Algebraic Expressions and Word Problems

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Examples, solutions, videos, worksheets, games and activities to help Algebra 1 or grade 7 students learn how to write algebraic expressions from word problems.

Beginning Algebra & Word Problem Steps

  • Name what x is.
  • Define everything in the problem in terms of x.
  • Write the equation.
  • Solve the equation.
  • Kevin’s age is 3 years more than twice Jane’s age. The sum of their ages is 39. How old is Kevin and Jane?
  • The difference between two numbers is 7. Find the two numbers if the larger number is three times the smaller.
  • Mary and Jim collect baseball cards, Mary has 5 more than 3 times as many cards as Jim. The total number of cards they both have is 253. How many cards does Mary have?

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How to Solve an Algebraic Expression

Last Updated: October 27, 2023 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 486,320 times.

An algebraic expression is a mathematical phrase that contains numbers and/or variables. Though it cannot be solved because it does not contain an equals sign (=), it can be simplified. You can, however, solve algebraic equations , which contain algebraic expressions separated by an equals sign. If you want to know how to master this mathematical concept, then see Step 1 to get started.

Understanding the Basics

Step 1 Understand the difference between an algebraic expression and an algebraic equation.

  • Algebraic expression : 4x + 2
  • Algebraic equation : 4x + 2 = 100

Step 2 Know how to combine like terms.

  • 3x 2 + 5 + 4x 3 - x 2 + 2x 3 + 9 =
  • 3x 2 - x 2 + 4x 3 + 2x 3 + 5 + 9 =
  • 2x 2 + 6x 3 + 14

Step 3 Know how to factor a number.

  • You can see that each coefficient can be divisible by 3. Just "factor out" the number 3 by dividing each term by 3 to get your simplified equation.
  • 3x/3 + 15/3 = 9x/3 + 30/3 =
  • x + 5 = 3x + 10

Step 4 Know the order of operations.

  • (3 + 5) 2 x 10 + 4
  • First, follow P, the operation in the parentheses:
  • = (8) 2 x 10 + 4
  • Then, follow E, the operation of the exponent:
  • = 64 x 10 + 4
  • Next, do multiplication:
  • And last, do addition:

Step 5 Learn how to isolate a variable.

  • 5x + 15 = 65 =
  • 5x/5 + 15/5 = 65/5 =
  • x + 3 = 13 =

Solve an Algebraic Equation

Step 1 Solve a basic linear algebraic equation.

  • 4x + 16 = 25 -3x =
  • 4x = 25 -16 - 3x
  • 4x + 3x = 25 -16 =
  • 7x/7 = 9/7 =

Step 2 Solve an algebraic equation with exponents.

  • First, subtract 12 from both sides.
  • 2x 2 + 12 -12 = 44 -12 =
  • Next, divide both sides by 2.
  • 2x 2 /2 = 32/2 =
  • Solve by taking the square root of both sides, since that will turn x 2 into x.
  • √x 2 = √16 =
  • State both answers:x = 4, -4

Step 3 Solve an algebraic expression with fractions.

  • First, cross multiply to get rid of the fraction. You have to multiply the numerator of one fraction by the denominator of the other.
  • (x + 3) x 3 = 2 x 6 =
  • Now, combine like terms. Combine the constant terms, 9 and 12, by subtracting 9 from both sides.
  • 3x + 9 - 9 = 12 - 9 =
  • Isolate the variable, x, by dividing both sides by 3 and you've got your answer.
  • 3x/3 = 3/3 =

Step 4 Solve an algebraic expression with radical signs.

  • First, move everything that isn't under the radical sign to the other side of the equation:
  • √(2x+9) = 5
  • Then, square both sides to remove the radical:
  • (√(2x+9)) 2 = 5 2 =
  • Now, solve the equation as you normally would by combining the constants and isolating the variable:
  • 2x = 25 - 9 =

Step 5 Solve an algebraic expression that contains absolute value.

  • |4x +2| - 6 = 8 =
  • |4x +2| = 8 + 6 =
  • |4x +2| = 14 =
  • 4x + 2 = 14 =
  • Now, solve again by flipping the sign of the term on the other side of the equation after you've isolated the absolute value:
  • 4x + 2 = -14
  • 4x = -14 -2
  • 4x/4 = -16/4 =
  • Now, just state both answers: x = -4, 3

Community Q&A

Donagan

  • The degree of a polynomial is the highest power within the terms. Thanks Helpful 9 Not Helpful 1
  • Once you're done, replace the variable with the answer, and solve the sum to see if it makes sense. If it does, then, congratulations! You just solved an algebraic equation! Thanks Helpful 7 Not Helpful 3
  • To cross-check your answer, visit wolfram-alpha.com. They give the answer and often the two steps. Thanks Helpful 8 Not Helpful 5

problem solving involving algebraic expressions

You Might Also Like

Evaluate an Algebraic Expression

  • ↑ https://www.math4texas.org/Page/527
  • ↑ https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-combining-like-terms/v/combining-like-terms-2
  • ↑ https://www.mathsisfun.com/algebra/factoring.html
  • ↑ https://www.mathsisfun.com/operation-order-pemdas.html
  • ↑ https://sciencing.com/tips-for-solving-algebraic-equations-13712207.html
  • ↑ https://www.mathsisfun.com/algebra/equations-solving.html
  • ↑ https://tutorial.math.lamar.edu/Classes/Alg/SolveExpEqns.aspx
  • ↑ https://www.mathsisfun.com/algebra/fractions-algebra.html
  • ↑ https://math.libretexts.org/Courses/Coastline_College/Math_C045%3A_Beginning_and_Intermediate_Algebra_(Chau_Duc_Tran)/10%3A_Roots_and_Radicals/10.07%3A_Solve_Radical_Equations
  • ↑ https://www.mathplanet.com/education/algebra-1/linear-inequalitites/solving-absolute-value-equations-and-inequalities

About This Article

David Jia

If you want to solve an algebraic expression, first understand that expressions, unlike equations, are mathematical phrase that can contain numbers and/or variables but cannot be solved. For example, 4x + 2 is an expression. To reduce the expression, combine like terms, for example everything with the same variable. After you've done that, factor numbers by finding the lowest common denominator. Then, use the order of operations, which is known by the acronym PEMDAS, to reduce or solve the problem. To learn how to solve algebraic equations, keep scrolling! Did this summary help you? Yes No

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1.4: Algebraic Expressions and Formulas

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  • Page ID 6229

Learning Objectives

  • Identify the parts of an algebraic expression.
  • Apply the distributive property.
  • Evaluate algebraic expressions.
  • Use formulas that model common applications.

Algebraic Expressions and the Distributive Property

In algebra, letters called variables are used to represent numbers. Combinations of variables and numbers along with mathematical operations form algebraic expressions 87 , or just expressions . The following are some examples of expressions with one variable, \(x\):

Terms 88 in an algebraic expression are separated by addition operators and factors 89 are separated by multiplication operators. The numerical factor of a term is called the coefficient 90 . For example, the algebraic expression \(x^{2} y^{2} + 6xy − 3\) can be thought of as \(x^{2} y^{2} + 6xy + (−3)\) and has three terms. The first term, \(x^{2} y^{2}\), represents the quantity \(1x^{2} y^{2} = 1 ⋅ x ⋅ x ⋅ y ⋅ y\) where \(1\) is the coefficient and x and y are the variables. All of the variable factors with their exponents form the variable part of a term 91 . If a term is written without a variable factor, then it is called a constant term 92 . Consider the components of \(x^{2} y^{2} + 6xy − 3\),

The third term in this expression, \(−3\), is called a constant term because it is written without a variable factor. While a variable represents an unknown quantity and may change, the constant term does not change.

Example \(\PageIndex{1}\):

List all coefficients and variable parts of each term: \(10a^{2}−5ab−b^{2}\).

We want to think of the third term in this example \(−b^{2}\) as  \(−1b^{2}\).

Answer : Coefficients: \(\{−5, −1, 10\}\); Variable parts: \(\{a^{2},  ab,  b^{2}\}\)

In our study of algebra, we will encounter a wide variety of algebraic expressions. Typically, expressions use the two most common variables, \(x\) and \(y\). However, expressions may use any letter (or symbol) for a variable, even Greek letters, such as alpha (\(\alpha\)) and beta (\(\beta\)). Some letters and symbols are reserved for constants, such as \(π ≈ 3.14159\) and \(e ≈ 2.71828\). Since there is only a limited number of letters, you will also use subscripts, \(x_{1} , x_{2} , x_{3} , x_{4} , …,\) to indicate different variables.

The properties of real numbers are important in our study of algebra because a variable is simply a letter that represents a real number. In particular, the distributive property 93 states that if given any real numbers \(a, b\) and \(c\), then,

\(\color{Cerulean}{a}\) \( ( b + c ) = \color{Cerulean}{a}\)\(b + \color{Cerulean}{a}\)\(c\)

This property is one that we apply often when simplifying algebraic expressions. To demonstrate how it will be used, we simplify \(2(5 − 3)\) in two ways, and observe the same correct result.

Certainly, if the contents of the parentheses can be simplified we should do that first. On the other hand, when the contents of parentheses cannot be simplified any further, we multiply every term within it by the factor outside of it using the distributive property. Applying the distributive property allows us to multiply and remove the parentheses.

Example \(\PageIndex{2}\):

Simplify: \(5(−2a+5b)−2c\).

Multiply only the terms grouped within the parentheses for which we are applying the distributive property.

de8d707ad6f92b32ffb9e5a5c6fcc2fd.png

\(=\color{Cerulean}{5}\)\(⋅(−2a)+\color{Cerulean}{5}\)\(⋅5b−2c\)

\(=−10a+25b−2c\)

Answer : \(−10a+25b−2c\)

Recall that multiplication is commutative and therefore we can write the distributive property in the following manner, \((b + c) a = ba + ca\).

Example \(\PageIndex{3}\):

Simplify:  \((3x−4y+1)⋅3\).

Multiply all terms within the parenthesis by \(3\).

\((3x−4y+1)⋅3=3x\color{Cerulean}{⋅3}\)\(−4y\color{Cerulean}{⋅3}\)\(+1\color{Cerulean}{⋅3}\)

\(=9x−12y+3\)

Answer : \(9x−12y+3\)

Terms whose variable parts have the same variables with the same exponents are called like terms 94 , or similar terms 95 . Furthermore, constant terms are considered to be like terms. If an algebraic expression contains like terms, apply the distributive property as follows:

\(5 \color{Cerulean}{x}\)\( + 7 \color{Cerulean}{x}\)\( = ( 5 + 7 ) \color{Cerulean}{x}\)\( = 12 \color{Cerulean}{x}\)

\(4 \color{Cerulean}{x ^ { 2 }}\)\( + 5 \color{Cerulean}{x ^ { 2 }}\)\( - 7 \color{Cerulean}{x ^ { 2 }}\)\( = ( 4 + 5 - 7 ) \color{Cerulean}{x ^ { 2 }}\)\( = 2 \color{Cerulean}{x ^ { 2 }}\)

In other words, if the variable parts of terms are exactly the same, then we can add or subtract the coefficients to obtain the coefficient of a single term with the same variable part. This process is called combining like terms 96 . For example,

\(12 x ^ { 2 } y ^ { 3 } + 3 x ^ { 2 } y ^ { 3 } = 15 x ^ { 2 } y ^ { 3 }\)

Notice that the variable factors and their exponents do not change. Combining like terms in this manner, so that the expression contains no other similar terms, is called simplifying the expression 97 . Use this idea to simplify algebraic expressions with multiple like terms.

Example \(\PageIndex{4}\):

\(x ^ { 2 } - 10 x + 8 + 5 x ^ { 2 } - 6 x - 1\).

Identify the like terms and add the corresponding coefficients.

\(\color{Cerulean}{\underline{1x^{2}}}\) \( - \color{OliveGreen}{\underline{\underline{10x}}}\)\( + \underline{\underline{\underline{8}}} + \color{Cerulean}{\underline{5 x ^ { 2 }}}\)\( -\color{OliveGreen}{\underline{\underline{6x}}}\)\( - \underline{\underline{\underline{1}}}\) \(\color{Cerulean}{Combine\: like\: terms.}\)

\(= 6 x ^ { 2 } - 16 x + 7\)

Answer : \(6 x ^ { 2 } - 16 x + 7\)

Example \(\PageIndex{5}\):

Simplify: \(a ^ { 2 } b ^ { 2 } - a b - 2 \left( 2 a ^ { 2 } b ^ { 2 } - 5 a b + 1 \right)\).

Distribute \(−2\) and then combine like terms.

\(\begin{aligned} a ^ { 2 } b ^ { 2 } - a b - 2 \left( 2 a ^ { 2 } b ^ { 2 } - 5 a b + 1 \right) & = a ^ { 2 } b ^ { 2 } - a b - 4 a ^ { 2 } b ^ { 2 } + 10 a b - 2 \\ & = - 3 a ^ { 2 } b ^ { 2 } + 9 a b - 2 \end{aligned}\)

Answer : \(- 3 a ^ { 2 } b ^ { 2 } + 9 a b - 2\)

Evaluating Algebraic Expressions

An algebraic expression can be thought of as a generalization of particular arithmetic operations. Performing these operations after substituting given values for variables is called evaluating 98 . In algebra, a variable represents an unknown value. However, if the problem specifically assigns a value to a variable, then you can replace that letter with the given number and evaluate using the order of operations.

Example \(\PageIndex{6}\):

  • \(5x − 2\) where \(x =\frac{2}{3}\)
  • \(y^{2} − y − 6\) where \(y = −4\)

To avoid common errors, it is a best practice to first replace all variables with parentheses, and then replace, or substitute 99 , the appropriate given value.

\(\begin{aligned} 5 x - 2 & = 5 (\:\: ) - 2 \\ & = 5 \left(\color{OliveGreen}{ \frac { 2 } { 3 }} \right) - 2 \\ & = \frac { 10 } { 3 } - \frac { 2 } { 1 } \cdot \color{Cerulean}{\frac { 3 } { 3 }} \\ & = \frac { 10 - 6 } { 3 } \\ & = \frac { 4 } { 3 } \end{aligned}\)

\(y ^ { 2 } - y - 6 = (\:\: ) ^ { 2 } - (\:\: ) - 6\)

\(= ( \color{OliveGreen}{- 4}\)\( ) ^ { 2 } - ( \color{OliveGreen}{- 4}\)\( ) - 6\)

\(\begin{array} { l } { = 16 + 4 - 6 } \\ { = 14 } \end{array}\)

a. \(\frac{4}{3}\)

Often algebraic expressions will involve more than one variable.

Example \(\PageIndex{7}\):

Evaluate \(a ^ { 3 } - 8 b ^ { 3 }\) where \(a = −1\) and \(b = \frac{1}{2}\).

After substituting in the appropriate values, we must take care to simplify using the correct order of operations.

\(a ^ { 3 } - 8 b ^ { 3 } = (\:\: ) ^ { 3 } - 8 (\:\: ) ^ { 3 } \color{Cerulean}{Replace\: variables\: with\: parentheses.}\)

\(= ( \color{OliveGreen}{- 1}\) \( )^{3} -8(\color{OliveGreen}{\frac{1}{2}}\)\()^{3} \color{Cerulean}{Substitute\: in\: the\: appropriate\: values.}\)

\(= - 1 - 8 \left( \frac { 1 } { 8 } \right) \color{Cerulean}{Simplify.}\)

\(\begin{array} { l } { = - 1 - 1 } \\ { = - 2 } \end{array}\)

Answer : \(-2\)

Example \(\PageIndex{8}\):

Evaluate \(\frac { x ^ { 2 } - y ^ { 2 } } { 2 x - 1 }\) where \(x = −\frac{3}{2}\) and \(y = −3\).

\(\frac { x ^ { 2 } - y ^ { 2 } } { 2 x - 1 } = \frac { (\:\: ) ^ { 2 } - ( \:\:) ^ { 2 } } { 2 ( \:\:) - 1 }\)

\(= \frac { \left( \color{OliveGreen}{- \frac { 3 } { 2 }} \right) ^ { 2 } - ( \color{OliveGreen}{- 3} \color{Black}{) ^ { 2 } }} { 2 \left( - \color{OliveGreen}{\frac { 3 } { 2 }} \right) - 1 }\)

\(= \frac { \frac { 9 } { 4 } - 9 } { - 3 - 1 }\)

At this point we have a complex fraction. Simplify the numerator and then multiply by the reciprocal of the denominator.

\(\begin{aligned} & = \frac { \frac { 9 } { 4 } - \frac { 9 } { 1 } \cdot \color{Cerulean}{\frac { 4 } { 4 } }} { - 4 } \\ & = \frac { \frac { - 27 } { 4 } } { { \frac { - 4 } { 1 } } } \\ & = \frac { - 27 } { 4 } \left( - \frac { 1 } { 4 } \right) \\ & = \frac { 27 } { 16 } \end{aligned}\)

Answer : \(\frac { 27 } { 16 }\)

The answer to the previous example can be written as a mixed number, \(\frac { 27 } { 16 } = 1 \frac { 11 } { 16 }\). Unless the original problem has mixed numbers in it, or it is an answer to a realworld application, solutions will be expressed as reduced improper fractions.

Example \(\PageIndex{9}\):

Evaluate \(\sqrt { b ^ { 2 } - 4 a c }\) where \(a = −1, b = −7\), and \(c = \frac{1}{4}\).

Substitute in the appropriate values and then simplify.

\(\sqrt { b ^ { 2 } - 4 a c } = \sqrt { ( \:\: ) ^ { 2 } - 4 ( \:\: ) \:\:(\:\:) }\)

\( = \sqrt { ( \color{OliveGreen}{- 7}\color{Black}{ ) ^ { 2 } - 4 (}\color{OliveGreen}{ - 1}\color{Black}{ ) (}\color{OliveGreen}{ \frac { 1 } { 4 }}\color{Black}{)} } \)

\(\begin{aligned} & =\sqrt { 49 + 4(\frac{1}{4}) } \\ & = \sqrt { 49 + 1 } \\ & =\sqrt{50} \\& = \sqrt { 25 \cdot 2 } \\ & = 5 \sqrt { 2 } \end{aligned}\)

Aligned : \(5 \sqrt { 2 }\)

Exercise \(\PageIndex{1}\)

Evaluate \(\frac { \sqrt { 3 \pi V h } } { \pi h }\) where \(V = 25\pi\) and \(h = 3\).

www.youtube.com/v/Y4RCMceThu4

Using Formulas

The main difference between algebra and arithmetic is the organized use of variables. This idea leads to reusable formulas 100 , which are mathematical models using algebraic expressions to describe common applications. For example, the volume of a right circular cone depends on its radius \(r\) and height \(h\) and is modeled by the formula:

\(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\)

458a624ccad311ca7b916123ca6f3aa4.png

In this equation, variables and constants are used to describe the relationship between volume and the length of the base and height. If the radius of the base measures \(3\) meters and the height measures \(5\) meters, then the volume can be calculated using the formula as follows:

\(\begin{aligned} V & = \frac { 1 } { 3 } \pi r ^ { 2 } h \\ & = \frac { 1 } { 3 } \pi ( 3 m ) ^ { 2 } ( 5 m ) \\ & = \frac { 1 } {\bcancel {3}} \pi \cdot \stackrel{\color{Cerulean}{3}}{\bcancel{9}} \cdot 5 m ^ { 3 } \\ & = 15 \pi \mathrm { m } ^ { 3 } \end{aligned}\)

Using \(π ≈ 3.14\), we can approximate the volume: \(V ≈ 15 (3.14) = 47.1\) cubic meters.

A list of formulas that describe the area and perimeter of common plane figures follows. The letter P represents perimeter and is measured in linear units. The letter A represents area and is measured in square units.

c322369730d452c465fea7d9501a5156.png

A list of formulas that describe the surface area and volume of common figures follows. Here SA represents surface area and is measured in square units. The letter V represents volume and is measured in cubic units.

6d0cb9d2d39d7326bc7e45de16ea11df.png

Example \(\PageIndex{10}\):

The diameter of a spherical balloon is \(10\) inches. Determine the volume rounded off to the nearest hundredth.

The formula for the volume of a sphere is

\(V = \frac { 4 } { 3 } \pi r ^ { 3 }\)

This formula gives the volume in terms of the radius, \(r\). Therefore, divide the diameter by \(2\) and then substitute into the formula. Here, \(r = \frac{10}{2} = 5\) inches and we have

\(\begin{aligned} V & = \frac { 4 } { 3 } \pi r ^ { 3 } \\ & = \frac { 4 } { 3 } \pi ( 5 \mathrm { in } ) ^ { 3 } \\ & = \frac { 4 } { 3 } \pi \cdot 125 \mathrm { in } ^ { 3 } \\ & = \frac { 500 \pi } { 3 } \mathrm { in } ^ { 3 } \approx 523.60 \mathrm { in } ^ { 3 } \end{aligned}\)

Answer : The volume of the balloon is approximately \(523.60\) cubic inches.

Formulas can be found in a multitude of subjects. For example, uniform motion 101 is modeled by the formula \(D = rt\), which expresses distance \(D\), in terms of the average rate, or speed, \(r\) and the time traveled at that rate, \(t\). This formula, \(D = rt\), is used often and is read, “ distance equals rate times time .”

Example \(\PageIndex{11}\):

Jim’s road trip took \(2\:\frac{1}{2}\) hours at an average speed of \(66\) miles per hour. How far did he travel?

Substitute the appropriate values into the formula and then simplify.

\(\begin{aligned} D & = r \cdot t \\ & = ( \color{Cerulean}{66 \frac { \mathrm { mi } } { \mathrm { hr } }}\color{Black}{ ) \cdot (}\color{Cerulean}{ 2 \frac { 1 } { 2 } \mathrm { hr }}\color{Black}{)} \\ & = \frac { 66 } { 1 } \cdot \frac { 5 } { 2 } \mathrm { mi } \\ & = 33 \cdot 5 \mathrm { mi } \\ & = 165 \mathrm { mi } \end{aligned}\)

Answer : Jim traveled \(165\) miles.

Simple interest 102 \(I\) is given by the formula \(I = prt\), where \(p\) represents the principal amount invested at an annual interest rate \(r\) for \(t\) years.

Example \(\PageIndex{12}\):

Calculate the simple interest earned on a \(2\)-year investment of \($1,250\) at an annual interest rate of \(3\:\frac{3}{4} %\).

Convert \(3\:\frac{3}{4}%\) to a decimal number before using it in the formula.

\(r = 3 \frac { 3 } { 4 } \% = 3.75 \% = 0.0375\)

Use this and the fact that \(p = $1,250\) and \(t = 2\) years to calculate the simple interest.

\(\begin{aligned} I & = p r t \\ & = ( \color{Cerulean}{1,250}\color{Black}{ ) (}\color{Cerulean}{ 0.0375}\color{Black}{ ) (}\color{Cerulean}{ 2}\color{Black}{ )} \\ & = 93.75 \end{aligned}\)

Answer : The simple interest earned is \($93.75\).

Key Takeaways

  • Think of algebraic expressions as generalizations of common arithmetic operations that are formed by combining numbers, variables, and mathematical operations.
  • The distributive property \(a (b + c) = ab + ac\), is used when multiplying grouped algebraic expressions. Applying the distributive property allows us to remove parentheses.
  • Combine like terms, or terms whose variable parts have the same variables with the same exponents, by adding or subtracting the coefficients to obtain the coefficient of a single term with the same variable part. Remember that the variable factors and their exponents do not change.
  • To avoid common errors when evaluating, it is a best practice to replace all variables with parentheses and then substitute the appropriate values.
  • The use of algebraic expressions allows us to create useful and reusable formulas that model common applications.

Exercise \(\PageIndex{2}\)

List all of the coefficients and variable parts of each term.

  • \(−5x^{2} + x − 1\)
  • \(y^{2} − 9y + 3\)
  • \(5x^{2} − 3xy + y^{2}\)
  • \(a^{2}b^{2} + 2ab − 4\)
  • \(x^{2}y + xy^{2} − 3xy + 9\)
  • \(x^{4} − x^{3} + x^{2} − x + 2\)

1. Coefficients: \(\{−5, 1, −1\}\) ; variable parts: \(\{x^{2} , x\}\)

3. Coefficients: \(\{5, −3, 1\}\) ; variable parts: \(\{x^{2} , xy, y^{2} \}\)

5. Coefficients: \(\{1, −3, 9\}\) ; variable parts: \(\{x^{2}y, xy^{2} , xy\}\)

Exercise \(\PageIndex{3}\)

  • \(5 (3x − 5) \)
  • \(3 (4x − 1) \)
  • \(−2 (2x^{2} − 5x + 1) \)
  • \(−5 (6x^{2} − 3x − 1)\)
  • \(\frac{2}{3} (9y^{2} + 12y − 3)\)
  • \(−\frac{3}{4} (8y^{2} + 20y + 4)\)
  • \(12(\frac{1}{3} a^{2} − \frac{5}{6} a + \frac{7}{12} )\)
  • \(−9 (\frac{1}{9} a^{2} − \frac{5}{3} a + 1 )\)
  • \(9 (a^{2} − 2b^{2} )\)
  • \(−5 (3x^{2} − y^{2} )\)
  • \((5a^{2} − 3ab + b^{2} ) ⋅ 6\)
  • \((a^{2}b^{2} − 9ab − 3) ⋅ 7\)
  • \(− (5x^{2} − xy + y^{2} )\)
  • \(− (x^{2}y^{2} − 6xy − 1)\)

1. \(15x − 25\)

3. \(−4x^{2} + 10x − 2\)

5. \(6y^{2} + 8y − 2\)

7. \(4a^{2} − 10a + 7\)

9. \(9a^{2} − 18b^{2}\)

11. \(30a^{2} − 18ab + 6b^{2}\)

13. \(−5x^{2} + xy − y^{2}\)

Exercise \(\PageIndex{4}\)

Combine like terms.

  • \(18x − 5x + 3x\)
  • \(30x − 50x + 10x\)
  • \(3y − 4 + 2y − 12\)
  • \(12y + 7 − 15y − 6\)
  • \(2x^{2} − 3x + 2 + 5x^{2} − 6x + 1\)
  • \(9x^{2} + 7x − 5 − 10x^{2} − 8x + 6\)
  • \(\frac{3}{5} a^{2} − \frac{1}{2} + \frac{1}{3} a^{2} + \frac{4}{5}\)
  • \(\frac{1}{6} a^{2} + \frac{2}{3} − \frac{4}{3} a^{2} − \frac{1}{9}\)
  • \(\frac{1}{2} y^{2} + \frac{2}{3} y − 3 + \frac{3}{5} y^{2} + \frac{1}{3} y − \frac{7}{3}\)
  • \(\frac{5}{6} x^{2} + \frac{1}{8} x − 1 − \frac{1}{2} x^{2} + \frac{3}{4} x − \frac{4}{5}\)
  • \(a^{2}b^{2} + 5ab − 2 + 7a^{2}b^{2} − 6ab + 12\)
  • \(a^{2} − 12ab + 4b^{2} − 6a^{2} + 10ab − 5b^{2}\)
  • \(3x^{2}y + 12xy − 5xy^{2} + 5xy − 8x^{2}y + 2xy^{2}\)
  • \(10x^{2}y + 2xy − 4xy^{2} + 2x^{2}y − 8xy + 5xy^{2}\)
  • \(7m^{2}n − 9mn + mn^{2} − 6m^{2}n + mn − 2mn^{2}\)
  • \(m^{2}n − 5mn + 5mn^{2} − 3m^{2}n + 5mn + 2mn^{2}\)
  • \(x^{2n} − 3x^{n} + 5 + 2x^{2n} − 4x^{n} − 3\)
  • \(5y^{2n} − 3y^{n} + 1 − 3y^{2n} − 2y^{n} − 1\)

3. \(5y − 16\)

5. \(7x^{2} − 9x + 3\)

7. \(\frac{14}{15}a^{2} + \frac{3}{10}\)

9. \(\frac{11}{10} y^{2} + y − \frac{16}{3}\)

11. \(8a^{2}b^{2} − ab + 10\)

13. \(−5x^{2}y + 17xy − 3xy^{2}\)

15. \(m^{2}n − 8mn − mn^{2}\)

17. \(3x^{2n} − 7x^{n} + 2\)

Exercise \(\PageIndex{5}\)

  • \(5 − 2 (4x + 8)\)
  • \(8 − 6 (2x − 1)\)
  • \(2 (x^{2} − 7x + 1) + 3x − 7\)
  • \(−5 (x^{2} + 4x − 1) + 8x^{2} − 5\)
  • \(5ab − 4 (ab + 5)\)
  • \(5 (7 − ab) + 2ab\)
  • \(2 − a^{2} + 3 (a^{2} + 4)\)
  • \(7 − 3y + 2 (y^{2} − 3y − 2)\)
  • \(8x^{2} − 3x − 5 (x^{2} + 4x − 1)\)
  • \(2 − 5y − 6 (y^{2} − y + 2)\)
  • \(a^{2}b^{2} − 5 + 3 (a^{2}b^{2} − 3ab + 2)\)
  • \(a^{2} − 3ab − 2 (a^{2} − ab + 1)\)
  • \(10y^{2} + 6 − (3y^{2} + 2y + 4)\)
  • \(4m^{2} − 3mn − (m^{2} − 3mn + n^{2} )\)
  • \(x^{2n} − 3x^{n} + 5 (x^{2n} − x^{n} + 1)\)
  • \(−3 (y^{2n} − 2y^{n} + 1) + 4y^{2n} − 5\)

1. \(−8x − 11\)

3. \(2x^{2} − 11x − 5\)

5. \(ab − 20\)

7. \(2a^{2} + 14\)

9. \(3x^{2} − 23x + 5\)

11. \(4a^{2}b^{2} − 9ab + 1\)

13. \(7y^{2} − 2y + 2\)

15. \(6x^{2n} − 8x^{n} + 5\)

Exercise \(\PageIndex{6}\)

  • \(−2x + 3\) where \(x = −2\)
  • \(8x − 5\) where \(x = −1\)
  • \(x^{2} − x + 5\) where \(x = −5\)
  • \(2x^{2} − 8x + 1\) where \(x = 3\)
  • \(\frac { x ^ { 2 } - x + 2 } { 2 x - 1 }\) where \(x = -\frac{1}{2}\)
  • \(\frac { 9 x ^ { 2 } + x - 2 } { 3 x - 4 }\) where \(x = -\frac{2}{3}\)
  • \(( 3 y - 2 ) ( y + 5 )\) where \(y = \frac { 2 } { 3 }\)
  • \((3x + 2) (5x + 1)\) where \(x = −\frac{1}{5}\)
  • \((3x − 1) (x − 8)\) where \(x = −1\)
  • \((7y + 5) (y + 1)\) where \(y = −2\)
  • \(y^{6} − y^{3} + 2\) where \(y = −1\)
  • \(y^{5} + y^{3} − 3\) where \(y = −2\)
  • \(a^{2} − 5b^{2}\) where \(a = −2\) and \(b = −1\)
  • \(a^{3} − 2b^{3}\) where \(a = −3\) and \(b = 2\)
  • \((x − 2y) (x + 2y)\) where \(x = 2\) and \(y = −5\)
  • \((4x − 3y) (x − y)\) where \(x = −4\) and \(y = −3\)
  • \(a^{2} − ab + b^{2}\) where \(a = −1\) and \(b = −2\)
  • \(x^{2}y^{2} − xy + 2\) where \(x = −3\) and \(y = −2\)
  • \(a^{4} − b^{4}\) where \(a = −2\) and \(b = −3\)
  • \(a^{6} − 2a^{3}b^{3} − b^{6}\) where \(a = 2\) and \(b = −1\)

5. \(−\frac{11}{8}\)

13. \(−1\)

15. \(−96\)

19. \(−65\)

Exercise \(\PageIndex{7}\)

Evaluate \(\sqrt { b ^ { 2 } - 4 a c }\) given the following values.

  • \(a = 6, b = 1\) and \(c = −1\)
  • \(a = 15, b = 4\) and \(c = −4\)
  • \(a = \frac{3}{4} , b = −2\) and \(c = −4\)
  • \(a = \frac{1}{2} , b = −2\) and \(c = −30\)
  • \(a = 1, b = 2\) and \(c = −1\)
  • \(a = 1, b = −4\) and \(c = −50\)
  • \(a = 1, b = −1\) and \(c = −\frac{1}{16}\)
  • \(a = −2, b = −\frac{1}{3}\) and \(c = 1\)

5. \(2\sqrt{2}\)

7. \(\frac { \sqrt { 5 } } { 2 }\)

Exercise \(\PageIndex{8}\)

Convert the following temperatures to degrees Celsius given \(C = \frac{5}{9} (F − 32)\), where F represents degrees Fahrenheit.

  • \(95°\)F
  • \(86°\)F
  • \(32°\)F
  • \(−40°\)F

1. \(35°\)C

3. \(0°\)C

Exercise \(\PageIndex{9}\)

  • Calculate the perimeter and area of a rectangle with dimensions \(12\) feet by \(5\) feet.
  • Calculate the perimeter and area of a rectangle with dimensions \(5\) meters by \(1\) meter.
  • Calculate the surface area and volume of a sphere with radius \(6\) centimeters.
  • The radius of the base of a right circular cylinder measures \(4\) inches and the height measures \(10\) inches. Calculate the surface area and volume.
  • Calculate the volume of a sphere with a diameter of \(18\) centimeters.
  • The diameter of the base of a right circular cone measures \(6\) inches. If the height is \(1\:\frac{1}{2}\) feet, then calculate its volume.
  • Given that the height of a right circular cylinder is equal to the radius of the base, derive a formula for the surface area in terms of the radius of the base.
  • Given that the area of the base of a right circular cylinder is \(25π\) square inches, find the volume if the height is \(1\) foot.
  • Jose was able to drive from Tucson to Phoenix in \(2\) hours at an average speed of \(58\) mph. How far is Phoenix from Tucson?
  • If a bullet train can average \(152\) mph, then how far can it travel in \(\frac{3}{4}\) of an hour?
  • Margaret traveled for \(1\:\frac{3}{4}\) hour at an average speed of \(68\) miles per hour. How far did she travel?
  • The trip from Flagstaff, AZ to the Grand Canyon national park took \(1\:\frac{1}{2}\) hours at an average speed of \(54\) mph. How far is the Grand Canyon national park from Flagstaff?
  • Calculate the simple interest earned on a \(3\)-year investment of \($2,500\) at an annual interest rate of \(5\:\frac{1}{4} \)%.
  • Calculate the simple interest earned on a \(1\)-year investment of \($5,750\) at an annual interest rate of \(2\:\frac{5}{8} \)%.
  • What is the simple interest earned on a \(5\)-year investment of \($20,000\) at an annual interest rate of \(6\)%?
  • What is the simple interest earned on a \(1\)-year investment of \($50,000\) at an annual interest rate of \(4.5\)%?
  • The time \(t\) in seconds an object is in free fall is given by the formula \(t = \frac { \sqrt { s } } { 4 }\), where s represents the distance in feet the object has fallen. How long does it take an object to fall \(32\) feet? (Give the exact answer and the approximate answer to the nearest hundredth.)
  • The current \(I\) measured in amperes, is given by the formula \(I = \sqrt { \frac { P } { R } }\), where \(P\) is the power usage measured in watts, and \(R\) is the resistance measured in ohms. If a light bulb uses \(60\) watts of power and has \(240\) ohms of resistance, then how many amperes of current are required?

1. \(P = 34\) feet; \(A = 60\) square feet

3. \(SA = 144π\) square centimeters; \(V = 288π\) cubic centimeters

5. \(972π\) cubic centimeters

7. \(SA = 4πr^{2}\)

9. \(116\) miles

11. \(119\) miles

13. \($393.75\)

15. \($6,000\)

17. \(\sqrt { 2 } \approx 1.41\) seconds

Exercise \(\PageIndex{10}\)

  • Find and post a useful mathematical model. Demonstrate its use with some values.
  • Research and discuss the history of the variable. What can we use if we run out of letters?
  • Find and post a link to a useful resource describing the Greek alphabet.
  • Given the algebraic expression \(5 − 3 (9x − 1)\), explain why we do not subtract \(5\) and \(3\) first.
  • Do we need a separate distributive property for more than two terms? For example, \(a (b + c + d) = ab + ac + ad\). Explain.
  • How can we check to see if we have simplified an expression correctly?

1. Answer may vary

3. Answer may vary

5. Answer may vary

87 Combinations of variables and numbers along with mathematical operations used to generalize specific arithmetic operations.

88 Components of an algebraic expression separated by addition operators.

89 Components of a term separated by multiplication operators.

90 The numerical factor of a term.

91 All the variable factors with their exponents.

92 A term written without a variable factor.

93 Given any real numbers \(a, b,\) and \(c, a (b + c) = ab + ac\) or \((b + c) a = ba + ca\).

94 Constant terms or terms whose variable parts have the same variables with the same exponents.

95 Used when referring to like terms.

96 Adding or subtracting like terms within an algebraic expression to obtain a single term with the same variable part.

97 The process of combining like terms until the expression contains no more similar terms.

98 The process of performing the operations of an algebraic expression for given values of the variables.

99 The act of replacing a variable with an equivalent quantity.

100 A reusable mathematical model using algebraic expressions to describe a common application.

101 The distance \(D\) after traveling at an average rate \(r\) for some time \(t\) can be calculated using the formula \(D = rt\).

102 Modeled by the formula \(I = prt\), where \(p\) represents the principal amount invested at an annual interest rate \(r\) for \(t\) years.

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Course: Algebra 2   >   Unit 10

  • Rational equations intro

Equations with rational expressions

  • Equations with rational expressions (example 2)
  • Rational equations
  • Finding inverses of rational functions
  • Find inverses of rational functions

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IMAGES

  1. GRADE 7 MATH : 👉Solving Problems Involving Algebraic Expressions

    problem solving involving algebraic expressions

  2. PROBLEMS INVOLVING ALGEBRAIC EXPRESSIONS

    problem solving involving algebraic expressions

  3. Solving Problems Involving Rational Algebraic Expressions

    problem solving involving algebraic expressions

  4. Solving Word Problem Involving Rational Algebraic Expression

    problem solving involving algebraic expressions

  5. How to Solve an Algebraic Expression: 10 Steps (with Pictures)

    problem solving involving algebraic expressions

  6. SOLVING PROBLEMS INVOLVING RATIONAL ALGEBRAIC EXPRESSIONS || GRADE 8 MATHEMATICS Q1

    problem solving involving algebraic expressions

VIDEO

  1. PROBLEM SOLVING INVOLVING RATIONAL ALGEBRAIC EXPRESSIONS

  2. Algebraic Manipulation

  3. math involving algebraic skills in a regular octoagon

  4. SAT Math

  5. Tip Use In Approaching Question in Quadratic Equation As A Mathematics Student

  6. algebraic problem solving strategy #maths #hsc #mathstricks #hscbangla #mathematics #tricks

COMMENTS

  1. Algebraic Expressions and Word Problems - Online Math Help ...

    Examples, solutions, videos, worksheets, games and activities to help Algebra 1 or grade 7 students learn how to write algebraic expressions from word problems. Beginning Algebra & Word Problem Steps. Name what x is. Define everything in the problem in terms of x. Write the equation.

  2. Algebraic word problems | Lesson (article) | Khan Academy

    Algebraic word problems are questions that require translating sentences to equations, then solving those equations. The equations we need to write will only involve. basic arithmetic operations. and a single variable. Usually, the variable represents an unknown quantity in a real-life scenario.

  3. Algebraic Sentences Word Problems | ChiliMath

    Almost always, the word “is” in an algebraic sentence denotes the symbol of equality. In our example above, the algebraic sentence, “ Five more than twice a number is forty-three “, is translated and written into its equation form: [latex]2x + 5 = 43 [/latex]. But before we delve into solving word problems that involve algebraic ...

  4. Grade 7 Mathematics Module: Solving Problems Involving ...

    In this lesson, we will use algebraic expression to find the values of the things or unknown in real life situations. This module was designed and written with you in mind. It is here to help you master Solving Problems Involving Algebraic Expressions. The scope of this module permits it to be used in many different learning situations.

  5. How to Solve an Algebraic Expression: 10 Steps (with Pictures)

    First, move everything that isn't under the radical sign to the other side of the equation: √ (2x+9) = 5. Then, square both sides to remove the radical: (√ (2x+9)) 2 = 5 2 =. 2x + 9 = 25. Now, solve the equation as you normally would by combining the constants and isolating the variable: 2x = 25 - 9 =. 2x = 16.

  6. 6.1: Evaluating Algebraic Expressions - Mathematics LibreTexts

    Evaluate the expression ( a − b) 2 If a = 3 and b = −5, at a = 3 and b = −5. Solution. Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression ( a − b) 2 with open parentheses. (a − b)2 = (() − ())2 ( a − b) 2 = ( () − ()) 2. Secondly, replace each variable with its ...

  7. 1.4: Algebraic Expressions and Formulas - Mathematics LibreTexts

    Table 1.4.1. Terms88 in an algebraic expression are separated by addition operators and factors89 are separated by multiplication operators. The numerical factor of a term is called the coefficient90. For example, the algebraic expression x2y2 + 6xy − 3 can be thought of as x2y2 + 6xy + ( − 3) and has three terms.

  8. Algebraic Expressions From Word Problems | Overview ...

    An algebraic expression is a series of terms - numbers, variables, or a product of numbers and/or variables - separated by signs of addition and subtraction. In some cases, an algebraic expression ...

  9. Algebraic expressions | Algebra basics | Math | Khan Academy

    The core idea in algebra is using letters to represent relationships between numbers without specifying what those numbers are! Let's explore the basics of communicating in algebraic expressions. If you're seeing this message, it means we're having trouble loading external resources on our website.

  10. Equations with rational expressions (video) | Khan Academy

    The problem with cross multiplication is that you end up with a much more difficult problem. Your polynomial takes a lot more work to try and find the factors. The technique shown in the video eliminates some factors in the denominators making the problem more manageable and simple to solve.