linear equations objective questions

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Questions on solving linear equations.

Multiple choice questions, with answers, on solving linear equations are presented. You may want to work through Solving Linear Equations - Tutorial before you start answering the questions below.

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Answers to the above questions., more references and links, popular pages.

• Solving Linear Equations - Tutorial
• Solve Equations, Systems of Equations and Inequalities
• Free Algebra Questions and Problems with Answers
• Applications of Linear Equations - Problems with Answers for Grade 8

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Linear equations

Here you will learn about linear equations, including how to graph linear equations and how to solve linear equations.

Students will first learn about linear equations as part of expression and equations in 8th grade math and will continue to use them throughout high school.

What is a linear equation?

A linear equation is an algebraic equation that contains variables that have an exponent that is not higher than one. A linear equation is called linear because when you plot the graph of the linear function, it will make a straight line graph.

Examples of equations

Nonlinear equations are a type of equations where terms can have variables raised to powers other than one, or terms that are multiplied or divided.

Examples of nonlinear equations:

• \cfrac{1}{x}+\cfrac{1}{y}=2

All linear equations have an equal sign, this means that everything on the left-hand side of the sign is exactly the same as everything on the right-hand side.

The linear equation formula is the way of expressing a linear equation and can be done in various ways, depending on the number of variables present.

• Standard form
• Slope-intercept form of a linear equation
• Point slope form of a linear equation

The standard form is ax+by=c, where a, b and c are constants.

For example,

\begin{aligned}& 2x-3y+4=0 \\\\ &3y-7x=8 \\\\ &4x=12 \end{aligned}

Slope-intercept form

The slope-intercept form of a linear equation can be written as y=mx+b, where m is the slope of a line and b is the y -intercept.

Point slope form

The point slope form of a linear equation can be written as y-y_1=m\left(x-x_1\right), where m is the slope of the line and \left(x_1, y_1\right) is a point on the line.

Graphing linear equations

Linear equations are related to coordinate grids, as they provide a way to graphically represent and solve linear equations.

Given the equation

\begin{aligned}&2x-3y+4=0 \\\\ &3y-7x=8 \\\\ &4x=12\end{aligned}

we can graphically represent this equation on a coordinate grid

A one variable linear equation with the variable x forms a vertical line that is parallel to the y -axis.

Given the equation x-4=0

You need to isolate the variable by adding 4 to both sides of the equation.

\begin{aligned}& x-4=0 \\\\ & +4+4 \\\\ & x=4\end{aligned}

To graph the line x=4, you will draw a straight, vertical line through point (4,0).

The slope in a vertical line is undefined.

A one variable linear equation with the variable y forms a horizontal line that is parallel to the y -axis.

Given the equation y+6=0

You need to isolate the variable by subtracting 6 from both sides of the equation.

\begin{gathered}y+6=0 \\\\ -6-6 \\\\ y=-6\end{gathered}

To graph the line y=-6, you will draw a straight, vertical line through point (0,-6).

The slope in a horizontal line is 0.

Common Core State Standards

How does this relate to 8 th grade math?

• Grade 8: Expressions and Equations (8.EE.C.7) Solve linear equations in one variable. a. Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers). b. Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
• High School: Algebra – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Linear Equations Worksheet (Grade 6 to 8)

Use this worksheet to check your grade 6 to 8 students’ understanding of linear equations. 15 questions with answers to identify areas of strength and support!

How to solve linear equations

In order to solve a linear equation, you will need to:

Simplify both sides of the equation.

Isolate the variable on one side of the equation.

Solve for the variable.

Linear equation examples

Example 1: linear equation with one unknown.

Solve 3 x+6=18.

Both sides of the equation are simplified.

2 Isolate the variable on one side of the equation.

Here, the opposite of +6 is -6.

3 Solve for the variable.

3 x=3\times x, so divide by 3.

Example 2: linear equation with an unknown on both sides

Solve 5 x+6=2 x+9.

Here, 2x is smaller than 5x, so eliminate it by -2x.

3x=3\times x, so we divide by 3.

Example 3: linear equation with an unknown on both sides

Solve -3 x-11=-12 x+16.

Here -3x is smaller than -12x, so eliminate it by adding 3x to both sides.

\begin{aligned} -3 x-11&=-12 x+16 \\\\ +3 x \hspace{0.2cm} & \hspace{0.5cm}+3 x \\\\ -11&=-9 x+16\end{aligned}

To isolate the variable, subtract 16 from both sides of the equal sign.

\begin{aligned} -11&=-9 x+16 \\\\ -16 & \hspace{0.5cm}-16 \\\\ -27&=-9 x\end{aligned}

To solve for the variable, divide each side by -9.

\begin{aligned} -27&=-9 x \\\\ \div(-9) & \hspace{0.2cm} \div(-9) \\\\ x&=3\end{aligned}

Example 4: solve a linear equation with parentheses

Solve 4(x-2)=12.

Start by distributing the 4 to the factors within the parentheses.

Here the opposite of -8 is +8, so you will add 8 to both sides.

\begin{aligned}4 x-8&=12 \\\\ +8 & \hspace{0.3cm} +8 \\\\ 4 x&=20\end{aligned}

Because 4x=4\times x, divide by 4 to solve for the variable.

Example 5: linear equation with parentheses

Solve -2(4 x+10)=12.

Start by distributing the -2 to the factors within the parentheses.

\begin{aligned}& -2(4 x+10)=12 \\\\ & -8 x-20=12\end{aligned}

The opposite of -20 is +20, so you will add 20 to both sides.

\begin{aligned}-8 x-20&=12 \\\\ +20 & \hspace{0.2cm} +20 \\\\ -8 x&=32\end{aligned}

To solve for the variable, we will divide both sides of the equation by -8.

\begin{aligned}-8 x& =32 \\\\ \div(-8) & \hspace{0.2cm} \div(-8) \\\\ x&=4\end{aligned}

Example 6: linear equation with fractions

Solve \cfrac{5 x-2}{4}=\cfrac{2 x+2}{2} \, .

Cross multiply, by multiplying each fraction by the denominator on the other side of the equal sign to get rid of the fractions.

\begin{aligned}& (5 x-2) \times(2)=(2 x+2) \times(4) \\\\ & 10 x-4=8 x+8\end{aligned}

Start by subtracting 8x from both sides of the equation.

\begin{aligned} 10 x-4&=8 x+8 \\\\ - \, 8 x \hspace{0.1cm} & \hspace{0.2cm}-8 x \\\\ 2 x-4&=8\end{aligned}

Then add 4 to both sides of the equation.

\begin{aligned}2 x-4&=8 \\\\ +4 \hspace{0.1cm} & \hspace{0.2cm} +4 \\\\ 2 x&=12\end{aligned}

Solve for x by dividing both sides of the equation by 2.

\begin{aligned} 2 x & =12 \\\\ \div 2 & \hspace{0.2cm} \div 2 \\\\ x&=6\end{aligned}

Teaching tips for linear equations

• Ensure students have a solid understanding of contents, variables, coefficients and terms before beginning linear equations.
• Use visual aids like graphs to help students understand the relationship between linear equations and lines. Show them how to plot linear equations on a coordinate plane and interpret the slope and y -intercept.
• Mistakes are opportunities to reinforce learning. Encourage students to make mistakes, as errors in problem solving can lead to class discussions about the mistakes and how to solve them.

Easy mistakes to make

• Making errors when using the distributive property When distributing a number to multiple terms, make sure to multiply by all of the numbers within parenthesis. For example, 5(x+6), you would multiply the 5 by both x and 6.

Related math equations lessons

• Math equations
• Rearranging equations
• How to find the equation of a line
• Solving equations
• Writing linear equations
• Substitution
• One step equations
• Identity math
• Solve equations with fractions

Practice linear equations problems

First, subtract 8 from both sides.

Solve for the variable by dividing both sides by 2.

First, add 3 to both sides.

Then divide both sides by 5.

First, subtract x from both sides.

Then add 7 to both sides to solve for the variable.

7 x-3=4 x+15

Start by subtracting 4x from both sides.

Then add 3 to both sides.

To solve for the variable, divide both sides by 3.

Start by dividing both sides by 5.

Add 3 to both sides of the equal sign to solve for x.

3(x-3)=6(x+5)

Start by distributing within the parenthesis.

3 x-9=6 x+30

Then, subtract 3x from both sides.

Subtract 30 from both sides.

\cfrac{3x+1}{2}+\cfrac{2x-2}{4}

(3 x+1) \times 2=(2 x-2) \times 4

From both sides of the equation, subtract 4x .

and subtract -4

Finally divide by 8.

Linear equation FAQs

Yes, a linear equation in two variables can have an infinite amount of solutions. The solutions are all of the points that lie on the line represented by the given equation.

Simultaneous linear equations are a system of linear equations, two or more, with two or more variables. Solving such systems involves finding values for the variables that satisfy all the equations simultaneously. See also: Systems of linear equations

Quadratic equations are an extension of linear equations, representing a more complex relationship between variables. Linear equations create straight lines, while quadratic equations create a parabola, or a U -shaped curve.

The general form of linear equations is another way to say “standard form”. The standard form, or general form, is ax+by=c, where a, b and c are constants.

The next lessons are

• Inequalities
• Types of graphs
• Coordinate plane

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5.1: Objectives

• Last updated
• Save as PDF
• Page ID 49366

• Denny Burzynski & Wade Ellis, Jr.
• College of Southern Nevada via OpenStax CNX

After completing this chapter, you should

Solving Equations

• be able to identify various types of equations
• understand the meaning of solutions and equivalent equations
• be able to solve equations of the form \(x+a=b\) and \(x−a=b\).
• be familiar with and able to solve linear equation

Solving Equations of the Form \(ax=b\) and \(\dfrac{x}{a}=b\)

• understand the equality property of addition and multiplication
• be able to solve equations of the form \(ax=b\) and \(\dfrac{x}{a}=b\)

Further Techniques in Equation Solving

• be comfortable with combining techniques in equation solving
• be able to recognize identities and contradictions

Applications I - Translating from Verbal to Mathematical Expressions

• be able to translate from verbal to mathematical expressions

Applications II - Solving Problems

• be able to solve various applied problems

Linear Inequalities in One Variable

• understand the meaning of inequalities
• be able to recognize linear inequalities
• know, and be able to work with, the algebra of linear inequalities and with compound inequalities

Linear Inequalities in Two Variables

• be able to identify the solution of a linear equation in two variables
• know that solutions to linear equations in two variables can be written as ordered pairs

Question and Answer forum for K12 Students

MCQ Questions for Class 8 Maths Chapter 2 Linear Equations in One Variable with Answers

We have compiled the NCERT MCQ Questions for Class 8 Maths Chapter 2 Linear Equations in One Variable with Answers Pdf free download covering the entire syllabus. Practice MCQ Questions for Class 8 Maths with Answers on a daily basis and score well in exams. Refer to the Linear Equations in One Variable Class 8 MCQs Questions with Answers here along with a detailed explanation.

Linear Equations in One Variable Class 8 MCQs Questions with Answers

Choose the correct option.

Question 1. In the equation \(\frac {x}{2}\) + \(\frac {5}{4}\) = \(\frac {-3}{7}\), transposing 5/4 to R.H.S. we get (a) \(\frac {x}{2}\) = –\(\frac {3}{7}\) + \(\frac {5}{4}\) (b) \(\frac {x}{2}\) = \(\frac {9}{7}\) – \(\frac {5}{4}\) (c) \(\frac {x}{2}\) = \(\frac {-3}{7}\) – \(\frac {5}{4}\) (d) None of these

Answer: (a) \(\frac {x}{2}\) = –\(\frac {3}{7}\) + \(\frac {5}{4}\)

Question 2. If \(\frac {x}{3}\) + 1 = \(\frac {7}{15}\) + 1 then which of the following is correct? (a) \(\frac {x}{3}\) = –\(\frac {7}{15}\) + 1 (b) \(\frac {x}{3}\) = \(\frac {-7}{15}\) + 1 (c) \(\frac {x}{3}\) = \(\frac {-7}{15}\) – 1 (d) \(\frac {x}{3}\) = \(\frac {7}{15}\) – 1

Answer: (d) \(\frac {x}{3}\) = \(\frac {7}{15}\) – 1

Question 3. If 7x + 3 = 17 then which of the following is the root of the equation. (a) 2 (b) – 2 (c) \(\frac {1}{2}\) (d) –\(\frac {1}{2}\)

Answer: (a) 2

Question 4. 5 years ago, I was ‘x’ years old. After 10 years, my age will be (a) (x + 15) years (b) (x – 15) years (c) (x + 10) years (d) (x – 10) years

Answer: (a) (x + 15) years

Question 5. If the sum of two consecutive numbers is 15 and the greater of them is x then the smaller number is (a) 16 (b) 14 (c) 8 (d) 7

Answer: (d) 7

Question 6. In a two digit number, the unit digit is x and the ten’s digit is y. Then, the number is (a) 10y + x (b) 10x + y (c) 10y – x (d) 10x – y

Answer: (a) 10y + x

Question 7. The value of x in \(\frac {3}{4}\)x = 7 – x is (a) 3 (b) 4 (c) \(\frac {7}{3}\) (d) 7

Answer: (b) 4

Question 8. \(\frac {3}{4}\) part of a number is 5 more than its \(\frac {2}{3}\) part. This statement in the form of and equation is (a) \(\frac {2}{3}\)x – \(\frac {3}{4}\)x = 5 (b) \(\frac {2}{3}\)x – 5 = \(\frac {3}{4}\)x (c) \(\frac {3}{4}\)x = \(\frac {2}{3}\)x + 5 (d) \(\frac {3}{4}\)x – 5 = \(\frac {-2}{3}\)x

Answer: (c) \(\frac {3}{4}\)x = \(\frac {2}{3}\)x + 5

Question 9. The root of the equation 3a + 4 = 5a – 4 is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (d) 4

Question 10. In this equation \(\frac {x}{4}\) + \(\frac {5}{2}\) = \(\frac {-3}{4}\) transposing \(\frac {5}{2}\) to RHS we get (a) \(\frac {x}{4}\) = \(\frac {-3}{4}\) + \(\frac {5}{2}\) (b) \(\frac {x}{4}\) = \(\frac {-5}{2}\) + \(\frac {3}{4}\) (c) \(\frac {x}{4}\) = \(\frac {-3}{4}\) + (\(\frac {-5}{2}\)) (d) None of these

Answer: (c) \(\frac {x}{4}\) = \(\frac {-3}{4}\) + (\(\frac {-5}{2}\))

Question 11. In this equation 3x = 4 – x transposing (-x) to LHS we get (a) 3x – x = 4 (b) 3x + x = 4 (c) -3x + x = 4 (d) -3x – x = 4

Answer: (b) 3x + x = 4

Fill in the blanks

Question 1. The value of y in \(\frac {5}{4}\)y = 8 – y is ……………….

Answer: \(\frac {32}{9}\)

Question 2. The root of the equation 5a + 6 = 7a – 8 is …………………

Question 3. In a two digit number the unit digit is x and the ten’s digit is y then the number obtained on reversing the digit is ………………..

Answer: 10x + y

Question 4. If the sum of two consecutive numbers is 25 and the smaller of them is y then the greater number is …………….

Question 5. Value of x in 2x = -x + 5 is …………………..

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

Hope the information shed above regarding NCERT MCQ Questions for Class 8 Maths Chapter 2 Linear Equations in One Variable with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 8 Maths Linear Equations in One Variable MCQs Multiple Choice Questions with Answers, feel free to reach us so that we can revert back to us at the earliest possible.

Linear Equation in One Variable Questions

Students will learn the concepts more effectively with the help of the linear equation in one variable questions and answers. Almost every class includes a discussion of linear equations.. The questions will be prepared in accordance with the NCERT norms. Both in daily life and in mathematics, linear equations are used. So, it’s important to understand the fundamentals of linear equations in one variable. The problems presented here will cover both the fundamentals and more difficult problems for students of all skill levels. Click here to read more about linear equations in one variable .

Go through the below problems to understand the linear equation in one variable and also solve the practice problems.

Linear Equation in One Variable Questions with Solutions

1. Convert the given statement into equation: An integer increased by 5 equals 20.

Given statement: An integer increased by 5 equals 20.

Assume that the unknown number is “x”.

Hence, x increased by 5 means x + 5.

Therefore, the equation x + 5 = 20

So, an integer increased by 5 equals 20 means x + 5 = 20, which is the required equation for the given statement.

2. Convert the given equation into a statement: 5x = 25.

The equation given is 5x = 25.

The statement equivalent to the equation 5x = 25 is “Five times the number x equals 25”.

3. Find the value of x for the equation 2x – 7 = 21. Also, verify the answer.

Given equation:2x – 7 = 21.

To solve the given equations, keep the variable x on one side and the constants on the other side.

Hence, the given equation becomes:

2x = 21 + 7

Therefore, the value of x is 14.

Verification:

Substitute x = 14 in the equation 2x – 7 = 21

Hence, 2(14) – 7 = 21

28 – 7 = 21

Hence, LHS = RHS.

4. Check whether x = 40 is the solution of the equation 5x/2 = 100.

To check whether x = 40 is the solution or root of the equation 5x/2 = 100, put the value x = 40 in the equation 5x/2 = 100

⇒ [5(40)]/2 = 100

⇒ 200/2 = 100

⇒ 100 = 100

Since, LHS = RHS, x = 40 is the solution of the equation 5x/2 = 100.

5. Find the number, if 15 is added to three times of the number results in 45. Also, justify your answer.

Given statement: 15 is added to three times of the number resulting in 45.

Let the unknown number be x.

Then according to the given statement, the equation formed is:

15 + 3x = 45

Now, keep the variable “x” on one side of the equation and constant on the other side of the equation.

⇒ 3x = 45 -15

Therefore, the required number is x = 10.

Substitute the value x =10 in the equation 15 + 3x = 45.

⇒ 15 + 3(10) = 45

⇒ 15 + 30 = 45

Therefore, LHS = RHS.

Hence, verified.

Also, read: Linear Equations in Two Variables.

6. Check whether the given statements are true or false.

• If x = 5, then 5x – 5 = 20
• If x = 7, then 4x – 4 = 20.

(a) Given equation: 5x – 5 = 20

= 5(5) – 5

= 25 – 5

Hence, if x = 5, then 5x – 5 = 20.

Therefore, the given equation is true.

(b) Given equation: 4x – 4 = 20

= 4(7) – 4

= 28 – 4

Hence, if x = 7, then 4x – 4 = 24.

Therefore, the given equation is false..

7. Determine the number, if the sum of two odd consecutive numbers is 56.

As we know that the difference between two odd consecutive numbers is 2.

Hence, let the small odd numbers be x and the successive odd numbers is x+2.

According to the given condition, we can write

⇒ x + x + 2 = 56

⇒ 2x + 2 = 56

⇒ 2x = 56 – 2

⇒ x = 27, which is the required odd number.

Hence, the other odd number is:

⇒ x + 2 = 27 + 2 = 29.

Therefore, the two odd numbers are 27 and 29.

8. Solve the given linear equation: 17 + 6p = 9

Given linear equation: 17 + 6p = 9

Now, keep the variable “p” on the left hand side and bring the constants on the right hand side.

Hence, we get

⇒ 6p = 9 – 17

Therefore, the value of p is -4/3.

Also, read: How to Solve Linear Equations?

9. Find the two numbers, if the numbers are in the ratio 5 : 3 and they differ by 18.

Let the unknown number be “p”.

Also, given that, the two numbers are in the ratio 5 : 3.

According to given conditions, we can write

⇒ 5p – 3p = 18

Therefore, the two numbers are:

5p = 5(9) = 45

3p = 3(9) = 27

Hence, the required two numbers are 45 and 27.

10. The ages of Rahul and Ramya are in the ratio 5: 7. After 4 years, the sum of their ages will be 56 years. Find their present ages.

Assume that the ages of Rahul and Ramya are 5p and 7p.

After 4 years, the ages of Rahul and Ramya will be 5p + 4 and 7p + 4.

According to the given condition, we get the following equation:

(5p + 4) + (7p + 4) = 56

⇒ 12p + 8 = 56

⇒ 12p = 56 – 8

⇒ p = 48/12

Therefore, Rahul’s present age = 5(4) = 20

Ramya’s present age = 7(4) = 28.

Hence, the present age of Rahul and Ramya are 20 and 28, respectively.

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Practice Questions

• Write the equation for the given statement: 3 decreased from 4x gives 17.
• Convert the given equation into a statement: 9x – 9 = 81.
• Determine the value of x for the given equation: 25x + 100 = 350.

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Mathematics > Analysis of PDEs

Title: on the derivation of the linear boltzmann equation from the nonideal rayleigh gas.

Abstract: This paper's objective is to improve the existing proof of the derivation of the Rayleigh--Boltzmann equation from the nonideal Rayleigh gas [6], yielding a far faster convergence rate. This equation is a linear version of the Boltzmann equation, describing the behavior of a small fraction of tagged particles having been perturbed from thermodynamic equilibrium. This linear equation, derived from the microscopic Newton laws as suggested by the Hilbert's sixth problem, is much better understood than the quadratic Boltzmann equation, and even enable results on long time scales for the kinetic description of gas dynamics.The present paper improves the physically poor convergence rate that had been previously proved, into a much more satisfactory rate which is more than exponentially better.

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