac{2-x}{x^{2}-4}\)
• \(\dfrac{a^{2}-b^{2}}{a} \div(b-a)^{2}\)
• \(\dfrac{(a−2b)^{2}}{2b}\div (2b^{2}+ab−a^{2})\)
• \(\dfrac{x^{2}−6x+9}{x^{2}+7x+12}\div \dfrac{9−x^{2}}{x^{2}+8x+16}\)
• \(\dfrac{2x^{2}−9x−5}{25−x^{2}}\div \dfrac{1−4x+4x^{2}}{−2x^{2}−9x+5}\)
• \(\dfrac{3x^{2}−16x+5}{100−4x^{2}}\div\dfrac{ 9x^{2}−6x+1}{3x^{2}+14x−5}\)
• \(\dfrac{10x^{2}−25x−15}{x^{2}−6x+9}\div\dfrac{9−x^{2}}{x^{2}+6x+9}\)

1. \(\dfrac{1}{6x}\)

3. \(\dfrac{3y^{2}}{5x}\)

5. \(\dfrac{x−4}{2x3}\)

7. \(\dfrac{x+3}{5x}\)

9. \(\dfrac{a−8}{2(a+5)}\)

11. \((x+5)(x−2)\)

13. \(\dfrac{y−3}{y(y−1)}\)

15. \(\dfrac{a−1}{a−2}\)

17. \(\dfrac{y−4}{y+1}\)

19. \(\dfrac{1}{2a−b}\)

21. \(−\dfrac{y}{10x}\)

23. \(−\dfrac{3x+4}{7x}\)

25. \(−\dfrac{7x−1}{4x+1}\)

27. \(\dfrac{a+b}{a(a-b)}\)

29. \(−\dfrac{(x−3)(x+4)}{(x+3)^{2}}\)

31. \(−\dfrac{1}{4}\)

Exercise \(\PageIndex{4}\) Dividing Rational Expressions

Recall that multiplication and division are to be performed in the order they appear from left to right. Simplify the following.

• \(\dfrac{1}{x^{2}} \cdot \dfrac{x-1}{x+3} \div \dfrac{x-1}{x^{3}}\)
• \(\dfrac{x−7}{x+9}\cdot\dfrac{1}{x^{3}}\div\dfrac{x−7}{x}\)
• \(\dfrac{x+1}{x−2}\div\dfrac{x}{x−5}\cdot\dfrac{x^{2}}{x+1}\)
• \(\dfrac{x+4}{2x+5}\div \dfrac{x−3}{2x+5}\cdot\dfrac{x+4}{x−3}\)
• \(\dfrac{2x−1}{x+1}\div\dfrac{x−4}{x^{2}+1}\cdot\dfrac{x−4}{2x−1}\)
• \(\dfrac{4x^{2}−1}{3x+2}\div\dfrac{2x−1}{x+5}\cdot\dfrac{3x+2}{2x+1}\)

1. \(\dfrac{x}{x+3}\)

3. \(\dfrac{x(x−5)}{x−2}\)

5. \(\dfrac{x^{2}+1}{x+1}\)

Exercise \(\PageIndex{5}\) Multiplying and Dividing Rational Functions

• \(f(x)=\dfrac{1}{x}\) and \(g(x)=\dfrac{1}{x−1}\)
• \(f(x)=\dfrac{x+1}{x−1}\) and \(g(x)=x^{2}−1\)
• \(f(x)=\dfrac{3x+2}{x+2}\) and \(g(x)=\dfrac{x^{2}−4}{(3x+2)^{2}}\)
• \(f(x)=\dfrac{(1−3x)}{2x−6}\) and \(g(x)=\dfrac{(x−6)^{2}}{9x^{2}−1}\)
• \(f(x)=\dfrac{25x^{2}−1}{x^{2}+6x+9}\) and \(g(x)=\dfrac{x^{2}−9}{5x+1}\)
• \(f(x)=\dfrac{x^{2}−49}{2x^{2}+13x−7}\) and \(g(x)=\dfrac{4x^{2}−4x+1}{7−x}\)

1. \((f⋅g)(x)=\dfrac{1}{x(x−1)} ; x≠0, 1 \)

3. \((f⋅g)(x)=\dfrac{x−2}{3x+2}; x≠−2, −\dfrac{2}{3}\)

5. \((f⋅g)(x)=\dfrac{(x−3)}{(5x−1)x+3}; x≠−3, −\dfrac{1}{5}\)

Exercise \(\PageIndex{6}\) Multiplying and Dividing Rational Functions

Calculate \((f/g)(x)\) and state the restrictions.

• \(f(x)=\dfrac{1}{x}\) and \(g(x)=\dfrac{x−2}{x−1}\)
• \(f(x)=\dfrac{(5x+3)^{2}}{x^{2}}\) and \(g(x)=\dfrac{5x+3}{6−x}\)
• \(f(x)=\dfrac{5−x}{(x−8)^{2}}\) and \(g(x)=\dfrac{x^{2}−2}{5x−8}\)
• \(f(x)=\dfrac{x^{2}−2x−1}{5x^{2}−3x−10}\) and \(g(x)=\dfrac{2x^{2}−5x−3}{x^{2}−7x+12}\)
• \(f(x)=\dfrac{3x^{2}+11x−4}{9x^{2}−6x+1}\) and \(g(x)=\dfrac{x^{2}−2x+1}{3x^{2}−4x+1}\)
• \(f(x)=\dfrac{36−x^{2}}{x^{2}+12x+36}\) and \(g(x)=\dfrac{x^{2}−12x+3}{6x^{2}+4x−12}\)

1. \((f/g)(x)=\dfrac{x−1}{x(x−2)}; x≠0, 1, 2\)

3. \((f/g)(x)=−\dfrac{1}{(x−8)(x+5)}; x≠±5, 8\)

5. \((f/g)(x)=\dfrac{(x+4)}{(x−1)}; x≠\dfrac{1}{3}, 1\)

Exercise \(\PageIndex{7}\) Discussion Board Topics

• In the history of fractions, who is credited for the first use of the fraction bar?
• How did the ancient Egyptians use fractions?
• Explain why \(x=7\) is a restriction to \(\dfrac{1}{x}\div\dfrac{x−7}{x−2}\).

1. Answer may vary

3. Answer may vary

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Relationship Between Multiplication and Division - Lesson Plan

In this interactive math lesson, students will explore the relationship between multiplication and division. they will discover how these two operations are connected and learn strategies for solving problems using their relationship. through hands-on activities and problem-solving tasks, students will develop a deep understanding of the connection between multiplication and division..

Know more about Relationship Between Multiplication and Division - Lesson Plan

Multiplication and division are inverse operations. Multiplication is repeated addition, while division is repeated subtraction or sharing equally. The product of a multiplication equation can be divided to find the original numbers.

Understanding the relationship between multiplication and division helps students solve problems more efficiently. It allows them to use known facts to find unknowns, simplify calculations, and check their answers.

Students can use fact families, number bonds, arrays, or models to relate multiplication and division. They can also think of real-world situations that involve equal sharing or grouping to understand how these operations are connected.

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Chapter 2, Lesson 3: Solving Equations by Using Multiplication and Division

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Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations

If you are finding the resource to learn and practice Grade 6 math then you reach the correct place. Gomathanswerkey provides the study materials for Grade 6 Chapter 8 which assist students to score full marks in their exams. Our material contains a brief explanation along with the graphs here. The answers are created by subject matter experts. Our Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations is helpful for quick revision and easy to understand.

Download Go Math Grade 6 Chapter 8 Solution of Equations Answer Key PDF for free and start your practice. If you keep practicing you will never end to love the maths. So, to get the best way of learning must refer to Go Math 6th grade 8th Chapter Solutions of Equations Answer Key.

Click on the below attached links and then go through the detailed stepwise Go Math Grade 6 Answer Key Chapter 8 Solution of Equations. Get the solutions for homework problems, Mid Chapter, and Review here.

Lesson 1: Solutions of Equations

Share and Show – Page No. 423

Problem solving + applications – page no. 424, solutions of equations – page no. 425, lesson check – page no. 426.

Lesson 2: Write Equations

Share and Show – Page No. 429

Problem solving + applications – page no. 430, write equations – page no. 431, lesson check – page no. 432.

Lesson 3: Investigate • Model and Solve Addition Equations

Share and Show – Page No. 435

Problem solving + applications – page no. 436, model and solve addition equations – page no. 437, lesson check – page no. 438.

Lesson 4: Solve Addition and Subtraction Equations

Share and Show – Page No. 441

Unlock the problem – page no. 442, solve addition and subtraction equations – page no. 443, lesson check – page no. 444.

Lesson 5: Investigate • Model and Solve Multiplication Equations

Share and Show – Page No. 447

Page no. 448, model and solve multiplication equations – page no. 449, lesson check – page no. 450.

Lesson 6: Solve Multiplication and Division Equations

Share and Show – Page No. 453

Problem solving + applications – page no. 454, solve multiplication and division equations – page no. 455, lesson check – page no. 456.

Lesson 7: Problem Solving • Equations with Fractions

Share and Show – Page No. 459

On your own – page no. 460, problem solving equations with fractions – page no. 461, lesson check – page no. 462.

Mid-Chapter Checkpoint

Mid-Chapter Checkpoint – Vocabulary – Page No. 463

• Mid-Chapter Checkpoint – Page No. 464

Lesson 8: Solutions of Inequalities

Share and Show – Page No. 467

• Problem Solving + Applications – Page No. 468

Solutions of Inequalities – Page No. 469

Lesson check – page no. 470.

Lesson 9: Write Inequalities

Share and Show – Page No. 473

Make generalizations – page no. 474, write inequalities – page no. 475, lesson check – page no. 476.

Lesson 10: Graph Inequalities

Share and Show – Page No. 479

Problem solving + applications – page no. 480, graph inequalities – page no. 481, lesson check – page no. 482.

Chapter 8 Review/Test

Chapter 8 Review/Test – Page No. 483

• Chapter 8 Review/Test – Page No. 484

Chapter 8 Review/Test Page No. 485

Chapter 8 review/test – page no. 486, chapter 8 review/test page no. 487, chapter 8 review/test page no. 488.

Determine whether the given value of the variable is a solution of the equation.

Question 1. x + 12 = 29; x = 7 The variable is __________

Answer: not a solution

Explanation: Substitute the value in the given equation x + 12 = 29 If x = 7 7 + 12 = 29 19 ≠ 29 Thus the variable is not a solution.

Question 2. n − 13 = 2; n = 15 The variable is __________

Answer: a solution

Explanation: Substitute the value in the given equation n = 15 n − 13 = 2 15 – 13 = 2 The variable is a solution.

Question 3. \(\frac{1}{2}\)c = 14; c = 28 The variable is __________

Explanation: Substitute the value in the given equation c = 28 \(\frac{1}{2}\)c = 14 \(\frac{1}{2}\) × 28 = 14 14 = 14 Thus the variable is a solution.

Question 4. m + 2.5 = 4.6; m = 2.9 The variable is __________

Explanation: Substitute the value in the given equation m + 2.5 = 4.6 m = 2.9 2.9 + 2.5 = 4.6 5.4 ≠ 4.6 Thus the variable is not a solution.

Question 5. d − 8.7 = 6; d = 14.7 The variable is __________

Explanation: Substitute the value in the given equation d = 14.7 d − 8.7 = 6 14.7 – 8.7 = 6 6 = 6 Thus the variable is a solution.

Question 6. k − \(\frac{3}{5}\) = \(\frac{1}{10}\); k = \(\frac{7}{10}\) The variable is __________

Explanation: Substitute the value in the given equation k = \(\frac{7}{10}\) k − \(\frac{3}{5}\) = \(\frac{1}{10}\) \(\frac{7}{10}\) – \(\frac{3}{5}\) = \(\frac{1}{10}\) \(\frac{7}{10}\) – \(\frac{6}{10}\) = \(\frac{1}{10}\) \(\frac{1}{10}\) = \(\frac{1}{10}\) Thus the variable is a solution.

On Your Own

Question 7. 17.9 + v = 35.8; v = 17.9 The variable is __________

Explanation: Substitute the value in the given equation 17.9 + v = 35.8 v = 17.9 17.9 + 17.9 = 35.8 35.8 = 35.8 Thus the variable is a solution.

Lesson 8.1 Answer Key 6th Grade Question 8. c + 35 = 57; c = 32 The variable is __________

Explanation: Substitute the value in the given equation c + 35 = 57 c = 32 32 + 35 = 57 67 ≠ 57 Thus the variable is not a solution.

Question 9. 18 = \(\frac{2}{3}\)h; h= 12 The variable is __________

Explanation: Substitute the value in the given equation 18 = \(\frac{2}{3}\)h h = 12 \(\frac{2}{3}\) × 12 = 8 18 ≠ 8 Thus the variable is not a solution.

Question 10. In the equation t + 2.5 = 7, determine whether t = 4.5, t = 5, or t = 5.5 is a solution of the equation. The solution is ________.

Answer: t = 4.5

Explanation: Substitute the value in the given equation t = 4.5 t + 2.5 = 7 4.5 + 2.5 = 7 7 = 7 t = 5 t + 2.5 = 7 5 + 2.5 = 7 7.5 ≠ 7 Not a solution t = 5.5 t + 2.5 = 7 5.5 + 2.5 = 7 8 ≠ 7 Not a solution

Question 11. Antonio ran a total of 9 miles in two days. The first day he ran 5 \(\frac{1}{4}\) miles. The equation 9 – d = 5 \(\frac{1}{4}\) can be used to find the distance d in miles Antonio ran the second day. Determine whether d = 4 \(\frac{3}{4}\), d = 4, or d = 3 \(\frac{3}{4}\) is a solution of the equation, and tell what the solution means. The solution is ________ \(\frac{□}{□}\)

Answer: 3 \(\frac{3}{4}\)

Explanation: 9 – d = 5 \(\frac{1}{4}\) Substitute d = 4 \(\frac{3}{4}\) in the above equation 9 – 4 \(\frac{3}{4}\) = 5 \(\frac{1}{4}\) 4 \(\frac{1}{4}\) ≠ 5 \(\frac{1}{4}\) Not a solution Substitute d = 4 9 – 4 = 5 \(\frac{1}{4}\) 5 ≠ 5 \(\frac{1}{4}\) Not a solution Substitute d = 3 \(\frac{3}{4}\) 9 – 3 \(\frac{3}{4}\) = 5 \(\frac{1}{4}\) 5 \(\frac{1}{4}\) = 5 \(\frac{1}{4}\) 9 – d = 5 \(\frac{1}{4}\); d = 3 \(\frac{3}{4}\) is a solution.

Question 12. Connect Symbols and Words The length of a day on Saturn is 14 hours less than a day on Mars. The equation 24.7 − s = 14 can be used to find the length in hours s of a day on Saturn. Determine whether s = 9.3 or s = 10.7 is a solution of the equation, and tell what the solution means. Type below: _____________

Answer: s = 10.7

Explanation: The length of a day on Saturn is 14 hours less than a day on Mars. The equation 24.7 − s = 14 can be used to find the length in hours s of a day on Saturn. 24.7 − s = 14 Substitute s = 9.3 in the equation 24.7 – 9.3 = 14 15.4 ≠ 14 Not a solution Substitute s = 10.7 in the equation 24.7 – 10.7 = 14 14 = 14 Therefore s = 10.7 is a solution to the equation.

Question 13. A storm on one of the planets listed in the table lasted for 60 hours, or 2.5 of the planet’s days. The equation 2.5h = 60 can be used to find the length in hours h of a day on the planet. Is the planet Earth, Mars, or Jupiter? Explain. Type below: _____________

Answer: Earth

Explanation: A storm on one of the planets listed in the table lasted for 60 hours, or 2.5 of the planet’s days. 2.5h = 60 h = 60/2.5 h = 24 hours By seeing the above table we can say that Earth is the answer.

Question 14. A day on Pluto is 143.4 hours longer than a day on one of the planets listed in the table. The equation 153.3 − p = 143.4 can be used to find the length in hours p of a day on the planet. What is the length of a storm that lasts \(\frac{1}{3}\) of a day on this planet? ________ hours

Answer: 3.3 hours

Explanation: A day on Pluto is 143.4 hours longer than a day on one of the planets listed in the table. 153.3 − p = 143.4 153.3 – 143.4 = p p = 153.3 – 143.4 p = 9.9 Now p with \(\frac{1}{3}\) to find the length of a storm that lasts of a day on this planet 9.9 × \(\frac{1}{3}\) = 3.3 hours

Question 15. What’s the Error? Jason said that the solution of the equation 2m = 4 is m = 8. Describe Jason’s error, and give the correct solution. Type below: _____________

Answer: m = 2

Explanation: Jason said that the solution of the equation 2m = 4 is m = 8. 2m = 4 m = 4/2 = 2 The error of Jason is he multiplied 2 and 4 but he should divide 4 by 2.

Question 16. The marking period is 45 school days long. Today is the twenty-first day of the marking period. The equation x + 21 = 45 can be used to find the number of days x left in the marking period. Using substitution, Rachel determines there are _____ days left in the marking period. Rachel determines there are _____________ days left.

Explanation: The marking period is 45 school days long. Today is the twenty-first day of the marking period. The equation x + 21 = 45 x = 45 – 21 = 24 days Using substitution, Rachel determines there are 24 days left in the marking period. Thus Rachel determines there are 24 days left.

Question 1. x − 7 = 15; x = 8 The variable is __________

Explanation: Substitute the value in the given equation. x = 8 8 – 7 = 15 1 ≠ 15 Therefore the variable is not a solution.

Question 2. c + 11 = 20; c = 9 The variable is __________

Explanation: Substitute the value in the given equation. c = 9 9 + 11 = 20 20 = 20 Therefore the variable is a solution.

Question 3. \(\frac{1}{3}\)h = 6; h = 2 The variable is __________

Explanation: Substitute the value in the given equation. \(\frac{1}{3}\)h = 6 h = 2 \(\frac{1}{3}\) × 2 = 6 \(\frac{2}{3}\) ≠ 6 Therefore the variable is not a solution.

Question 4. 16.1 + d = 22; d = 6.1 The variable is __________

Explanation: Substitute the value in the given equation. 16.1 + d = 22 d = 6.1 16.1 + 6.1 = 22 22.2 ≠ 22 Therefore the variable is not a solution.

Question 5. 9 = \(\frac{3}{4}\)e; e = 12 The variable is __________

Explanation: Substitute the value in the given equation. 9 = \(\frac{3}{4}\)e e = 12 9 = \(\frac{3}{4}\)(12) 9 = 3 × 3 9 = 9 Therefore the variable is a solution.

Question 6. 15.5 – y = 7.9; y = 8.4 The variable is __________

Explanation: Substitute the value in the given equation. 15.5 – y = 7.9 y = 8.4 15.5 – 8.4 = 7.9 7.1 ≠ 7.9 Therefore the variable is not a solution.

Problem Solving

Question 7. Terrance needs to score 25 points to win a game. He has already scored 18 points. The equation 18 + p = 25 can be used to find the number of points p that Terrance still needs to score. Determine whether p = 7 or p = 13 is a solution of the equation, and tell what the solution means. Type below: _____________

Answer: p = 7

Explanation: Terrance needs to score 25 points to win a game. He has already scored 18 points. The equation is 18 + p = 25 Substitute p = 7 in the above equation. 18 + 7 = 25 25 = 25 The variable is a solution. Substitute p = 13 18 + p = 25 18 + 13 = 25 31 ≠ 25 The variable is not a solution. Therefore p = 7 is a solution for the equation.

Lesson 8 Problem Set 8.1 Answer Key Question 8. Madeline has used 50 sheets of a roll of paper towels, which is \(\frac{5}{8}\) of the entire roll. The equation \(\frac{5}{8}\)s = 50 can be used to find the number of sheets s in a full roll. Determine whether s = 32 or s = 80 is a solution of the equation, and tell what the solution means. Type below: _____________

Answer: Madeline has used 50 sheets of a roll of paper towels, which is \(\frac{5}{8}\) of the entire roll. \(\frac{5}{8}\)s = 50 s = 50 × \(\frac{8}{5}\) s = 80 because 80 × 5 = 400 400 ÷ 8 = 50

Question 9. Use mental math to find the solution to 4x = 36. Then use substitution to check your answer. Type below: _____________

Answer: x = 9

Explanation: 4x = 36 x = 36/4 x = 9

Question 1. Sheena received a gift card for $50. She has already used it to buy a lamp for $39.99. The equation 39.99 + x = 50 can be used to find the amount x that is left on the gift card. What is the solution of the equation? _____

Answer: 10.01

Explanation: Given: Sheena received a gift card for $50. She has already used it to buy a lamp for $39.99. The equation 39.99 + x = 50 39.99 + x = 50 x = 50 – 39.99 x = 50.00 – 39.99 x = 10.01 Thus $10.01 is left on the gift card.

Question 2. When Pete had a fever, his temperature was 101.4°F. After taking some medicine, his temperature was 99.2°F. The equation 101.4 – d = 99.2 can be used to find the number of degrees d that Pete’s temperature decreased. What is the solution of the equation? _____

Answer: 2.2

Explanation: Given, When Pete had a fever, his temperature was 101.4°F. After taking some medicine, his temperature was 99.2°F. The equation 101.4 – d = 99.2 104.4 – 99.2 = d d = 104.4 – 99.2 d = 2.2

Spiral Review

Question 3. Melanie has saved $60 so far to buy a lawn mower. This is 20% of the price of the lawn mower. What is the full price of the lawn mower that she wants to buy? $ _____

Answer: 300

Explanation: Melanie has saved $60 so far to buy a lawn mower. This is 20% of the price of the lawn mower. 60 ÷ 20% 60 ÷ 20/100 60 × 100/20 = 6000/20 = 300 She wants to buy a $300 price of the lawn mower.

Chapter 8 Review Answer Key Question 4. A team of scientists is digging for fossils. The amount of soil in cubic feet that they remove is equal to 6³. How many cubic feet of soil do the scientists remove? _____ cubic feet

Answer: 216

Explanation: A team of scientists is digging for fossils. The amount of soil in cubic feet that they remove is equal to 6³. 6 × 6 × 6 = 216 Thus the scientists removed 216 cubic feet of soil.

Question 5. Andrew made p picture frames. He sold 2 of them at a craft fair. Write an expression that could be used to find the number of picture frames Andrew has left. Type below: _____________

Answer: p – 2

Explanation: Andrew made p picture frames. He sold 2 of them at a craft fair. The expression is the difference of 9 and 2 The equation is p – 2

Question 6. Write an expression that is equivalent to 4 + 3(5 + x). Type below: _____________

Answer: 4 + 15 + 3x

Explanation: 4 + 3(5 + x) = 4 + 15 + 3x 3x + 19 Thus the expression 4 + 3(5 + x) is equivalent to 4 + 15 + 3x or 3x + 19

Question 1. Write an equation for the word sentence “25 is 13 more than a number.” Type below: _____________

Answer: Let n represent the unknown number. The phrase ‘more than’ indicates an addition operation. Thus the equation is 25 = 13 + n.

Write an equation for the word sentence.

Question 2. The difference of a number and 2 is 3 \(\frac{1}{3}\). Type below: _____________

Answer: Let n represent the unknown number. The phrase “difference” indicates the subtraction operation. The equation is n – 2 = 3 \(\frac{1}{3}\)

Question 3. Ten times the number of balloons is 120. Type below: _____________

Answer: Let n represent the unknown number. The phrase “times” indicates multiplication operation. The equation is 10 × n = 120

Write a word sentence for the equation.

Question 4. x − 0.3 = 1.7 Type below: _____________

Answer: The difference between x and 0.3 is 1.7

Question 5. 25 = \(\frac{1}{4}\)n Type below: _____________

Answer: 25 is n times \(\frac{1}{4}\)

Question 6. The quotient of a number and 20.7 is 9. Type below: _____________

Answer: Let n represent the unknown number. The phrase “quotient” indicates the division operation. Thus the equation is n ÷ 20.7 = 9.

Question 7. 24 less than the number of snakes is 35. Type below: _____________

Answer: Let n represent the unknown number. The phrase “less than” indicates a subtraction operation. Thus the equation is n – 24 = 35

Question 8. 75 is 18 \(\frac{1}{2}\) more than a number. Type below: _____________

Answer: Let n represent the unknown number. The phrase “more than” indicates addition operation. 75 = 18 \(\frac{1}{2}\) + n

Question 9. d degrees warmer than 50 degrees is 78 degrees. Type below: _____________

Answer: Let n represent the unknown number. The phrase “warmer than” indicates additional operation. The equation is d + 50 = 78 degrees

Question 10. 15g = 135 Type below: _____________

Answer: g times 15 is 135

Question 11. w ÷ 3.3 = 0.6 Type below: _____________

Answer: The quotient of w and 3.3 is 0.6

Question 12. Write an equation that could be used to find how many miles a hybrid SUV can travel in the city on 20 gallons of gas. Type below: _____________

Answer: From table 36 miles per gallon in the city. A hybrid SUV uses 36 miles per gallon in the city. So, no. of miles = y x = no. of gallons So, y = 36 × x x = 20 gallons Thus y = 36 × 20

Question 13. A sedan traveled 504 miles on the highway on a full tank of gas. Write an equation that could be used to find the number of gallons the tank holds. Type below: _____________

Answer: A sedan uses 28 miles per gallon on the highway. The equation that could be used to find the number of gallons the tank holds is 504 = 28g

Question 14. Connect Symbols to Words Sonya was born in 1998. Carmen was born 11 years after Sonya. If you wrote an equation to find the year in which Carmen was born, what operation would you use in your equation? Type below: _____________

Answer: In this equation, I would use addition or subtraction operations.

Question 15. A magazine has 110 pages. There are 23 full-page ads and 14 half-page ads. The rest of the magazine consists of articles. Write an equation that can be used to find the number of pages of articles in the magazine. Type below: _____________

Answer: The equation that can be used to find the number of pages of articles in the magazine is 23 + 14/2 + a = 110 where a represents the number of articles.

Question 16. What’s the Error? Tony is traveling 560 miles to visit his cousins. He travels 313 miles on the first day. He says that he can use the equation m − 313 = 560 to find the number of miles m he has left on his trip. Describe and correct Tony’s error. Type below: _____________

Answer: Tony subtracted the number of miles traveled from the number of miles left. Tony should have written m + 313 = 560

Question 17. Jamie is making cookies for a bake sale. She triples the recipe in order to have enough cookies to sell. Jamie uses 12 cups of flour to make the triple batch. Write an equation that can be used to find out how much flour f is needed for one batch of cookies. Type below: _____________

Answer: The equation that can be used to find out how much flour f is needed for one batch of cookies is 3f = 12

Question 1. 18 is 4.5 times a number. Type below: _____________

Answer: Let n represents the unknown number. The phrase “times” indicates the multiplication operation. The equation is 18 = 4.5n

Question 2. Eight more than the number of children is 24. Type below: _____________

Answer: Let c represents the number of children. The phrase “more than” indicates addition operation. Thus the equation is 8 + c = 24.

Question 3. The difference of a number and \(\frac{2}{3}\) is \(\frac{3}{8}\). Type below: _____________

Answer: Let n represents the unknown number. The phrase “difference” indicates a subtraction operation. The equation is n – \(\frac{2}{3}\) = \(\frac{3}{8}\)

Question 4. A number divided by 0.5 is 29. Type below: _____________

Answer: Let n represents the unknown number. The phrase divided by indicates division operation. The equation is n ÷ 0.5 = 29

Question 5. x − 14 = 52 Type below: _____________

Answer: 14 less than x is 52 the difference of x and 14 is 52 14 fewer than a number is 52.

Question 6. 2.3m = 0.46 Type below: _____________

Answer: The product of 2.3 and m is 0.46 2.3 times m is .46 2.3 of m is 0.46

Question 7. 25 = k ÷ 5 Type below: _____________

Answer: 25 is the quotient of k and 5.

Question 8. \(4 \frac{1}{3}+q=5 \frac{1}{6}\) Type below: _____________

Answer: The sum of \(4 \frac{1}{3}\) and q is [/latex]5 \frac{1}{6}[/latex] q is more than \(4 \frac{1}{3}\) and [/latex]5 \frac{1}{6}[/latex] \(4 \frac{1}{3}\) increased by a number is [/latex]5 \frac{1}{6}[/latex]

Question 9. An ostrich egg weighs 2.9 pounds. The difference between the weight of this egg and the weight of an emu egg is 1.6 pounds. Write an equation that could be used to find the weight w, in pounds, of the emu egg. Type below: _____________

Answer: 2.9 – w = 1.6

Explanation: An ostrich egg weighs 2.9 pounds. The difference between the weight of this egg and the weight of an emu egg is 1.6 pounds. The phrase “difference” indicates the subtraction operation. The equation will be 2.9 – w = 1.6

Question 10. In one week, the number of bowls a potter made was 6 times the number of plates. He made 90 bowls during the week. Write an equation that could be used to find the number of plates p that the potter made. Type below: _____________

Answer: 6p = 90

Explanation: Given, In one week, the number of bowls a potter made was 6 times the number of plates. He made 90 bowls during the week. The phrase “times” indicates the multiplication operation. The equation to find the number of plates p that the potter made will be 6p = 90

Question 11. When writing a word sentence as an equation, explain when to use a variable. Type below: _____________

Answer: In a word sentence, a variable represents “a number.” The sum of a number and three = n + 3 The difference of five times a number and four = 5n – 4

Question 1. Three friends are sharing the cost of a bucket of popcorn. The total cost of the popcorn is $5.70. Write an equation that could be used to find the amount a in dollars that each friend should pay. Type below: _____________

Answer: 3a = 5.70

Explanation: Three friends are sharing the cost of a bucket of popcorn. The total cost of the popcorn is $5.70. The expression will be “5.70 is the product of 3 and a. The equation is 3a = 5.70

Question 2. Salimah had 42 photos on her phone. After she deleted some of them, she had 23 photos left. What equation could be used to find the number of photos p that Salimah deleted? Type below: _____________

Answer: p + 23 = 42

Explanation: Salimah had 42 photos on her phone. After she deleted some of them, she had 23 photos left. The expression is the sum of p and 23 is 42. Thus the equation is p + 23 = 42

Chapter 8 Test Answer Key Question 3. A rope is 72 feet long. What is the length of the rope in yards? ______ yards

Answer: 24 yard

Explanation: A rope is 72 feet long. Convert from feet to yards. 1 yard = 3 feet 1 foot = 1/3 yards 72 feet = 72 × 1/3 = 24 yards Thus the length of the rope is 24 yards.

Question 4. Julia evaluated the expression 3 3 + 20 ÷ 2 2 . What value should she get as her answer? ______

Explanation: The equation is 3 3 + 20 ÷ 2 2 . 3 3 = 3 × 3 × 3 = 27 2 2 = 2 × 2 = 4 27 + (20 ÷ 4) 27 + 5 = 32 The answer for the above equation is 32.

Question 5. The sides of a triangle have lengths s, s + 4, and 3s. Write an expression in the simplest form that represents the perimeter of the triangle. Type below: _____________

Answer: 5s + 4

Explanation: The perimeter of the triangle is a + b + c P = a + b + c P = s + s + 4 + 3s P = 5s + 4 Thus the perimeter of the triangle is 5s + 4

Question 6. Gary knows that p = 2 \(\frac{1}{2}\) is a solution to one of the following equations. Which one has p = 2 \(\frac{1}{2}\) as its solution? \(p+2 \frac{1}{2}=5\)        \(p-2 \frac{1}{2}=5\) \(2+p=2 \frac{1}{2}\)       4 – p = 2 \(\frac{1}{2}\) Type below: _____________

Answer: p + 2 \(\frac{1}{2}\) = 5

Explanation: \(p+2 \frac{1}{2}=5\) p + 2 \(\frac{1}{2}\) = 5 p = 5 – 2 \(\frac{1}{2}\) p = 2 \(\frac{1}{2}\) \(p-2 \frac{1}{2}=5\) p – 2 \(\frac{1}{2}\) = 5 p = 5 + 2 \(\frac{1}{2}\) p = 7 \(\frac{1}{2}\) \(2+p=2 \frac{1}{2}\) 2 + p = 2 \(\frac{1}{2}\) p = 2 \(\frac{1}{2}\) – 2 p = \(\frac{1}{2}\) 4 – p = 2 \(\frac{1}{2}\) p = 4 – 2 \(\frac{1}{2}\) p = 1 \(\frac{1}{2}\)

Model and solve the equation by using algebra tiles or iTools.

Question 1. x + 5 = 7 x = ______

Explanation:

• Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
• Use algebra tiles to model the equation. Model x + 5 in the left rectangle, and model 7 in the right rectangle.
• To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
• Remove five 1 tiles on the left side and five 1 tiles on the right side.
• The remaining titles will be two 1 tiles on the right sides.

Question 2. 8 = x + 1 x = ______

• Use algebra tiles to model the equation. Model x + 1 in the left rectangle, and model 8 in the right rectangle.
• Remove one 1 tiles on the left side and one 1 tiles on the right side.
• The remaining titles will be seven 1 tiles on the right sides.

Question 3. x + 2 = 5 x = ______

• Use algebra tiles to model the equation. Model x + 2 in the left rectangle, and model 5 in the right rectangle.
• Remove two 1 tiles on the left side and five 1 tiles on the right side.
• The remaining titles will be three 1 tiles on the right sides.

Question 4. x + 6 = 8 x = ______

• Use algebra tiles to model the equation. Model x + 6 in the left rectangle, and model 8 in the right rectangle.
• Remove six 1 tiles on the left side and six 1 tiles on the right side.

Question 5. 5 + x = 9 x = ______

• Use algebra tiles to model the equation. Model x + 5 in the left rectangle, and model 9 in the right rectangle.
• The remaining titles will be four 1 tiles on the right sides.

Question 6. 5 = 4 + x x = ______

• Use algebra tiles to model the equation. Model x + 4 in the left rectangle, and model 5 in the right rectangle.
• Remove four 1 tiles on the left side and four 1 tiles on the right side.
• The remaining titles will be one 1 tiles on the right sides.

Solve the equation by drawing a model.

Question 7. x + 1 = 5 x = ______

• Use algebra tiles to model the equation. Model x + 1 in the left rectangle, and model 5 in the right rectangle.

Question 8. 3 + x = 4 x = ______

• Use algebra tiles to model the equation. Model x + 3 in the left rectangle, and model 4 in the right rectangle.
• Remove three 1 tiles on the left side and three 1 tiles on the right side.

Question 9. 6 = x + 4 x = ______

• Use algebra tiles to model the equation. Model x + 4 in the left rectangle, and model 6 in the right rectangle.

Question 10. 8 = 2 + x x = ______

• Use algebra tiles to model the equation. Model x + 2 in the left rectangle, and model 8 in the right rectangle.
• Remove two 1 tiles on the left side and two 1 tiles on the right side.
• The remaining titles will be six 1 tiles on the right sides.

Go Math Chapter 8 Review Test Answers Question 11. Describe a Method Describe how you would draw a model to solve the equation x + 5 = 10. Type below: _____________

Answer: x = 5

• Use algebra tiles to model the equation. Model x + 5 in the left rectangle, and model 10 in the right rectangle.
• The remaining titles will be five 1 tiles on the right sides.

Question 12. Interpret a Result The table shows how long several animals have lived at a zoo. The giraffe has lived at the zoo 4 years longer than the mountain lion. The equation 5 = 4 + y can be used to find the number of years y the mountain lion has lived at the zoo. Solve the equation. Then tell what the solution means. Type below: _____________

Answer: The table shows how long several animals have lived in a zoo. The giraffe has lived at the zoo 4 years longer than the mountain lion. 5 = 4 + y y = 5 – 4 y = 1 The solution is y = 1 The solution means that the mountain lion has lived at the zoo for 1 year.

Question 13. Carlos walked 2 miles on Monday and 5 miles on Saturday. The number of miles he walked on those two days is 3 miles more than the number of miles he walked on Friday. Write and solve an addition equation to find the number of miles Carlos walked on Friday Type below: _____________

Answer: Given that, Carlos walked 2 miles on Monday and 5 miles on Saturday. The number of miles he walked on those two days is 3 miles more than the number of miles he walked on Friday. The equation is f + 3 = 2 + 5 f + 3 = 7 f = 7 – 3 f = 4 The solution is f = 4 The solution means that Carlos walked 4 miles on Friday.

Question 14. Sense or Nonsense? Gabriela is solving the equation x + 1 = 6. She says that the solution must be less than 6. Is Gabriela’s statement sense or nonsense? Explain. Type below: _____________

Answer: Gabriela’s statement makes sense. x + 1 = 6 x = 6 – 1 x = 5 Thus the solution is less than 6.

Answer: Remove 5 squares from each side. The rectangle is by itself on the left and 7 squares are on the right side. So, the solution is x = 7

Model and solve the equation by using algebra tiles.

Question 1. x + 6 = 9 x = ________

• Use algebra tiles to model the equation. Model x + 6 in the left rectangle, and model 9 in the right rectangle.

Question 2. 8 + x = 10 x = ________

• Use algebra tiles to model the equation. Model x + 8 in the left rectangle, and model 10 in the right rectangle.
• Remove eight 1 tiles on the left side and eight 1 tiles on the right side.

8 + x = 10 x = 10 – 8 = 2 x = 2

Question 3. 9 = x + 1 x = ________

• Use algebra tiles to model the equation. Model x + 1 in the left rectangle, and model 9 in the right rectangle.
• Remove 1 tile on the left side and 1 tile on the right side.
• The remaining titles will be eight 1 tiles on the right sides.

Question 4. x + 4 = 7 x = ________

Question 5. x + 6 = 10 x = ________

Question 6. The temperature at 10:00 was 10°F. This is 3°F warmer than the temperature at 8:00. Model and solve the equation x + 3 = 10 to find the temperature x in degrees Fahrenheit at 8:00. Type below: _____________

Answer: x = 7

Explanation: The temperature at 10:00 was 10°F. This is 3°F warmer than the temperature at 8:00. The equation is x + 3 = 10 x = 10 – 3 = 7

Go Math Chapter 8 Answer Key Grade 6 Question 7. Jaspar has 7 more checkers left than Karen does. Jaspar has 9 checkers left. Write and solve an addition equation to find out how many checkers Karen has left. Type below: _____________

Answer: c = 2

Explanation: Jaspar has 7 more checkers left than Karen does. Jaspar has 9 checkers left. The expression is c + 7 = 9 The equation to find out how many checkers Karen has left is c + 7 = 9.

Question 8. Explain how to use a drawing to solve an addition equation such as x + 8 = 40. Type below: _____________

• Use algebra tiles to model the equation. Model x + 8 in the left rectangle, and model 40 in the right rectangle.
• Remove eight 1 tile on the left side and eight 1 tile on the right side.
• The remaining titles will be 32 1 tiles on the right side.

x + 8 = 40 x = 40 – 8 x = 32

The equation is x + 6 = 7 x = 7 – 6 x = 1

Question 2. Alice has played soccer for 8 more years than Sanjay has. Alice has played for 12 years. The equation y + 8 = 12 can be used to find the number of years y Sanjay has played. How long has Sanjay played soccer? ________ years

Answer: 4 years

Explanation: Alice has played soccer for 8 more years than Sanjay has. Alice has played for 12 years. the equation is y + 8 = 12 y = 12 – 8 y = 4 years Sanjay played soccer games for 4 years.

Question 3. A car’s gas tank has a capacity of 16 gallons. What is the capacity of the tank in pints? ________ pints

Answer: 128 pints

Explanation: A car’s gas tank has a capacity of 16 gallons. Convert from gallons to pints. 1 gallon = 8 pints 16 gallons = 16 × 8 = 128 pints Thus the capacity of the tank is 128 pints.

Question 4. Craig scored p points in a game. Marla scored twice as many points as Craig but 5 fewer than Nelson scored. How many points did Nelson score? Type below: _____________

Answer: 2p + 5

Explanation: Craig scored p points in a game. Marla scored twice as many points as Craig but 5 fewer than Nelson scored. The equation will be 2p + 5.

Question 5. Simplify 3x + 2(4y + x). Type below: _____________

Answer: 5x + 8y

Explanation: The expression is 3x + 2(4y + x) 3x + 2 × 4y + 2 × x 3x + 8y + 2x Combine the like terms. 5x + 8y 3x + 2(4y + x) = 5x + 8y

Question 6. The Empire State Building in New York City is 443.2 meters tall. This is 119.2 meters taller than the Eiffel Tower in Paris. Write an equation that can be used to find the height h in meters of the Eiffel Tower. Type below: _____________

Answer: 119.2 + h = 443.2

Explanation: The Empire State Building in New York City is 443.2 meters tall. This is 119.2 meters taller than the Eiffel Tower in Paris. Here we have to use the addition operation. The equation is 119.2 + h = 443.2

Question 1. Solve the equation n + 35 = 80. n = ________

Explanation: The given equation is n + 35 = 80 n = 80 – 35 n = 45

Solve the equation, and check the solution.

Question 2. 16 + x = 42 x = ________

Explanation: Given the equation 16 + x = 42 x + 16 = 42 x = 42 – 16 x = 26

Question 3. y + 6.2 = 9.1 y = ________

Answer: 2.9

Explanation: The given equation is y + 6.2 = 9.1 y = 9.1 – 6.2 y = 2.9

Question 4. m + \(\frac{3}{10}=\frac{7}{10}\) m = \(\frac{□}{□}\)

Answer: \(\frac{4}{10}\)

Explanation: The given equation is m + \(\frac{3}{10}=\frac{7}{10}\) m = \(\frac{7}{10}\) – \(\frac{3}{10}\) The denominators are common so subtract the numerators m = \(\frac{4}{10}\)

Question 5. z – \(\frac{1}{3}=1 \frac{2}{3}\) z = ________

Explanation: The given equation is z – \(\frac{1}{3}=1 \frac{2}{3}\) z = \(\frac{1}{3}\) + 1 \(\frac{2}{3}\) z = 1 + \(\frac{1}{3}\) + \(\frac{2}{3}\) z = 1 + \(\frac{3}{3}\) z = 1 + 1 = 2 Thus the value of z is 2.

Go Math Grade 6 Chapter 8 Answer Key Question 6. 12 = x − 24 x = ________

Explanation: The given equation is 12 = x − 24 x – 24 = 12 x = 12 + 24 x = 36 Thus the value of x is 36.

Question 7. 25.3 = w − 14.9 w = ________

Answer: 40.2

Explanation: The given equation is 25.3 = w − 14.9 w – 14.9 = 25.3 w = 25.3 + 14.9 w = 40.2 The value of w is 40.2

Practice: Copy and Solve Solve the equation, and check the solution.

Question 8. y − \(\frac{3}{4}=\frac{1}{2}\) y = _______ \(\frac{□}{□}\)

Answer: 1 \(\frac{1}{4}\)

Explanation: The given equation is y − \(\frac{3}{4}=\frac{1}{2}\) y = \(\frac{1}{2}\) + \(\frac{3}{4}\) y = 1 \(\frac{1}{4}\) Therefore the value of y is 1 \(\frac{1}{4}\).

Question 9. 75 = n + 12 n = ________

Explanation: The given equation is 75 = n + 12 n + 12 = 75 n = 75 – 12 n = 63 The value of n is 63.

Question 10. m + 16.8 = 40 m = ________

Answer: 23.2

Explanation: The given equation is m + 16.8 = 40 m = 40 – 16.8 m = 23.2 The value of m is 23.2

Question 11. w − 36 = 56 w = ________

Explanation: The given equation is w − 36 = 56 w = 56 + 36 w = 92 The value of  is 92.

Question 12. 8 \(\frac{2}{5}\) = d + 2\(\frac{2}{5}\) d = ________

Explanation: The given equation is 8 \(\frac{2}{5}\) = d + 2\(\frac{2}{5}\) d + 2\(\frac{2}{5}\) = 8 \(\frac{2}{5}\) d = 8 \(\frac{2}{5}\) – 2\(\frac{2}{5}\) d = 8 + \(\frac{2}{5}\) – 2 – \(\frac{2}{5}\) d = 8 – 2 = 6 Thus the value of d is 6.

Question 13. 8.7 = r − 1.4 r = ________

Answer: 10.1

Explanation: The given equation is 8.7 = r − 1.4 r − 1.4 = 8.7 r = 8.7 + 1.4 r = 10.1 The value of r is 10.1

Question 14. The temperature dropped 8 degrees between 6:00 p.m. and midnight. The temperature at midnight was 26ºF. Write and solve an equation to find the temperature at 6:00 p.m. ________ ºF

Answer: 34ºF

Explanation: The temperature dropped 8 degrees between 6:00 p.m. and midnight. The temperature at midnight was 26ºF. 26ºF + 8ºF = 34ºF The equation to find the temperature at 6:00 p.m is 34ºF

Question 15. Reason Abstractly Write an addition equation that has the solution x = 9. Type below: _____________

Answer: x + 4 = 13

Explanation: Let the equation be x + 4 = 13 x = 13 – 4 x = 9

Answer: We need to find Kimberly’s balance at the beginning of July.

Question 16. b. What information do you need from the bank statement? Type below: _____________

Answer: We need the information about the deposit on July 12 and July 25 from the bank statement.

Question 16. c. Write an equation you can use to solve the problem. Explain what the variable represents. Type below: _____________

Answer: x = bank account balance y = deposit 1 z = deposit 2 x = y + z

Question 16. d. Solve the equation. Show your work and describe each step. Type below: _____________

Answer: 120.62 = y + z Where y is the deposit 1 and z represents the deposit 2. y = $45.50, z = $43.24 45.50 + 43.24 = 88.74 x + 88.74 = 120.62

Question 16. e. Write Kimberly’s balance at the beginning of July. $ _______

Answer: 31.88

Explanation: x + 88.74 = 120.62 x = 120.62 – 88.74 x = $31.88 Kimberly’s balance at the beginning of July is $31.88

Go Math Grade 6 Chapter 8 Review Test Answer Key Question 17. If x + 6 = 35, what is the value of x + 4? Explain how to find the value without solving the equation. Type below: _____________

Answer: x + 6 = 35 x + 4 + 2 = 35 x  + 4 = 35 – 2 x + 4 = 33 Thus the value of x + 4 = 33

Question 18. Select the equations that have the solution n = 23. Mark all that apply. Options: a. 16 + n = 39 b. n – 4 = 19 c. 25 = n – 2 d. 12 = n – 11

Answer: A, B, D

Explanation: a. 16 + n = 39 n = 23 16 + 23 = 39 39 = 39 The variable is a solution. b. n – 4 = 19 n = 23 23 – 4 = 19 19 = 19 The variable is a solution. c. 25 = n – 2 25 = 23 – 2 25 ≠ 21 The variable is not a solution. d. 12 = n – 11 n = 23 12 = 23 – 11 12 = 12 The variable is a solution. Thus the correct answers are options A, B, D.

Question 1. y − 14 = 23 y = _______

Explanation: y − 14 = 23 y = 23 + 14 y = 37 Thus the solution is 37.

Question 2. x + 3 = 15 x = _______

Explanation: The equation is x + 3 = 15 x = 15 – 3 x = 12 The solution is 12.

Question 3. n + \(\frac{2}{5}=\frac{4}{5}\) n = _______ \(\frac{□}{□}\)

Answer: \(\frac{2}{5}\)

Explanation: The equation is n + \(\frac{2}{5}=\frac{4}{5}\) n + \(\frac{2}{5}\) = \(\frac{4}{5}\) n = \(\frac{4}{5}\) – \(\frac{2}{5}\) n = (4 – 2)/5 n = \(\frac{2}{5}\) Thus the solution is \(\frac{2}{5}\)

Question 4. 16 = m − 14 m = _______

Explanation: The equation is 16 = m − 14 m – 14 = 16 m = 16 + 14 m = 30 The solution is m = 30

Question 5. w − 13.7 = 22.8 w = _______

Answer: 36.5

Explanation: The equation is w − 13.7 = 22.8 w = 22.8 + 13.7 w = 36.5 The solution is w = 36.5

Question 6. s + 55 = 55 s = _______

Explanation: The equation is s + 55 = 55 s = 55 – 55 s = 0 The solution is s = 0

Question 7. 23 = x − 12 x = _______

Explanation: The given equation is 23 = x – 12 x – 12 = 23 x = 23 + 12 x = 35 The solution is x = 35.

Question 8. p − 14 = 14 p = _______

Explanation: The given equation is p − 14 = 14 p = 14 + 14 p = 28 The solution is p = 28.

Question 9. m − \(2 \frac{3}{4}=6 \frac{1}{2}\) m = _______ \(\frac{□}{□}\)

Answer: 9 \(\frac{1}{4}\)

Explanation: The given equation is m − \(2 \frac{3}{4}=6 \frac{1}{2}\) m – 2 \(\frac{3}{4}\) = 6 \(\frac{1}{2}\) m = 6 \(\frac{1}{2}\) + 2 \(\frac{3}{4}\) m = 6 + 2 + \(\frac{1}{2}\) + \(\frac{3}{4}\) m = 8 + 1 \(\frac{1}{4}\) m = 9 \(\frac{1}{4}\)

Question 10. A recipe calls for 5 \(\frac{1}{2}\) cups of flour. Lorenzo only has 3 \(\frac{3}{4}\) cups of flour. Write and solve an equation to find the additional amount of flour Lorenzo needs to make the recipe. Type below: _____________

Answer: 1 \(\frac{3}{4}\)

Explanation: A recipe calls for 5 \(\frac{1}{2}\) cups of flour. Lorenzo only has 3 \(\frac{3}{4}\) cups of flour. x + 3 \(\frac{3}{4}\) = 5 \(\frac{1}{2}\) x = 5 \(\frac{1}{2}\) – 3 \(\frac{3}{4}\) x =  1 \(\frac{3}{4}\)

Question 11. Jan used 22.5 gallons of water in the shower. This amount is 7.5 gallons less than the amount she used for washing clothes. Write and solve an equation to find the amount of water Jan used to wash clothes. Type below: _____________

Explanation: Jan used 22.5 gallons of water in the shower. This amount is 7.5 gallons less than the amount she used for washing clothes. Let the amount of water Jan used to wash clothes be x x – 7.5 = 22.5 x = 22.5 + 7.5 x = 30 Therefore the amount of water Jan used to wash clothes is 30 gallons.

Question 12. Explain how to check if your solution to an equation is correct. Type below: _____________

Answer: i. Evaluate the left-hand side expression at the given value to get a number. ii. Evaluate the right-hand side expression at the given value to get a number. iii. See if the numbers match.

Question 1. The price tag on a shirt says $21.50. The final cost of the shirt, including sales tax, is $23.22. The equation 21.50 + t = 23.22 can be used to find the amount of sales tax t in dollars. What is the sales tax? $ _______

Answer: 1.72

Explanation: The price tag on a shirt says $21.50. The final cost of the shirt, including sales tax, is $23.22. The equation is 21.50 + t = 23.22 t = 23.22 – 21.50 t = 1.72 Therefore the sales tax is $1.72 dollars.

Question 2. The equation l – 12.5 = 48.6 can be used to find the original length l in centimeters of a wire before it was cut. What was the original length of the wire? _______ centimeters

Answer: 61.1 centimeters

Explanation: The equation l – 12.5 = 48.6 can be used to find the original length l in centimeters of a wire before it was cut. l – 12.5 = 48.6 l = 48.6 + 12.5 l = 61.1 centimeters Thus the original length of the wire is 61.1 centimeters.

Question 3. How would you convert a mass in centigrams to a mass in milligrams? Type below: _____________

Answer: The conversion factor is 10; so 1 centigram = 10 milligrams. In other words, the value in cg multiplies by 10 to get a value in mg.

Question 4. In the expression 4 + 3x + 5y, what is the coefficient of x? The coefficient is _______

Answer: A numerical or constant quantity is placed before and multiplied by the variable in an algebraic expression. Thus the coefficient of 3x is 3.

Question 5. Write an expression that is equivalent to 10c. Type below: _____________

Answer: -2(-5c) expand the brackets -2 × -5c = 10c

Question 6. Miranda bought a $ 7 movie ticket and popcorn for a total of $10. The equation 7 + x = 10 can be used to find the cost x in dollars of the popcorn. How much did the popcorn cost? $ _______

Explanation: Miranda bought a $ 7 movie ticket and popcorn for a total of $10. The equation is 7 + x = 10 x = 10 – 7 x = 3 Therefore the cost of the popcorn is $3.

Question 1. 4x = 16 x = _______

• Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
• Use algebra tiles to model the equation. Model 4x in the left rectangle, and model 16 in the right rectangle.
• There are four x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
• To do this, divide each side of your model into 4 equal groups.

Question 2. 3x = 12 x = _______

• Use algebra tiles to model the equation. Model 3x in the left rectangle, and model 12 in the right rectangle.
• There are three x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
• To do this, divide each side of your model into 3 equal groups.

Question 3. 4 = 4x x = _______

• Use algebra tiles to model the equation. Model 4x in the left rectangle, and model 4 in the right rectangle.

Question 4. 3x = 9 x = _______

• Use algebra tiles to model the equation. Model 3x in the left rectangle, and model 9 in the right rectangle.

Go Math Grade 6 Chapter 8 Test Pdf Question 5. 2x = 10 x = _______

• Use algebra tiles to model the equation. Model 2x in the left rectangle, and model 10 in the right rectangle.
• There are two x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
• To do this, divide each side of your model into two equal groups.

Question 6. 15 = 5x x = _______

• Use algebra tiles to model the equation. Model 5x in the left rectangle, and model 15 in the right rectangle.
• There are five x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
• To do this, divide each side of your model into five equal groups.

Question 7. 4x = 8 x = _______

Question 8. 3x = 18 x = _______

Problem Solving + Applications

Question 9. Communicate Explain the steps you use to solve a multiplication equation with algebra tiles. Type below: _____________

Answer: To solve an equation, model the terms of the equation on both sides of an equals sign. Isolate the variable on one side by adding opposites and creating zero pairs. To remove a factor from the variable, divide the sides into rows equal to the factor, and distribute the terms equally among all the rows.

Question 10. Naomi is doing a report about the 1900 and 1904 Olympic Games. Each page will contain info7rmation about 4 of the countries that competed each year. Write and solve an equation to find the number of pages Naomi will need. _______ pages

Answer: 9 pages

Explanation: By seeing the above table we can say that the equation is 4x = 36 The number of countries that competed in the 1900 summer Olympic games is 24. The number of countries that competed in the 1904 summer Olympic games is 12. The total number of countries competed in total is 36. Each page of Naomi’s report contains information about 4 of the countries that competed each year. 4x = 36 x = 36/4 x = 9 Thus Naomi would require 9 pages to complete her report.

Question 11. Pose a Problem Use the information in the bar graph to write and solve a problem involving a multiplication equation. Type below: _____________

Answer: By seeing the above table we can say that the equation is 4x = 72 The number of countries that competed in the 1900 summer Olympic games is 24. The number of countries that competed in the 1904 summer Olympic games is 12. The number of countries that competed in the 1896 summer Olympic games is 14. The number of countries that competed in the 1908 summer Olympic games is 22. The total number of countries competed in total is 72. 4x = 72 x = 72/4 x = 18

Question 12. The equation 7s = 21 can be used to find the number of snakes s in each cage at a zoo. Solve the equation. Then tell what the solution means. s = _______

Explanation: The equation 7s = 21 can be used to find the number of snakes s in each cage at a zoo. Solve the equation. 7 × s = 21 s = 21/7 = 3 The solution s is 3.

Question 13. A choir is made up of 6 vocal groups. Each group has an equal number of singers. There are 18 singers in the choir. Solve the equation 6p = 18 to find the number of singers in each group. Use a model. _______ singers

Answer: 3 singers

Explanation: A choir is made up of 6 vocal groups. Each group has an equal number of singers. There are 18 singers in the choir. The equation 6p = 18 p = 18/6 = 3 p = 3 The solution p is 3.

Question 1. 2x = 8 x = _______

• Use algebra tiles to model the equation. Model 2x in the left rectangle, and model 8 in the right rectangle.

Question 2. 5x = 10 x = _______

• Use algebra tiles to model the equation. Model 5x in the left rectangle, and model 10 in the right rectangle.

Question 3. 21 = 3x x = _______

• Use algebra tiles to model the equation. Model 3x in the left rectangle, and model 21 in the right rectangle.
• To do this, divide each side of your model into three equal groups.

Question 4. 6 = 3x

Question 5. 4x = 12 x = _______

Question 6. A chef used 20 eggs to make 5 omelets. Model and solve the equation 5x = 20 to find the number of eggs x in each omelet. _______ eggs

Explanation: A chef used 20 eggs to make 5 omelets. The equation is 5x = 20 x = 50/5 = 4 Thus there are 4 eggs in each omelet.

Question 7. Last month, Julio played 3 times as many video games as Scott did. Julio played 18 video games. Write and solve an equation to find the number of video games Scott played. _______ video games

Explanation: Last month, Julio played 3 times as many video games as Scott did. Julio played 18 video games. The equation will be 3x = 18 x = 18/3 = 6 x = 6 The number of video games Scott played is 6.

Question 8. Write a multiplication equation, and explain how you can solve it by using a model. Type below: _____________

Answer: 15 = 5x Explanation:

Explanation: The equation for the above figure is 3x = 3 Substitute x = 1 3(1) = 3 3/3 = 1 Thus the solution is 1.

Question 2. Carlos bought 5 tickets to a play for a total of $20. The equation 5c = 20 can be used to find the cost c in dollars of each ticket. How much does each ticket cost? $ _______

Explanation: Carlos bought 5 tickets to a play for a total of $20. The equation is 5c = 20 c = 20/5 = 4 c = 4 The cost of each ticket is $4.

Question 3. A rectangle is 12 feet wide and 96 inches long. What is the area of the rectangle? _______ square feet

Answer: 1152

Explanation: A rectangle is 12 feet wide and 96 inches long. Area of rectangle is l × w A = 12 × 96 A = 1152 square feet. Thus the area of the rectangle is 1152 square feet.

Question 4. Evaluate the algebraic expression 24 – x ÷ y for x = 8 and y = 2. _______

Explanation: 24 – x ÷ y for x = 8 and y = 2. Substitute the value of x and y in the equation. 24 – (8 ÷ 2) 24 – 4 = 20

Go Math Grade 6 Chapter 8 Review Test Question 5. Ana bought a 15.5-pound turkey at the grocery store this month. The equation p – 15.5 = 2.5 can be used to find the weight p, in pounds, of the turkey she bought last month. What is the solution of the equation? p = _______

Explanation: Ana bought a 15.5-pound turkey at the grocery store this month. The equation is p – 15.5 = 2.5 p = 2.5 + 15.5 p = 18 The solution for the equation is 18.

Question 6. A pet store usually keeps 12 birds per cage, and there are 7 birds in the cage now. The equation 7 + x = 12 can be used to find the remaining number of birds x that can be placed in the cage. What is the solution of the equation? x = _______

Explanation: A pet store usually keeps 12 birds per cage, and there are 7 birds in the cage now. The equation is 7 + x = 12 x = 12 – 7 x = 5 Thus the solution of the equation is 5.

Question 1. Solve the equation 2.5m = 10. m = _______

Explanation: 2.5m = 10 m = 10/2.5 m = 4

Question 2. 3x = 210 x = _______

Explanation: 3x = 210 x = 210/3 x = 70

Question 3. 2.8 = 4t t = _______

Answer: 0.7

Explanation: 2.8 = 4t 4t = 2.8 t = 2.8/4 t = 0.7

Question 4. \(\frac{1}{3}\)n = 15 n = _______

Explanation: \(\frac{1}{3}\)n = 15 n = 15 × 3 n = 45

Question 5. \(\frac{1}{2}\)y = \(\frac{1}{10}\) y = _______

Answer: \(\frac{1}{5}\)

Explanation: \(\frac{1}{2}\)y = \(\frac{1}{10}\) y = \(\frac{1}{10}\) × 2 y = \(\frac{1}{5}\)

Question 6. 25 = \(\frac{a}{5}\) a = _______

Answer: 125

Explanation: 25 = \(\frac{a}{5}\) a = 25 × 5 a = 125

Question 7. 1.3 = \(\frac{c}{4}\) c = _______

Answer: 5.2

Explanation: 1.3 = \(\frac{c}{4}\) c = 1.3 × 4 c = 5.2

Question 8. 150 = 6m m = _______

Explanation: 6m = 150 m = 150/6 m = 25

Question 9. 14.7 = \(\frac{b}{7}\) b = _______

Answer: 102.9

Explanation: 14.7 = \(\frac{b}{7}\) b = 14.7 × 7 b = 102.9

Question 10. \(\frac{1}{4}\) = \(\frac{3}{5}\)s s = \(\frac{□}{□}\)

Answer: \(\frac{5}{12}\)

Explanation: \(\frac{1}{4}\) = \(\frac{3}{5}\)s \(\frac{1}{4}\) × \(\frac{5}{3}\) = s s = \(\frac{5}{12}\)

Question 11. There are 100 calories in 8 fluid ounces of orange juice and 140 calories in 8 fluid ounces of pineapple juice. Tia mixed 4 fluid ounces of each juice. Write and solve an equation to find the number of calories in each fluid ounce of Tia’s juice mixture. _______ calories

Answer: 15 calories

Explanation: Number of calories in 8 ounces of orange juice = 100 Number of calories in 1 ounce of juice = 100/8 Number of calories in 4 ounces of juice 100/8 × 4 = 50 calories Number of calories in 8 ounces of pineapple juice = 140 Number of calories in 1 ounce of juice = 140/8 Number of calories in 4 ounces of pineapple juice = 140/8 × 4 =70 calories Now the mixture has 50 + 70 calories = 120 calories in 8 ounces So, 1 ounce of the mixture has 120/8 = 15 calories.

Question 12. Write a division equation that has the solution x = 16. Type below: _____________

Answer: 2x = 32 x = 32/2 x = 16 Thus the equation is x = 16.

What’s the Error?

Question 13. Melinda has a block of clay that weighs 14.4 ounces. She divides the clay into 6 equal pieces. To find the weight w in ounces of each piece, Melinda solved the equation 6w = 14.4. Look at how Melinda solved the equation. Find her error. 6w = 14.4 \(\frac{6 w}{6}\) = 6 × 14.4 w = 86.4 Correct the error. Solve the equation, and explain your steps. Describe the error that Melinda made Type below: _____________

Answer: Melinda has a block of clay that weighs 14.4 ounces. She divides the clay into 6 equal pieces. The equation is 6w = 14.4 The error of Melinda is she used the multiplication equation to solve the equation. She must have used the division equation to get the solution. 6w = 14.4 w = 14.4/6 w = 2.4

Question 14. For numbers 14a−14d, choose Yes or No to indicate whether the equation has the solution x = 15. 14a. 15x = 30 14b. 4x = 60 14c. \(\frac{x}{5}\) = 3 14d. \(\frac{x}{3}\) = 5 14a. _____________ 14b. _____________ 14c. _____________ 14d. _____________

Answer: Given the value of x is 15 14a. 15x = 30 15 × 15 = 30 225 ≠ 30 The answer is No. 14b. 4x = 60 4 × 15 = 60 60 = 60 The answer is yes. 14c. \(\frac{x}{5}\) = 3 x/5 = 3 15/5 = 3 3 = 3 The answer is yes. 14d. \(\frac{x}{3}\) = 5 x/3 = 5 15/3 = 5 5 = 5 The answer is yes.

Question 1. 8p = 96 p = ________

Explanation: 8p = 96 8 × p = 96 p = 96/8 p = 12 The solution is 12

Question 2. \(\frac{z}{16}\) = 8 z = ________

Answer: 128

Explanation: The given equation is \(\frac{z}{16}\) = 8 z = 8 × 16 z = 128 The solution is 128.

Question 3. 3.5x = 14.7 x = ________

Answer: 4.2

Explanation: The given equation is 3.5x = 14.7 x = 14.7/3.5 x = 4.2 The solution x is 4.2

Question 4. 32 = 3.2c c = ________

Explanation: The given equation is 32 = 3.2c 3.2 × c = 32 c = 32/3.2 c = 1/0.1 = 10 The solution c is 10.

Question 5. \(\frac{2}{5}\)w = 40 w = ________

Answer: 100

Explanation: The given equation is \(\frac{2}{5}\)w = 40 \(\frac{2}{5}\) × w = 40 w = 40 × 5/2 w = 200/2 w = 100

Question 6. \(\frac{a}{14}\) = 6.8 a = ________

Answer: 95.2

Explanation: The given equation is \(\frac{a}{14}\) = 6.8 a = 6.8 × 14 a = 95.2

Question 7. 1.6x = 1.6 x = ________

Explanation: The given equation is 1.6x = 1.6 x = 1.6/1.6 x = 1 The solution x is 1

Question 8. 23.8 = 3.5b b = ________

Answer: 6.8

Explanation: The given equation is 23.8 = 3.5b 3.5b = 23.8 b = 23.8/3.5 b = 6.8 Thus the solution of the variable b is 6.8

Question 9. \(\frac{3}{5}\) = \(\frac{2}{3}\)t t = \(\frac{□}{□}\)

Answer: \(\frac{9}{10}\)

Explanation: The given equation is \(\frac{3}{5}\) = \(\frac{2}{3}\)t t = \(\frac{3}{5}\) × \(\frac{3}{2}\) t = \(\frac{9}{10}\) Thus the solution of the variable t is \(\frac{9}{10}\)

Question 10. Anne runs 6 laps on a track. She runs a total of 1 mile, or 5,280 feet. Write and solve an equation to find the distance, in feet, that she runs in each lap. ________ feet

Answer: 880

Explanation: Anne runs 6 laps on a track. She runs a total of 1 mile, or 5,280 feet. Let the l represents the runs in each lap. 6 × l = 5280 feet l = 5280/6 l = 880 feet Therefore Anne runs 880 feets in each lap.

Question 11. In a serving of 8 fluid ounces of pomegranate juice, there are 32.8 grams of carbohydrates. Write and solve an equation to find the amount of carbohydrates in each fluid ounce of the juice. ________ grams

Answer: 4.1

Explanation: Given, In a serving of 8 fluid ounces of pomegranate juice, there are 32.8 grams of carbohydrates. Let c represents the amount of carbohydrates in each fluid ounce of the juice 8 × c = 32.8 grams c = 32.8/8 c = 4.1 grams

Question 12. Write and solve a word problem that can be solved by solving a multiplication equation. Type below: _____________

Answer: The quotient of 6 and p is 12 6 ÷ p = 12 p = 6/12 p = 1/2

Question 1. Estella buys 1.8 pounds of walnuts for a total of $5.04. She solves the equation 1.8p = 5.04 to find the price p in dollars of one pound of walnuts. What does one pound of walnuts cost? $ ________

Answer: 2.8

Explanation: Given that, Estella buys 1.8 pounds of walnuts for a total of $5.04. p represents the price in dollars of one pound of walnuts. The equation to find one pound of walnuts cost is 1.8p = 5.04 1.8p = 5.04 p = 5.04/1.8 p = 2.8 Therefore the cost of one pound of walnuts is $2.8

Question 2. Gabriel wants to solve the equation \(\frac{5}{8}\)m = 25. What step should he do to get m by itself on one side of the equation? Type below: _____________

Explanation: Gabriel wants to solve the equation \(\frac{5}{8}\)m = 25. \(\frac{5}{8}\)m = 25 5m = 25 × 8 5 × m = 200 m = 200/5 = 40 Thus m = 40

Question 3. At top speed, a coyote can run at a speed of 44 miles per hour. If a coyote could maintain its top speed, how far could it run in 15 minutes? ________ miles

Explanation: At top speed, a coyote can run at a speed of 44 miles per hour. Convert from minutes to hour. 60 minutes = 1 hour 15 minutes = 15 × 1/60 = 0.25 = 1/4 44 × 1/4 = 11 miles A coyote can run at a speed of 11 miles for 15 minutes.

Question 4. An online store sells DVDs for $10 each. The shipping charge for an entire order is $5.50. Frank orders d DVDs. Write an expression that represents the total cost of Frank’s DVDs. Type below: _____________

Answer: 10d + $5.50

Explanation: An online store sells DVDs for $10 each. The shipping charge for an entire order is $5.50. Frank orders d DVDs. The expression will be the product of 10 and d more than 5.50 The expression is 10d + $5.50

Question 5. A ring costs $27 more than a pair of earrings. The ring costs $90. Write an equation that can be used to find the cost c in dollars of the earrings. Type below: _____________

Answer: $90 – $27 = c

Explanation: A ring costs $27 more than a pair of earrings. The ring costs $90. c represents the cost in dollars of the earrings. Thus the equation is c + $27 = $90 c = $90 – $27.

Question 6. The equation 3s = 21 can be used to find the number of students s in each van on a field trip. How many students are in each van? ________ students

Answer: 7 students

Explanation: The equation 3s = 21 can be used to find the number of students s in each van on a field trip. 3s = 21 s = 21/3 = 7 s = 7 Thus there are 7 students in each van.

Question 1. Connor ran 3 kilometers in a relay race. His distance represents \(\frac{3}{10}\) of the total distance of the race. The equation \(\frac{3}{10}\)d = 3 can be used to find the total distance d of the race in kilometers. What was the total distance of the race? ________ kilometers

Explanation: Connor ran 3 kilometers in a relay race. His distance represents \(\frac{3}{10}\) of the total distance of the race. \(\frac{3}{10}\)d = 3 3 × d = 3 × 10 3 × d = 30 d = 30/3 = 10 kilometers Therefore the total distance of the race is 10 kilometers.

Question 2. What if Connor’s distance of 3 kilometers represented only \(\frac{2}{10}\) of the total distance of the race. What would the total distance of the race have been? ________ kilometers

Explanation: Connor’s distance of 3 kilometers represented only \(\frac{2}{10}\) of the total distance of the race. \(\frac{2}{10}\) × d = 3 2 × d = 3 × 10 d = 30/2 d = 15 kilometers Therefore the total distance of the race has been 15 kilometers.

Question 3. The lightest puppy in a litter weighs 9 ounces, which is \(\frac{3}{4}\) of the weight of the heaviest puppy. The equation \(\frac{3}{4}\)w = 9 can be used to find the weight w in ounces of the heaviest puppy. How much does the heaviest puppy weigh? ________ ounces

Explanation: The lightest puppy in a litter weighs 9 ounces, which is \(\frac{3}{4}\) of the weight of the heaviest puppy. \(\frac{3}{4}\)w = 9 3 × w = 9 × 4 3 × w = 36 w = 36/3 w = 12 The heaviest puppy weighs 12 ounces.

Question 4. Sophia took home \(\frac{2}{5}\) of the pizza that was left over from a party. The amount she took represents \(\frac{1}{2}\) of a whole pizza. The equation \(\frac{2}{5}\)p = \(\frac{1}{2}\) can be used to find the number of pizzas p left over from the party. How many pizzas were left over? _______ \(\frac{□}{□}\) pizzas

Answer: 1 \(\frac{1}{4}\) pizzas

Explanation: Sophia took home \(\frac{2}{5}\) of the pizza that was left over from a party. The amount she took represents \(\frac{1}{2}\) of a whole pizza. \(\frac{2}{5}\)p = \(\frac{1}{2}\) p = \(\frac{1}{2}\) × \(\frac{5}{2}\) p = \(\frac{5}{4}\) p = 1 \(\frac{1}{4}\) pizzas 1 \(\frac{1}{4}\) pizzas were leftover.

Question 5. A city received \(\frac{3}{4}\) inch of rain on July 31. This represents \(\frac{3}{10}\) of the total amount of rain the city received in July. The equation \(\frac{3}{10}\)r = \(\frac{3}{4}\) can be used to find the amount of rain r in inches the city received in July. How much rain did the city receive in July? _______ \(\frac{□}{□}\) inches of rain

Answer: 2 \(\frac{1}{2}\) inches of rain

Explanation: A city received \(\frac{3}{4}\) inch of rain on July 31. This represents \(\frac{3}{10}\) of the total amount of rain the city received in July. \(\frac{3}{10}\)r = \(\frac{3}{4}\) r = \(\frac{3}{4}\) × \(\frac{10}{3}\) r = \(\frac{30}{12}\) r = \(\frac{5}{2}\) r = 2 \(\frac{1}{2}\) The city received 2 \(\frac{1}{2}\) inches of rain in July.

Question 6. Carole ordered 4 dresses for $80 each, a $25 sweater, and a coat. The cost of the items without sales tax was $430. What was the cost of the coat? $ _______

Explanation: Carole ordered 4 dresses for $80 each, a $25 sweater, and a coat. The cost of the items without sales tax was $430. Cost of 4 dresses is 4 × 80 = $320 $320 + $25 = $345 c + 345 = 430 c = 430 – 345 c = 85 Therefore the cost of the coat is $85

Question 7. A dog sled race is 25 miles long. The equation \(\frac{5}{8}\)k = 25 can be used to estimate the race’s length k in kilometers. Approximately how many hours will it take a dog sled team to finish the race if it travels at an average speed of 30 kilometers per hour? _______ \(\frac{□}{□}\) hours

Answer: 1 \(\frac{1}{3}\) hours

Explanation: A dog sled race is 25 miles long. The equation \(\frac{5}{8}\)k = 25 k represents race length in kilometers. \(\frac{5}{8}\)k = 25 5 × k = 25 × 8 5k = 200 k = 200/5 = 40 k = 40 Average speed is k/30 40/30 = 4/3 The average speed of 30 kilometers per hour is 1 \(\frac{1}{3}\) hours.

Question 8. Explain a Method Explain how you could use the strategy solve a simpler problem to solve the equation \(\frac{3}{4}\)x = \(\frac{3}{10}\). Type below: _____________

Answer: x = \(\frac{2}{5}\)

Explanation: \(\frac{3}{4}\)x = \(\frac{3}{10}\) x = \(\frac{3}{10}\) × \(\frac{4}{3}\) x = \(\frac{12}{30}\) x = \(\frac{2}{5}\)

Question 9. In a basket of fruit, \(\frac{5}{6}\) of the pieces of fruit are apples. There are 20 apples in the display. The equation \(\frac{5}{6}\)f = 20 can be used to find how many pieces of fruit f are in the basket. Use words and numbers to explain how to solve the equation to find how many pieces of fruit are in the basket. _______ pieces of fruit

Explanation: In a basket of fruit, \(\frac{5}{6}\) of the pieces of fruit are apples. There are 20 apples in the display. \(\frac{5}{6}\)f = 20 5 × f = 20 × 6 5 × f = 120 f = 120/5 f = 24 There are 24 pieces of friut in the basket.

Read each problem and solve.

Question 1. Stu is 4 feet tall. This height represents \(\frac{6}{7}\) of his brother’s height. The equation \(\frac{6}{7}\)h = 4 can be used to find the height h, in feet, of Stu’s brother. How tall is Stu’s brother? ______ \(\frac{□}{□}\) feet

Answer: 4 \(\frac{2}{3}\) feet

Explanation: Stu is 4 feet tall. This height represents \(\frac{6}{7}\) of his brother’s height. The equation \(\frac{6}{7}\)h = 4 6/7 × h = 4 6 × h = 4 × 7 6 × h =28 h = 28/6 h = 14/3 h = 4 \(\frac{2}{3}\) feet Thus the height of Stu’s brother in feet is 4 \(\frac{2}{3}\) feet.

Question 2. Bryce bought a bag of cashews. He served \(\frac{7}{8}\) pound of cashews at a party. This amount represents \(\frac{2}{3}\) of the entire bag. The equation \(\frac{2}{3}\)n = \(\frac{7}{8}\) can be used to find the number of pounds n in a full bag. How many pounds of cashews were in the bag that Bryce bought? ______ \(\frac{□}{□}\) pounds

Answer: 1 \(\frac{5}{16}\)

Explanation: Bryce bought a bag of cashews. He served \(\frac{7}{8}\) pound of cashews at a party. This amount represents \(\frac{2}{3}\) of the entire bag. \(\frac{2}{3}\)n = \(\frac{7}{8}\) n = \(\frac{7}{8}\) × \(\frac{3}{2}\) n = \(\frac{21}{16}\) n = 1 \(\frac{5}{16}\) Bryce bought 1 \(\frac{5}{16}\) pounds of cashews were in the bag.

Question 3. In Jaime’s math class, 9 students chose soccer as their favorite sport. This amount represents \(\frac{3}{8}\) of the entire class. The equation \(\frac{3}{8}\)s = 9 can be used to find the total number of students s in Jaime’s class. How many students are in Jaime’s math class? ______ students

Answer: 24 students

Explanation: In Jaime’s math class, 9 students chose soccer as their favorite sport. This amount represents \(\frac{3}{8}\) of the entire class. \(\frac{3}{8}\)s = 9 3 × s = 9 × 8 3 × s = 72 s = 72/3 s = 24 students 24 students are in Jaime’s math class.

Question 4. Write a math problem for the equation \(\frac{3}{4}\)n = \(\frac{5}{6}\). Then solve a simpler problem to find the solution. Type below: _____________

Answer: 1 \(\frac{1}{9}\)

Explanation: \(\frac{3}{4}\)n = \(\frac{5}{6}\) n = \(\frac{5}{6}\) × \(\frac{4}{3}\) n = \(\frac{20}{18}\) n = \(\frac{10}{9}\) n = 1 \(\frac{1}{9}\)

Question 1. Roger served \(\frac{5}{8}\) pound of crackers, which was \(\frac{2}{3}\) of the entire box. What was the weight of the crackers originally in the box? \(\frac{□}{□}\) pounds

Answer: \(\frac{15}{16}\) pounds

Explanation: Roger served \(\frac{5}{8}\) pound of crackers, which was \(\frac{2}{3}\) \(\frac{2}{3}\) × p = \(\frac{5}{8}\) p = \(\frac{5}{8}\) × \(\frac{3}{2}\) p = \(\frac{15}{16}\) pounds \(\frac{15}{16}\) was the weight of the crackers originally in the box.

Question 2. Bowser ate 4 \(\frac{1}{2}\) pounds of dog food. That amount is \(\frac{3}{4}\) of the entire bag of dog food. How many pounds of dog food were originally in the bag? ______ pounds

Answer 6 pounds

Explanation: Bowser ate 4 \(\frac{1}{2}\) pounds of dog food. That amount is \(\frac{3}{4}\) of the entire bag of dog food. 4 \(\frac{1}{2}\) = \(\frac{9}{2}\) \(\frac{3}{4}\) p = \(\frac{9}{2}\) p = \(\frac{9}{2}\) × \(\frac{4}{3}\) p = 6 pounds 6 pounds of dog food were originally in the bag.

Question 3. What is the quotient 4 \(\frac{2}{3}\) ÷ 4 \(\frac{1}{5}\) _______ \(\frac{□}{□}\)

Explanation: 4 \(\frac{2}{3}\) ÷ 4 \(\frac{1}{5}\) \(\frac{14}{3}\) ÷ \(\frac{21}{5}\) = \(\frac{70}{63}\) The mixed fraction of \(\frac{70}{63}\) is 1 \(\frac{1}{9}\) 4 \(\frac{2}{3}\) ÷ 4 \(\frac{1}{5}\) = 1 \(\frac{1}{9}\)

Question 4. Miranda had 4 pounds, 6 ounces of clay. She divided it into 10 equal parts. How heavy was each part? _______ ounces

Answer: 7 ounces

Explanation: Miranda had 4 pounds, 6 ounces of clay. She divided it into 10 equal parts. Convert from pounds to ounces We know that 1 pound = 16 ounces 4 pounds = 4 × 16 ounces = 64 ounces 64 ounces + 6 ounces = 70 ounces Now divide 70 ounces into 10 equal parts. 70 ÷ 10 = 7 ounces. Thus each part was 7 ounces.

Question 5. The amount Denise charges to repair computers is $50 an hour plus a $25 service fee. Write an expression to show how much she will charge for h hours of work. Type below: _____________

Answer: 50h + 25

Explanation: The amount Denise charges to repair computers is $50 an hour plus a $25 service fee. The expression will be product of 50 and h more than 25. The expression is 50h + 25.

Question 6. Luis has saved $14 for a skateboard that costs $52. He can use the equation 14 + m = 52 to find how much more money m he needs. How much more does he need? $ _______

Explanation: Luis has saved $14 for a skateboard that costs $52. He can use the equation 14 + m = 52 14 + m = 52 m = 52 – 14 m = 38 He needs $38 more.

Question 1. A(n) _____ is a statement that two mathematical expressions are equal. Type below: _____________

Answer: An equation is a mathematical statement that two expressions are equal.

Question 2. Adding 5 and subtracting 5 are _____. Type below: _____________

Answer: Solution of an equation.

Concepts and Skills

Question 3. The sum of a number and 4.5 is 8.2. Type below: _____________

Answer: The phrase “sum” indicates an addition operation. So, the equation is n + 4.5 = 8.2

Question 4. Three times the cost is $24. Type below: _____________

Answer: The phrase “times” indicates multiplication. Multiply 3 with c. 3c = 24

Question 5. x − 24 = 58; x = 82 The variable is _____________

Explanation: 82 – 24 = 58 58 = 58 Thus the variable is a solution.

Question 6. \(\frac{1}{3}\)c = \(\frac{3}{8}\), c = \(\frac{3}{4}\) The variable is _____________

Explanation: \(\frac{1}{3}\)c = \(\frac{3}{8}\) c = \(\frac{3}{4}\) \(\frac{1}{3}\) × \(\frac{3}{4}\) = \(\frac{3}{8}\) \(\frac{3}{12}\) ≠ \(\frac{3}{8}\)

Question 7. a + 2.4 = 7.8 a = _____

Answer: 5.4

Explanation: Given the equation is a + 2.4 = 7.8 a + 2.4 = 7.8 a = 7.8 – 2.4 a = 5.4

Question 8. \(b-\frac{1}{4}=3 \frac{1}{2}\) b = _______ \(\frac{□}{□}\)

Explanation: Given the equation is \(b-\frac{1}{4}=3 \frac{1}{2}\) b – \(\frac{1}{4}\) = 3 \(\frac{1}{2}\) b = 3 \(\frac{1}{2}\) + \(\frac{1}{4}\) b = 3 + \(\frac{1}{4}\) + \(\frac{1}{2}\) b = 3 \(\frac{3}{4}\)

Question 9. 3x = 27 x = _______

Explanation: Given the equation is 3x = 27 x = 27/3 x = 9

Question 10. \(\frac{1}{3} s=\frac{1}{5}\) s = \(\frac{□}{□}\)

Answer: \(\frac{3}{5}\)

Explanation: Given the equation is \(\frac{1}{3} s=\frac{1}{5}\) \(\frac{1}{3}\)s = \(\frac{1}{5}\) s = \(\frac{3}{5}\)

Question 11. \(\frac{t}{4}\) = 16 t = _______

Explanation: Given the equation is \(\frac{t}{4}\) = 16 t = 16 × 4 t = 64

Question 12. \(\frac{w}{7}\) = 0.3 w = _______

Answer: 2.1

Explanation: \(\frac{w}{7}\) = 0.3 w/7 = 0.3 w = 0.3 × 7 w = 2.1

Page No. 464

Question 13. A stadium has a total of 18,000 seats. Of these, 7,500 are field seats, and the rest are grandstand seats. Write an equation that could be used to find the number of grandstand seats s. Type below: _____________

Answer: s + 7500 = 18000

Explanation: A stadium has a total of 18,000 seats. Of these, 7,500 are field seats, and the rest are grandstand seats. Let s be the number of grandstand seats. s + 7,500 = 18,000

Question 14. Aaron wants to buy a bicycle that costs $128. So far, he has saved $56. The equation a + 56 = 128 can be used to find the amount a in dollars that Aaron still needs to save. What is the solution of the equation? The solution is _______

Explanation: Aaron wants to buy a bicycle that costs $128. So far, he has saved $56. The equation a + 56 = 128 a = 128 – 56 a = 72 The solution of the equation a + 56 = 128 is 72.

Question 15. Ms. McNeil buys 2.4 gallons of gasoline. The total cost is $7.56. Write and solve an equation to find the price p in dollars of one gallon of gasoline. $ _______

Answer: $3.15

Explanation: Ms. McNeil buys 2.4 gallons of gasoline. The total cost is $7.56. 2.4p = 7.56 p = 7.56/2.4 p = $3.15 The price of one gallon of gasoline is $3.15

Question 16. Crystal is picking blueberries. So far, she has filled \(\frac{2}{3}\) of her basket, and the blueberries weigh \(\frac{3}{4}\) pound. The equation \(\frac{2}{3}\)w = \(\frac{3}{4}\) can be used to estimate the weight w in pounds of the blueberries when the basket is full. About how much will the blueberries in Crystal’s basket weigh when it is full? ______ \(\frac{□}{□}\) pounds

Answer: 1 \(\frac{1}{8}\) pounds

Explanation: Crystal is picking blueberries. So far, she has filled \(\frac{2}{3}\) of her basket, and the blueberries weigh \(\frac{3}{4}\) pound. The equation \(\frac{2}{3}\)w = \(\frac{3}{4}\) w = \(\frac{3}{4}\) × \(\frac{3}{2}\) w = \(\frac{9}{8}\) The mixed fraction of \(\frac{9}{8}\) is 1 \(\frac{1}{8}\) pounds

Determine whether the given value of the variable is a solution of the inequality.

Question 1. a ≥ −6, a = −3 The variable is _____________

Explanation: Substitute the solution a in the inequality. a = -3 -3 ≥ -6 -3 is greater than -6 Thus the variable is a solution.

Question 2. y < 7.8, y = 8 The variable is _____________

Explanation: Substitute the solution y in the inequality. y = 8 8 is less than 7.8 8<7.8 The variable is not the solution.

Question 3. c > \(\frac{1}{4}\), c = \(\frac{1}{5}\) The variable is _____________

Explanation: Substitute the solution c in the inequality. c = \(\frac{1}{5}\) \(\frac{1}{5}\) > \(\frac{1}{4}\) \(\frac{1}{5}\) is greater than \(\frac{1}{4}\) \(\frac{1}{5}\) > \(\frac{1}{4}\) Thus the variable is a solution.

Question 4. x ≤ 3, x = 3 The variable is _____________

Explanation: Substitute the solution x in the inequality. x = 3 3 ≤ 3 3 is less than or equal to 3. Thus the variable is a solution.

Question 5. d < – 0.52, d = – 0.51 The variable is _____________

Explanation: Substitute the solution d in the inequality. -0.51 < -0.52 -0.51 is greater than -0.52 The variable is not the solution.

Question 6. t ≥ \(\frac{2}{3}\), t = \(\frac{3}{4}\) The variable is _____________

Explanation: Substitute the solution t in the inequality. t = \(\frac{3}{4}\) \(\frac{3}{4}\) ≥ \(\frac{2}{3}\) \(\frac{3}{4}\) is greater than \(\frac{2}{3}\) Thus the variable is a solution.

Practice: Copy and Solve Determine whether s = \(\frac{3}{5}\), s = 0, or s = 1.75 are solutions of the inequality.

Question 7. s > – 1 Type below: _____________

Answer: s > – 1 s = \(\frac{3}{5}\) \(\frac{3}{5}\) > -1 \(\frac{3}{5}\) is greater than -1. The variable is the solution. s = 0 0 > -1 0 is greater than -1 Thus the variable is a solution. s = 1.75 1.75 > -1 1.75 is greater than -1 s > -1 Thus the variable is a solution.

Question 8. s ≤ 1 \(\frac{2}{3}\) Type below: _____________

Answer: s ≤ 1 \(\frac{2}{3}\) s = \(\frac{3}{5}\) \(\frac{3}{5}\) ≤ 1 \(\frac{2}{3}\) \(\frac{3}{5}\) is less than but not equal to 1 \(\frac{2}{3}\) The variable is not the solution. s ≤ 1 \(\frac{2}{3}\) s = 0 0 ≤ 1 \(\frac{2}{3}\) The variable is not the solution. s = 1.75 1.75 ≤ 1 \(\frac{2}{3}\) The variable is not the solution.

Question 9. s < 0.43 Type below: _____________

Answer: s < 0.43 \(\frac{3}{5}\) < 0.43 \(\frac{3}{5}\) = 0.6 0.6 is not less than 0.43 Thus the variable is not the solution. s = 0 0 < 0.43 0 is less than 0.43 Thus the variable is the solution. s = 1.75 1.75 < 0.43 1.75 is greater than 0.43 Thus the variable is not the solution.

Give two solutions of the inequality.

Question 10. e < 3 Type below: _____________

Answer: The solution to the inequality must be whole numbers less than 3. e = 1 and 2 are the solutions because 1 and 2 are less than 3. Thus the 2 solutions are 1 and 2.

Question 11. p > – 12 Type below: _____________

Answer: The solution to the inequality must be whole numbers greater than -12 p = 0 and -5 are the solutions because 0 and -5 are greater than -12. Thus the 2 solutions are 0 and -5.

Question 12. y ≥ 5.8 Type below: _____________

Answer: The solution to the inequality must be whole numbers greater than or equal to 5.8 y = 5.8 and 5.9 are the solutions because 5.8 and 5.9 greater than or equal to 5.8 Thus the 2 solutions are 5.8 and 5.9

Question 13. Connect Symbols and Words A person must be at least 18 years old to vote. The inequality a ≥ 18 represents the possible ages a in years at which a person can vote. Determine whether a = 18, a = 17\(\frac{1}{2}\), and a = 91.5 are solutions of the inequality, and tell what the solutions mean. Type below: _____________

Answer: a ≥ 18 Substitute the values of a in the inequality a = 18 18 ≥ 18 Thus the variable is the solution. a = 17\(\frac{1}{2}\) 17\(\frac{1}{2}\) ≥ 18 17\(\frac{1}{2}\) is less than 18. The variable is not the solution. a = 91.5 91.5 > 18 The solution is mean.

Problem Solving + Applcations – Page No. 468

Question 14. The inequality p < 4.75 represents the prices p in dollars that Paige is willing to pay for popcorn. The inequality p < 8.00 represents the prices p in dollars that Paige is willing to pay for a movie ticket. At how many theaters would Paige be willing to buy a ticket and popcorn? ______ theater

Explanation: The inequality p < 4.75 represents the prices p in dollars that Paige is willing to pay for popcorn. The inequality p < 8.00 represents the prices p in dollars that Paige is willing to pay for a movie ticket. From the above table, we can see that there is the only theatre with 8.00 and 4.75 So, Paige is willing to buy a ticket and popcorn from 1 theatre.

Question 15. Sense or Nonsense? Edward says that inequality d ≥ 4.00 represents the popcorn prices in the table, where d is the price of popcorn in dollars. Is Edward’s statement sense or nonsense? Explain. Type below: _____________

Answer: Edward’s statement makes sense because all of the popcorn prices in the table are greater than or equal to $4.00.

Question 16. Use Math Vocabulary Explain why the statement t > 13 is an inequality. Type below: _____________

Answer: The statement is equality because it compares two amounts t and 13 using an inequality symbol.

Question 17. The minimum wind speed for a storm to be considered a hurricane is 74 miles per hour. The inequality w ≥ 74 represents the possible wind speeds of a hurricane. Two possible solutions for the inequality w ≥ 74 are _____ and _____. Two possible solutions for the inequality w ≥ 74 are _____ and _____

Answer: 75 and 80

Explanation: Given that w is greater than or equal to 74. The two possible solutions for the inequality w ≥ 74 are 75 and 80.

Question 1. s ≥ – 1, s = 1 The variable is _____________

Explanation: The inequality is s ≥ – 1 s = 1 1 ≥ – 1 1 is a positive number so 1 will be greater than or equal to -1 Thus the variable is a solution.

Question 2. p < 0, p = 4 The variable is _____________

Explanation: The inequality is p < 0 Given p = 4 Substitute p = 4 in the inequality. 4 < 0 4 is not less than 0 Thus the variable is not a solution.

Question 3. y ≤ – 3, y = – 1 The variable is _____________

Explanation: The inequality is y ≤ – 3 y = -1 -1 ≤ – 3 – 1 is greater than -3 Thus the variable is not a solution.

Question 4. u > \(\frac{-1}{2}\), u = 0 The variable is _____________

Explanation: The inequality is u > \(\frac{-1}{2}\) u = 0 0 > \(\frac{-1}{2}\) 0 is greater than \(\frac{-1}{2}\) Thus the variable is a solution.

Question 5. q ≥ 0.6, q = 0.23 The variable is _____________

Explanation: The inequality is q ≥ 0.6 q = 0.23 0.23 is less than 0.6 Thus the variable is a solution.

Question 6. b < 2 \(\frac{3}{4}\), b = \(\frac{2}{3}\) The variable is _____________

Explanation: The inequality is b < 2 \(\frac{3}{4}\) b = \(\frac{2}{3}\) \(\frac{2}{3}\) < 2 \(\frac{3}{4}\) \(\frac{2}{3}\) is less than 2 \(\frac{3}{4}\) Thus the variable is a solution.

Question 7. k < 2 Type below: _____________

Answer: k = 0 and 1 because they are less than 2. Thus the two possible inequalities for k < 2 are 0 and 1.

Question 8. z ≥ – 3 Type below: _____________

Answer: z = -3 and -2 because -3 and -2 are greater than or equal to -3 Thus the two solutions of the inequality are -3 and -2

Question 9. f ≤ – 5 Type below: _____________

Answer: f = -5 and -6 because -5 and -6 are less than or equal to -5 Thus the two solutions of the inequality are -5 and -6.

Question 10. The inequality s ≥ 92 represents the score s that Jared must earn on his next test to get an A on his report card. Give two possible scores that Jared could earn to get the A. Type below: _____________

Answer: Two possible scores that Jared could earn to get the A are 92 and 100.

Question 11. The inequality m ≤ $20 represents the amount of money that Sheila is allowed to spend on a new hat. Give two possible money amounts that Sheila could spend on the hat. Type below: _____________

Answer: Two possible money amounts that Sheilla could spend on the hat are $15 or $10.

Question 12. Describe a situation and write an inequality to represent the situation. Give a number that is a solution and another number that is not a solution of the inequality. Type below: _____________

Answer: In the United States, the minimum age required to run for president is 35. This can be represented by the inequality a ≥ 35. A number that is a solution is 55 and a number that is not a solution is 29.

Question 1. Three of the following are solutions of g < – 1\(\frac{1}{2}\). Which one is not a solution? g = – 4     g = – 7\(\frac{1}{2}\)   g = 0    g = – 2\(\frac{1}{2}\) Type below: _____________

Answer: g = 0

Explanation: g < – 1\(\frac{1}{2}\). g = – 4 -4 < – 1\(\frac{1}{2}\) g = – 7\(\frac{1}{2}\) – 7\(\frac{1}{2}\) < – 1\(\frac{1}{2}\). g = – 2\(\frac{1}{2}\) – 2\(\frac{1}{2}\) < – 1\(\frac{1}{2}\) g = 0 0 < – 1\(\frac{1}{2}\) Thus 0 is not the solution.

Question 2. The inequality w ≥ 3.2 represents the weight of each pumpkin, in pounds, that is allowed to be picked to be sold. The weights of pumpkins are listed. How many pumpkins can be sold? Which pumpkins can be sold? 3.18 lb, 4 lb, 3.2 lb, 3.4 lb, 3.15 lb Type below: _____________

Answer: 3.2 lb, 3.4 lb

Explanation: The inequality w ≥ 3.2 represents the weight of each pumpkin, in pounds, that is allowed to be picked to be sold. Substitute the solutions in the inequality. w = 3.18 3.18 ≥ 3.2 3.18 is less than 3.2 3.18 < 3.2 lb w = 4 lb 4 ≥ 3.2 4 is greater than 3.2 4 > 3.2 w = 3.2 lb 3.2 ≥ 3.2 3.2 lb is greater than 3.2 lb w = 3.4 lb 3.4 ≥ 3.2 3.4 lb is greater than 3.2 lb w = 3.15 lb 3.15 < 3.2 Thus 3.2 lb, 3.4 lb pumpkins can be sold.

Question 3. What is the value of 8 + (27 ÷ 9) 2 ? _______

Explanation: 8 + (27 ÷ 9) 2 ? 8 + (3) 2 8 + 9 = 17

Question 4. Write an expression that is equivalent to 5(3x + 2z). Type below: _____________

Answer: 15x + 10z

Explanation: 5(3x + 2z) 5 × 3x + 5 × 2z 15x + 10z The expression equivalent to 5(3x + 2z) is 15x + 10z

Question 5. Tina bought a t-shirt and sandals. The total cost was $41.50. The t-shirt cost $8.95. The equation 8.95 + c = 41.50 can be used to find the cost c in dollars of the sandals. How much did the sandals cost? $ _______

Answer: $32.55

Explanation: Tina bought a t-shirt and sandals. The total cost was $41.50. The t-shirt cost $8.95. The equation is 8.95 + c = 41.50 c = 41.50 – 8.95 c = $32.55 The cost of the sandal is 32.55

Question 6. Two-thirds of a number is equal to 20. What is the number? _______

Explanation: 2/3 × n = 20 n = 3/2 × 20 n =  3 × 10 n = 30 The number is 30.

Write an inequality for the word sentence. Tell what type of numbers the variable in the inequality can represent.

Question 1. The elevation e is greater than or equal to 15 meters. Type below: _____________

Answer: The phrase greater than or equal to represents “≥” Thus the inequality is e ≥ 15

Question 2. A passenger’s age a must be more than 4 years. Type below: _____________

Answer: The phrase more than represents the greater than symbol “>” Thus the inequality is a > 4

Write a word sentence for the inequality.

Question 3. b < \(\frac{1}{2}\) Type below: _____________

Answer: By seeing the above inequality we can write the word sentence for inequality as, b is less than \(\frac{1}{2}\)

Question 4. m ≥ 55 Type below: _____________

Answer: By seeing the above inequality we can write the word sentence for inequality as, m is greater than or equal to 55.

Question 5. Compare Explain the difference between t ≤ 4 and t < 4. Type below: _____________

Answer: t ≤ 4 is t is less than or equal to 4 which means t is equal to 4 or 3.9. t < 4 is t is less than 4 which means t is equal to 3, 2, or 1 or 0.

Question 6. A children’s roller coaster is limited to riders whose height is at least 30 inches and at most 48 inches. Write two inequalities that represent the height h of riders for the roller coaster. Type below: _____________

Answer: h represents the height of riders for the roller coaster. A children’s roller coaster is limited to riders whose height is at least 30 inches and at most 48 inches. ar least 30 inches means h must be greater than or equal to 30 inches. i.e., h ≥ 30 inches at most 48 inches means h must be less than 48 inches. i.e., h < 48 inches

The reading skill make generalizations can help you write inequalities to represent situations. A generalization is a statement that is true about a group of facts.

Question 8. Write two inequalities that represent generalizations about the sea otter weights. Type below: _____________

Answer: First, list the weights in pounds in order from least to greatest. 50, 51, 54, 58, 61, 61, 62, 62, 66, 68, 69, 71 Next, write an inequality to describe the weights by using the least weight on the list. Let w represent weights of the otters in the pounds. The least weight is 50 pounds, so all of the weights are greater than or equal to 50 pounds. w ≥ 50 Now write an inequality to describe the weights by using the greatest weights in the list. The greatest weight is 71 pounds, so all of the weights are less than or equal to 71 pounds. w ≤ 71

Answer: First, list the number of pups in order from least to greatest. 6, 6, 7, 10, 15, 16, 20, 23 Next, write an inequality to describe the number of pups by using the least number of pups on the list. Let n represent the number of pups. The least weight is 6 pups. So all of the pups will be greater than or equal to 6. n ≥ 6 Now write an inequality to describe the number of pups by using the greatest weights in the list. The greatest weight is 23 pups so all of the weights are less than or equal to 23 pups. n ≤ 23 pups

Question 1. The width w is greater than 4 centimeters. Type below: _____________

Answer: The inequality symbol for “greater than” is >. w > 4, where w is the width in centimeters. w is a positive number.

Question 2. The score s in a basketball game is greater than or equal to 10 points Type below: _____________

Answer: The inequality symbol for “greater than or equal to” is ≥. s ≥ 10, where s is the score in the basketball game. s is a positive number.

Question 3. The mass m is less than 5 kilograms Type below: _____________

Answer: The inequality symbol for “less than” is <. m < 5, where m is the mass in kilograms. m is a positive number.

Question 4. The height h is greater than 2.5 meters Type below: _____________

Answer: The inequality symbol for “greater than” is >. h > 2.5, where h is the height in meters. h is a positive number.

Question 5. The temperature t is less than or equal to −3°. Type below: _____________

Answer: The inequality symbol for “less than or equal to” is ≤. t ≤  −3° where t is the temperature in degrees. t is a negative number.

Question 6.4 k < – 7 Type below: _____________

Answer: The word sentence for the inequality is k is less than -7.

Question 7. z ≥ 2 \(\frac{3}{5}\) Type below: _____________

Answer: The word sentence for the inequality is z is greater than or equal to 2 \(\frac{3}{5}\).

Question 8. Tabby’s mom says that she must read for at least 30 minutes each night. If m represents the number of minutes reading, what inequality can represent this situation? Type below: _____________

Answer: m ≥ 30

Explanation: Tabby’s mom says that she must read for at least 30 minutes each night. m represents the number of minutes of reading. m is greater than or equal to 30. Thus the inequality is m ≥ 30.

Question 9. Phillip has a $25 gift card to his favorite restaurant. He wants to use the gift card to buy lunch. If c represents the cost of his lunch, what inequality can describe all of the possible amounts of money, in dollars, that Phillip can spend on lunch? Type below: _____________

Answer: c ≤ 25

Explanation: Phillip has a $25 gift card to his favorite restaurant. He wants to use the gift card to buy lunch. c represents the cost of his lunch c is less than or equal to 25. Thus the inequality is c ≤ 25.

Question 10. Write a short paragraph explaining to a new student how to write an inequality. Type below: _____________

Answer: Inequality is a statement that two quantities are not equal. To know which direction to shade a graph, I write inequalities with the variable on the left side of the inequality symbol. I know that the symbol has to point to the same number after I rewrite the inequality. For example, I write 4 < y as y > 4 Now the inequality symbol points in the direction that I should draw the shaded arrow on my graph.

Question 1. At the end of the first round in a quiz show, Jeremy has at most −20 points. Write an inequality that means “at most −20”. Type below: _____________

Answer: The phrase at most refers to less than or equal to. Thus the inequality is J ≤ -20

Question 2. Describe the meaning of y ≥ 7.9 in words. Type below: _____________

Answer: y ≥ 7.9 means y is greater than or equal to 7.9

Question 3. Let y represent Jaron’s age in years. If Dawn were 5 years older, she would be Jaron’s age. Which expression represents Dawn’s age? Type below: _____________

Answer: y – 5

Explanation: Let y represent Jaron’s age in years. If Dawn were 5 years older, she would be Jaron’s age. We have to subtract 5 years to know the age of Jaron. Thus the expression is y – 5.

Question 4. Simplify the expression 7 × 3g. Type below: _____________

Answer: 21g

Question 5. What is the solution of the equation 8 = 8f? f = ________

Answer: 8 = 8f f = 8/8 = 1 f = 1 The solution for the equation 8 = 8f is 1.

Question 6. Which of the following are solutions of the inequality k ≤ – 2? k = 0   k = – 2   k = – 4   k = 1   k = – 1 \(\frac{1}{2}\) Type below: _____________

Answer: k = -2 k = -4

Explanation: k = 0 in the inequality k ≤ – 2 0 ≤ – 2 0 is less than but not equal to -2 Thus 0 is not the solution. k = – 2 k ≤ – 2 -2 ≤ – 2 Thus -2 is the solution. k = – 4 k ≤ – 2 -4 ≤ – 2 Thus -4 is the solution. k = 1 1 ≤ – 2 1 ≤ – 2 1 is greater than but not equal to -2 Thus 1 is not the solution. k = – 1 \(\frac{1}{2}\) – 1 \(\frac{1}{2}\) ≤ – 2 – 1 \(\frac{1}{2}\) ≤ – 2 – 1 \(\frac{1}{2}\) is less than but not equal to -2 Thus – 1 \(\frac{1}{2}\) is not the solution.

Graph the inequality.

Question 1. m < 15 Type below: _____________

Question 2. c ≥ – 1.5 Type below: _____________

Question 3. b ≤ \(\frac{5}{8}\) Type below: _____________

Practice: Copy and Solve Graph the inequality.

Question 4. a < \(\frac{2}{3}\) Type below: _____________

Question 5. x > – 4 Type below: _____________

Question 6. k ≥ 0.3 Type below: _____________

Question 7. t ≤ 6 Type below: _____________

Write the inequality represented by the graph.

Answer: m < 6

Answer: n ≥ -7

Question 10. Model Mathematics The inequality w ≥ 60 represents the wind speed w in miles per hour of a tornado. Graph the solutions of the inequality on the number line. Type below: _____________

Question 11. Graph the solutions of the inequality c < 12 ÷ 3 on the number line Type below: _____________

Question 12. Write an inequality representing t, the heights in inches of people who can go on Twirl & Whirl. Type below: _____________

Answer: The minimum height of people who can go on Twirl and Whirl is 48 inches. So, inequality is t ≥ 48.

Question 13. Graph your inequality from Exercise 12. Type below: _____________

Answer: Draw a full circle at 48 to show that 48 is a solution. Shade to the right of 48 to show that values greater than or equal to 48 are solutions.

Question 14. Write an inequality representing r, the heights in inches of people who can go on Race Track. Type below: _____________

Answer: The minimum height of people who can go on Race track is 24 inches. So, the inequality is r ≥ 42.

Question 15. Graph your inequality from Exercise 14. Type below: _____________

Answer: Draw a full circle at 42 to show that 42 is a solution. Shade to the right of 42 to show that values greater than or equal to 48 are solutions.

Question 16. Write an inequality representing b, the heights in inches of people who can go on both River Rapids and Mighty Mountain. Explain how you determined your answer. Type below: _____________

Answer: You need to be at least 38 inches tall to go on River Rapids and at least 44 inches tall to go on Mighty mountain. So, you need to be at least 44 inches tall to go on both rides. The inequality is b ≥ 44.

Answer: Yes I agree with Darius. That dark circle and the arrow to the left indicates that c ≤ 25

Question 1. h ≥ 3 Type below: _____________

Question 2. x < \(\frac{-4}{5}\) Type below: _____________

Question 3. y > – 2 Type below: _____________

Question 4. n ≥ 1 \(\frac{1}{2}\) Type below: _____________

Question 5. c ≤ – 0.4 Type below: _____________

Answer: n > 3

Answer: n > -5

Question 8. The inequality x ≤ 2 represents the elevation x of a certain object found at a dig site. Graph the solutions of the inequality on the number line. Type below: _____________

Question 9. The inequality x ≥ 144 represents the possible scores x needed to pass a certain test. Graph the solutions of the inequality on the number line. Type below: _____________

Question 10. Write an inequality and graph the solutions on a number line. Type below: _____________

Answer: x ≥ -2 The number line at right shows the solutions of the inequality x ≥ -2

Question 2. Describe the graph of g < 0.6. Type below: _____________

Question 3. Write an expression that shows the product of 5 and the difference of 12 and 9. Type below: _____________

Answer: The equation for the product of 5 and the difference of 12 and 9 5 × 12 – 9 The equation is 5(12 – 9).

Question 4. What is the solution of the equation 8.7 + n = 15.1? n = ________

Answer: 6.4

Explanation: The equation is 8.7 + n = 15.1 n + 8.7 = 15.1 n = 15.1 – 8.7 n = 6.4

Question 5. The equation 12x = 96 gives the number of egg cartons x needed to package 96 eggs. Solve the equation to find the number of cartons needed. ________ cartons

Explanation: Given, The equation 12x = 96 gives the number of egg cartons x needed to package 96 eggs. 12x = 96 x = 96/12 = 8 Thus 8 number of cartons are needed.

Question 6. The lowest price on an MP3 song is $0.35. Write an inequality that represents the cost c of an MP3 song. Type below: _____________

Answer: Given that, The lowest price on an MP3 song is $0.35. c ≥ 0.35 That is an inequality to represent the cost of an MP3 song.

Question 1. For numbers 1a–1c, choose Yes or No to indicate whether the given value of the variable is a solution of the equation. 1a. \(\frac{2}{5}\)v=10; v = 25 1b. n + 5 = 15; n = 5 1c. 5z = 25; z = 5 1a. _____________ 1b. _____________ 1c. _____________

Answer: 1a. \(\frac{2}{5}\)v=10; v = 25 \(\frac{2}{5}\) × 25=10 2 × 5 = 10 10 = 10 The variable is a solution. Thus the answer is yes. 1b. n + 5 = 15; n = 5 Substitute n = 5 5 + 5 = 15 10 ≠ 15 The variable is not a solution. The answer is no. 1c. 5z = 25; z = 5 Substitute z = 5 5 × 5 = 25 25 = 25 The variable is a solution. Thus the answer is yes.

Question 2. The distance from third base to home plate is 88.9 feet. Romeo was 22.1 feet away from third base when he was tagged out. The equation 88.9 − t = 22.1 can be used to determine how far he needed to run to get to home plate. Using substitution, the coach determines that Romeo needed to run _____ feet to get to home plate. Using substitution, the coach determines that Romeo needed to run _____________ feet to get to home plate

Answer: 66.8 feet

Explanation: The distance from third base to home plate is 88.9 feet. Romeo was 22.1 feet away from third base when he was tagged out. The equation is 88.9 − t = 22.1 88.9 − t = 22.1 88.9 – 22.1 = t t = 66.8 feet Thus Using substitution, the coach determines that Romeo needed to run 66.8 feet to get to the home plate.

Question 3. There are 84 grapes in a bag. Four friends are sharing the grapes. Write an equation that can be used to find out how many grapes g each friend will get if each friend gets the same number of grapes. Type below: _____________

Answer: 84 = 4g 84 is the total amount of grapes 4 is the number of friends g = how many grapes each friend will get

Chapter 8 Review/Test Page No. 484

Question 5. Frank’s hockey team attempted 15 more goals than Spencer’s team. Frank’s team attempted 23 goals. Write and solve an equation that can be used to find how many goals Spencer’s team attempted. ______ goals

Answer: 8 goals

Explanation: Frank’s hockey team attempted 15 more goals than Spencer’s team. Frank’s team attempted 23 goals. Let x be the Spencer’s team The phrase more than indicates addition operation. x + 15 = 23 x = 23 – 15 x = 8 goals

Answer: y = 7

• Draw 11 rectangles on your MathBoard to represent the two sides of the equation.
• Use algebra tiles to model the equation. Model y + 10 in the left rectangle, and model 17 in the right rectangle.
• To solve the equation, get the y tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
• Remove ten 1 tiles on the left side and ten 1 tiles on the right side.

Thus 10 + y = 17 y = 17 – 10 = 7 y = 7

Question 7. Gabriella and Max worked on their math project for a total of 6 hours. Max worked on the project for 2 hours by himself. Solve the equation x + 2 = 6 to find out how many hours Gabriella worked on the project. ______ hours

Answer: 4 hours

Explanation: Gabriella and Max worked on their math project for a total of 6 hours. Max worked on the project for 2 hours by himself. x + 2 = 6 x = 6 – 2 x = 4 Gabriella worked 4 hours on the project.

Question 8. Select the equations that have the solution m = 17. Mark all that apply. Options: a. 3 + m = 21 b. m − 2 = 15 c. 14 = m − 3 d. 2 = m − 15

Answer: B, C, D

Explanation: a. 3 + m = 21 3 + 17 = 21 20 ≠ 21 b. m − 2 = 15 17 – 2 = 15 15 = 15 c. 14 = m − 3 14 = 17 – 3 14 = 14 d. 2 = m − 15 2 = 17 – 15 2 = 2 Thus the correct answers are B, C and D.

Question 9. Describe how you could use algebra tiles to model the equation 4x = 20. Type below: _____________

Question 10. For numbers 10a–10d, choose Yes or No to indicate whether the equation has the solution x = 12. 10a. \(\frac{3}{4}\)x = 9 10b. 3x = 36 10c. 5x = 70 10d. \(\frac{x}{3}\) = 4 10a. _____________ 10b. _____________ 10c. _____________ 10d. _____________

Answer: 10a. Yes 10b. Yes 10c. No 10d. Yes

Explanation: 10a. \(\frac{3}{4}\)x = 9 \(\frac{3}{4}\) × 12 = 9 3 × 3 = 9 9 = 9 Thus the answer is yes. 10b. 3x = 36 x = 12 3 × 12 = 36 36 = 36 Thus the answer is yes. 10c. 5x = 70 x = 12 5 × 12 = 70 60 ≠ 70 Thus the answer is no. 10d. \(\frac{x}{3}\) = 4 x/3 = 4 x = 4 × 3 x = 12 Thus the answer is yes.

Question 11. Bryan rides the bus to and from work on the days he works at the library. In one month, he rode the bus 24 times. Solve the equation 2x = 24 to find the number of days Bryan worked at the library. Use a model. Type below: _____________

Question 12. Betty needs \(\frac{3}{4}\) of a yard of fabric to make a skirt. She bought 9 yards of fabric. Part A Write and solve an equation to find how many skirts x she can make from 9 yards of fabric. ________ skirts

Answer: 12 skirts

Explanation: Betty needs \(\frac{3}{4}\) of a yard of fabric to make a skirt. She bought 9 yards of fabric. x × \(\frac{3}{4}\) = 9 x = 9 × \(\frac{4}{3}\) x = 3 × 4 = 12 x = 12 she can make 12 skirts from 9 yards of fabric.

Question 12. Part B Explain how you determined which operation was needed to write the equation Type below: _____________

Answer: Division operation is needed to write the equation to know how many x skirts she can make from 9 yards of fabric.

Question 13. Karen is working on her math homework. She solves the equation \(\frac{b}{8}\) = 56 and says that the solution is b = 7. Do you agree or disagree with Karen? Use words and numbers to support your answer. If her answer is incorrect, find the correct answer. Type below: _____________

Answer: Karen is working on her math homework. She solves the equation \(\frac{b}{8}\) = 56 and says that the solution is b = 7. I Disagree with Karen. b/8 = 56; multiply both sides by 8 to solve for b, and you get b = 448

Question 14. There are 70 historical fiction books in the school library. Historical fiction books make up \(\frac{1}{10}\) of the library’s collection. The equation \(\frac{1}{10}\)b = 70 can be used to find out how many books the library has. Solve the equation to find the total number of books in the library’s collection. Use numbers and words to explain how to solve \(\frac{1}{10}\)b = 70. Type below: _____________

Answer: Given Number of historical books = 70 The equation used to find the totals number of books in the library collection. \(\frac{1}{10}\)b = 70 b = 70 × 10 b = 700 Hence there are 700 books in the library collection.

Question 15. Andy drove 33 miles on Monday morning. This was \(\frac{3}{7}\) of the total number of miles he drove on Monday. Solve the equation \(\frac{3}{7}\)m = 33 to find the total number of miles Andy drove on Monday. ______ miles

Answer: 77 miles

Explanation: Andy drove 33 miles on Monday morning. This was \(\frac{3}{7}\) of the total number of miles he drove on Monday. \(\frac{3}{7}\)m = 33 3 × m = 33 × 7 3 × m = 231 m = 231/3 m = 77 miles Therefore the total number of miles Andy drove on Monday is 77 miles.

Question 16. The maximum number of players allowed on a lacrosse team is 23. The inequality t≤23 represents the total number of players t allowed on the team. Two possible solutions for the inequality are _____ and _____. Two possible solutions for the inequality are _____ and _____

Answer: The maximum number of players allowed on a lacrosse team is 23. t ≤ 23 Thus the two possible solutions for the inequality are 22 and 23.

Question 17. Mr. Charles needs to have at least 10 students sign up for homework help in order to use the computer lab. The inequality h ≥ 10 represents the number of students h who must sign up. Select possible solutions of the inequality. Mark all that apply. Options: a. 7 b. 8 c. 9 d. 10 e. 11 f. 12

Answer: D, E

Explanation: Mr. Charles needs to have at least 10 students sign up for homework help in order to use the computer lab. h ≥ 10 The number near to 10 is 10 and 11 Thus the correct answers are options D and E.

Question 18. The maximum capacity of the school auditorium is 420 people. Write an inequality for the situation. Tell what type of numbers the variable in the inequality can represent. Type below: _____________

Answer: The maximum capacity of the school auditorium is 420 people Let x be the maximum people The inequality is x is less than or equal to 420. x ≤ 420

Answer: Agree with Dylan. Because the dark circle shows that it is not the solution.

Question 20. Part B Suppose Cydney’s graph had an empty circle at 14. Write the inequality represented by this graph. Type below: _____________

Conclusion:

Grow through a daily set of highly focused practice questions present on Go Math Grade 6 Chapter 8 Solution Key. We have included the advice of expert professionals to help the students for a better understanding of solving problems. Test your knowledge by solving the problems in the Review test. Stay tuned to our Go Math Answer Key to get the study materials of all grade 6 chapters.

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Into Math Grade 6 Module 9 Lesson 3 Answer Key Use Multiplication and Division Equations to Solve Problems

We included  HMH Into Math Grade 6 Answer Key PDF   Module 9 Lesson 3 Use Multiplication and Division Equations to Solve Problems to make students experts in learning maths.

HMH Into Math Grade 6 Module 9 Lesson 3 Answer Key Use Multiplication and Division Equations to Solve Problems

I Can write and solve equations using multiplication and division to represent real-world situations involving an unknown.

Spark Your Learning

Turn and Talk If ham costs $5.79 per pound and cheese costs $4.29 per pound, how can you find the total amount you pay for the ham and cheese in each sandwich? Can you stay within a budget of $1.50 per sandwich? Explain. Answer: ham costs $5.79 per pound and cheese costs $4.29 per pound 5.79 + 4.29 = 10.08 We need to add the price to find the total amount you pay for the ham and cheese in each sandwich. 0.15 × 5.79 = 0.86 0.08 × 4.29 = 0.34 0.86 + 0.34 = 1.2 Yes, You stay within a budget of $1.50 because it costs $1.20 to make a sandwich.

Build Understanding

In the previous lesson, you learned to solve addition equations with a variable by subtracting. How can you solve a multiplication equation?

B. Write an equation to represent this problem. Answer: 4x = 20

C. Model the equation using algebra tiles. How many x-tiles are in your model? How many 1-tiles? Answer: 4x = 20 x = 20/4 x = 5

D. How many equal groups do you need to divide your model into to solve for x? Explain. Answer: You need to divide by 4 to get equal groups of apples.

E. Draw circles around the tiles you drew for Part C to separate the tiles into equal groups. Answer:

F. How many 1-tiles are in each group? How many pounds of apples does each person get? Answer:

Turn and Talk What other method can you use to find how many pounds of apples each person gets? Explain.

Step It Out

The Division Property of Equality states that you can divide both sides of an equation by the same nonzero number and the two sides will remain equal.

The Multiplication Property of Equality states that you can multiply both sides of an equation by the same number and the two sides will remain equal.

C. Find the value of x if you know that \(\frac{x}{10}\) = 9. Describe how you found the solution. Answer: \(\frac{x}{10}\) = 9 x = 9 × 10 x = 90

Turn and Talk Why can’t you divide both sides of an equation by zero? Explain. Answer: We can’t divide both sides of an equation by zero. Because any number divided by 0 will be infinity.

E. Interpret your answer. Answer:

Turn and Talk How is solving a multiplication equation with decimals like solving a multiplication equation with whole numbers? Answer: Multiplying decimals is the same as multiplying whole numbers except for the placement of the decimal point in the answer. When you multiply decimals, the decimal point is placed in the product so that the number of decimal places in the product is the sum of the decimal places in the factors.

Check Understanding

On Your Own

Question 3. Preston earns $33.00 for babysitting his neighbor’s children for 2.4 hours. A. Model with Mathematics Write an equation you can use to find out how much Preston earns per hour. Answer: Given, Preston earns $33.00 for babysitting his neighbor’s children for 2.4 hours. x × 2.4 = 33 x = 33/2.4 x = 13.75

B. How much does Preston earn in one hour? Answer: Preston earn $13.75 in one hour.

Question 4. Henri has exactly enough quarters to wash 3 loads of laundry. It takes 8 quarters to wash one load of laundry. How many quarters does Henri have? Use the equation \(\frac{x}{3}\) = 8. Answer: Given, Henri has exactly enough quarters to wash 3 loads of laundry. It takes 8 quarters to wash one load of laundry. By using the equation \(\frac{x}{3}\) = 8. x = 8 × 3 x = 24 Henri have 24 quarters.

For Problems 6-9, find each solution and graph it on the number line.

Question 10. Glenn is 5 feet tall. He is \(\frac{4}{5}\) as tall as his brother. How tall is Glenn’s brother? A. Model with Mathematics Write an equation to find Glenn’s brother’s height, using x to represent the height. Answer: Let the height of his brother be x. \(\frac{4}{5}\)x = 5

B. How can you solve this equation for x? Answer: \(\frac{4}{5}\)x = 5 4x = 5 × 5 x = 25/4 x = 5 \(\frac{5}{4}\)

C. How tall is Glenn’s brother in feet? Answer: Glenn’s brother is 5 \(\frac{5}{4}\) ft tall.

Question 11. Iris made bracelets with string and beads. She used x centimeters of string and made 7 bracelets. She used 17.8 centimeters of string for each bracelet. How much string did she use in all? Write an equation and use it to solve this problem. Answer: Given, Iris made bracelets with string and beads. She used x centimeters of string and made 7 bracelets. She used 17.8 centimeters of string for each bracelet. The equation would be x ÷ 17.8 = 7 x = 7 × 17.8 x = 124.6 Thus she use 124.6 centimeters in all.

For Problems 13-16, find each solution.

Question 13. 9x = 171 Answer: 9x = 171 x = 171/9 x = 19

Question 14. \(\frac{x}{6}\) = 15 Answer: \(\frac{x}{6}\) = 15 x = 15 × 6 x = 90

Question 15. \(\frac{9}{5}\)x = 2\(\frac{7}{10}\) Answer: \(\frac{9}{5}\)x = 2\(\frac{7}{10}\) \(\frac{9}{5}\)x = \(\frac{27}{10}\) x = \(\frac{27}{10}\) × \(\frac{5}{9}\) x = \(\frac{3}{2}\) or 1 \(\frac{1}{2}\)

Question 16. \(\frac{x}{4}\) = 28.80 Answer: \(\frac{x}{4}\) = 28.80 x = 28.80 × 4 x = 115.2

I’m in a Learning Mindset!

How can I help a classmate develop a growth-mindset response? Answer:

Lesson 9.3 More Practice/Homework

Question 2. STEM A 120-volt battery is connected to a circuit with 30 ohms of resistance. The current in amperes c flowing through the circuit can be found using the equation 30c = 120. How many amperes of current are in the wire? Answer: Given, A 120-volt battery is connected to a circuit with 30 ohms of resistance. The current in amperes c flowing through the circuit can be found using the equation 30c = 120. 30c = 120 c = 120/30 c = 4 There are 4 amperes of current in the wire.

Math on the Spot For Problems 3-4, solve each equation. Check your answer.

Question 3. \(\frac{x}{7}\) = 2 Answer: \(\frac{x}{7}\) = 2 x = 2 × 7 x = 14

Question 4. 36 = \(\frac{w}{4}\) Answer: 36 = \(\frac{w}{4}\) w/4 = 36 w = 36 × 4 w = 144

Question 6. Model with Mathematics One ride on a city bus costs $1.50. Martina has $18 on her bus pass. Write and solve an equation to find how many rides she can take without loading more money on her bus pass. Explain. Answer: Given, One ride on a city bus costs $1.50. Martina has $18 on her bus pass. The equation would be 1.5x = 18 x = 18/1.5 x = 12 She can take 12 rides without loading more money on her bus pass.

For Problems 7-10, find each solution.

Question 11. Bonnie buys 4 ounces of loose-leaf tea as a gift. The total cost of the tea is $25.00. Use the equation 4t = 25. How much does each ounce of tea cost? Answer: Given, Bonnie buys 4 ounces of loose-leaf tea as a gift. The total cost of the tea is $25.00. By using the equation 4t = 25 4t = 25 t = 25/4 t = 6.25 Thus each ounce of tea cost $6.25.

Question 12. Santos cuts a length of ribbon into 2-inch long pieces. He has enough ribbon to make 10 pieces. What is the total length of the ribbon that Santos has? Use the equation \(\frac{x}{2}\) = 10. Answer: Given, Santos cuts a length of ribbon into 2-inch long pieces. He has enough ribbon to make 10 pieces. \(\frac{x}{2}\) = 10. x = 10 × 2 x = 20 Thus the total length of the ribbon that Santos has is 20 inches.

Question 13. What is the solution to the equation 5x = 6? (A) 1.2 (B) 0.83 (C) 1 (D) 30 Answer: 5x = 6 x = 6/5 x = 1.2 Option A is the correct answer.

Spiral Review

Question 15. What is the quotient of 13.5 ÷ 0.75? Answer: Change the divisor 0.75 to a whole number by moving the decimal point 2 places to the right. Then move the decimal point in the dividend the same, 2 places to the right. We then have the equations: 1350 ÷ 75 = 18.000 and therefore: 13.5 ÷ 0.75 = 18.000 Both are calculated to 3 decimal places. Thus the quotient of 13.5 ÷ 0.75 is 18

Question 16. A grocery store sells long-grain rice for $2.49 per pound. Write an expression for the cost of y pounds of long grain rice. Answer: Given, A grocery store sells long-grain rice for $2.49 per pound. We need to write an expression for the cost of y pounds of long grain rice. The expression is a product of the cost of y and price per pound 2.49y

Question 17. Order the fractions from least to greatest. \(\frac{3}{8}\), \(\frac{4}{12}\), \(\frac{4}{9}\), \(\frac{3}{6}\), \(\frac{1}{4}\) Answer: Given fractions, \(\frac{3}{8}\), \(\frac{4}{12}\), \(\frac{4}{9}\), \(\frac{3}{6}\), \(\frac{1}{4}\) The denominator with the greatest number will be the smallest fraction Let us make all the denominators the same. LCM is 72 \(\frac{3}{8}\) = \(\frac{27}{72}\) \(\frac{4}{12}\) = \(\frac{24}{72}\) \(\frac{4}{9}\) = \(\frac{32}{72}\) \(\frac{3}{6}\) = \(\frac{36}{72}\) \(\frac{1}{4}\) = \(\frac{18}{72}\) \(\frac{27}{72}\), \(\frac{24}{72}\), \(\frac{32}{72}\), \(\frac{36}{72}\) and \(\frac{18}{72}\) Now we can write the fractions from least to the greatest. \(\frac{18}{72}\), \(\frac{24}{72}\), \(\frac{27}{72}\), \(\frac{32}{72}\) and \(\frac{36}{72}\) ×

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