Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Mathematics LibreTexts

8.E: Solving Linear Equations (Exercises)

  • Last updated
  • Save as PDF
  • Page ID 5024

8.1 - Solve Equations using the Subtraction and Addition Properties of Equality

In the following exercises, determine whether the given number is a solution to the equation.

  • x + 16 = 31, x = 15
  • w − 8 = 5, w = 3
  • −9n = 45, n = 54
  • 4a = 72, a = 18

In the following exercises, solve the equation using the Subtraction Property of Equality.

  • y + 2 = −6
  • a + \(\dfrac{1}{3} = \dfrac{5}{3}\)
  • n + 3.6 = 5.1

In the following exercises, solve the equation using the Addition Property of Equality.

  • u − 7 = 10
  • x − 9 = −4
  • c − \(\dfrac{3}{11} = \dfrac{9}{11}\)
  • p − 4.8 = 14

In the following exercises, solve the equation.

  • n − 12 = 32
  • y + 16 = −9
  • f + \(\dfrac{2}{3}\) = 4
  • d − 3.9 = 8.2
  • y + 8 − 15 = −3
  • 7x + 10 − 6x + 3 = 5
  • 6(n − 1) − 5n = −14
  • 8(3p + 5) − 23(p − 1) = 35

In the following exercises, translate each English sentence into an algebraic equation and then solve it.

  • The sum of −6 and m is 25.
  • Four less than n is 13.

In the following exercises, translate into an algebraic equation and solve.

  • Rochelle’s daughter is 11 years old. Her son is 3 years younger. How old is her son?
  • Tan weighs 146 pounds. Minh weighs 15 pounds more than Tan. How much does Minh weigh?
  • Peter paid $9.75 to go to the movies, which was $46.25 less than he paid to go to a concert. How much did he pay for the concert?
  • Elissa earned $152.84 this week, which was $21.65 more than she earned last week. How much did she earn last week?

8.2 - Solve Equations using the Division and Multiplication Properties of Equality

In the following exercises, solve each equation using the Division Property of Equality.

  • 13a = −65
  • 0.25p = 5.25
  • −y = 4

In the following exercises, solve each equation using the Multiplication Property of Equality.

  • \(\dfrac{n}{6}\) = 18
  • y −10 = 30
  • 36 = \(\dfrac{3}{4}\)x
  • \(\dfrac{5}{8} u = \dfrac{15}{16}\)

In the following exercises, solve each equation.

  • −18m = −72
  • \(\dfrac{c}{9}\) = 36
  • 0.45x = 6.75
  • \(\dfrac{11}{12} = \dfrac{2}{3} y\)
  • 5r − 3r + 9r = 35 − 2
  • 24x + 8x − 11x = −7−14

8.3 - Solve Equations with Variables and Constants on Both Sides

In the following exercises, solve the equations with constants on both sides.

  • 8p + 7 = 47
  • 10w − 5 = 65
  • 3x + 19 = −47
  • 32 = −4 − 9n

In the following exercises, solve the equations with variables on both sides.

  • 7y = 6y − 13
  • 5a + 21 = 2a
  • k = −6k − 35
  • 4x − \(\dfrac{3}{8}\) = 3x

In the following exercises, solve the equations with constants and variables on both sides.

  • 12x − 9 = 3x + 45
  • 5n − 20 = −7n − 80
  • 4u + 16 = −19 − u
  • \(\dfrac{5}{8} c\) − 4 = \(\dfrac{3}{8} c\) + 4

In the following exercises, solve each linear equation using the general strategy.

  • 6(x + 6) = 24
  • 9(2p − 5) = 72
  • −(s + 4) = 18
  • 8 + 3(n − 9) = 17
  • 23 − 3(y − 7) = 8
  • \(\dfrac{1}{3}\)(6m + 21) = m − 7
  • 8(r − 2) = 6(r + 10)
  • 5 + 7(2 − 5x) = 2(9x + 1) − (13x − 57)
  • 4(3.5y + 0.25) = 365
  • 0.25(q − 8) = 0.1(q + 7)

8.4 - Solve Equations with Fraction or Decimal Coefficients

In the following exercises, solve each equation by clearing the fractions.

  • \(\dfrac{2}{5} n − \dfrac{1}{10} = \dfrac{7}{10}\)
  • \(\dfrac{1}{3} x + \dfrac{1}{5} x = 8\)
  • \(\dfrac{3}{4} a − \dfrac{1}{3} = \dfrac{1}{2} a + \dfrac{5}{6}\)
  • \(\dfrac{1}{2}\)(k + 3) = \(\dfrac{1}{3}\)(k + 16)

In the following exercises, solve each equation by clearing the decimals.

  • 0.8x − 0.3 = 0.7x + 0.2
  • 0.36u + 2.55 = 0.41u + 6.8
  • 0.6p − 1.9 = 0.78p + 1.7
  • 0.10d + 0.05(d − 4) = 2.05

PRACTICE TEST

  • \(\dfrac{23}{5}\)
  • n − 18 = 31
  • 4y − 8 = 16
  • −8x − 15 + 9x − 1 = −21
  • −15a = 120
  • \(\dfrac{2}{3}\)x = 6
  • x + 3.8 = 8.2
  • 10y = −5y + 60
  • 8n + 2 = 6n + 12
  • 9m − 2 − 4m + m = 42 − 8
  • −5(2x + 1) = 45
  • −(d + 9) = 23
  • 2(6x + 5) − 8 = −22
  • 8(3a + 5) − 7(4a − 3) = 20 − 3a
  • \(\dfrac{1}{4} p + \dfrac{1}{3} = \dfrac{1}{2}\)
  • 0.1d + 0.25(d + 8) = 4.1
  • Translate and solve: The difference of twice x and 4 is 16.
  • Samuel paid $25.82 for gas this week, which was $3.47 less than he paid last week. How much did he pay last week?

Contributors and Attributions

Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at http://cnx.org/contents/[email protected] ."

Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Elimination Method of solving linear equations

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Graphical Method of Solving Linear Equations

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Cross Multiplication Method of solving linear equations

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

  • Linear Equations
  • Application of Linear Equations
  • Two-Variable Linear Equations
  • Linear Equations and Half Planes
  • One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Solving Linear Equations Example

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

go to slide go to slide go to slide

solving linear equations practice questions

Book a Free Trial Class

Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

  • Substitution method
  • Elimination method
  • Cross multiplication method
  • Graphical method

Practice Test on Linear Equations

In math equations, students can practice test on linear equations in one variable showing step-by-step solution using addition, subtraction, multiplication and division. While practicing this sheet keep in mind the basic concept on linear equations.

Notes to practice test on linear equations:  An equation which involves only one variable whose highest power is 1 is known as linear equation in that variable.  We can add or subtract the same number to both sides of the equation.  We can divide or multiply both sides of the equation by the same non-zero integer.  Transposition is a process in which any term in an equation can be shifted to the other side of equal to sign by simply changing the sign from (+ to -) ,  (- to +) ,  (× to ÷) , and  (÷ to ×) . The process of multiplying the numerator on L.H.S. with denominator on R.H.S. and multiplying denominator on L.H.S. with numerator on R.H.S. is called cross multiplication. 

Solve the following equations:  

(a) 5x - 11 = 3x + 9 (b) 3y + 4 = 7 - 2y (c) 9 - 2(x - 5) = x + 10 (d) 5(y - 1) = 3(2y - 5) - (1 - 3y) (e) 2(x - 1) - 6x = 10 - 2(x - 4) (f) x/3 - (x - 2)/2 = 7/3 (g) (x - 3)/4 + (x - 1)/5 - (x - 2)/3 = 1 (h) (3y - 2)/3 + (2y + 3)/3 = (y + 7)/6 (i) (8x - 5)/(7x + 1) = -4/5 (j) (5 - 7x)/(2 + 4x) = -8/7 (k) (x - 2)/(x - 3) = (x - 1)/(x + 1) (l) (2x - 5)/(3x - 1) = (2x - 1)/(3x + 2) (m) (3 - 7x)/(15 + 2x) = 0 (n) (0.4y - 3)/(1.5y + 9) = -7/5 (o) 2/(3x - 1) + 3/(3x + 1) = 5/3x (p) 2/(x - 3) + 1/(x - 1) = 5/(x - 1) - 2/(x - 2) (q) 15(x - y) - 3(x - 9) + 5(x + 6) = 0 (r) y/2 - 1/2 = y/3 + 1/4 (s) (0.5y - 9)/0.25 = 4y - 3 (t) [17(2 - y) - 5(y + 12)]/(1 - 7y) = 8

Answers for the practice test on linear equations in one variable are given below to check the exact answers of the equations.

(b) 3/5 (c) 3 (d) 11/4 (e) -10 (f) -8 (g) 11 (h) 5/9 (i) 21/68 (j) 3 (k) 5/3 (l) -11/6 (m) 3/7 (n) - 96/25 (o) 5/3 (p) 7/3 (q) -1/6 (r) 9/2 (s) -16.5 (t) 1

●   Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Word Problems on Linear Equations

●   Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

7th Grade Math Problems 8th Grade Math Practice   From Practice Test on Linear Equations to HOME PAGE

New! Comments

Didn't find what you were looking for? Or want to know more information about Math Only Math . Use this Google Search to find what you need.

  • Preschool Activities
  • Kindergarten Math
  • 1st Grade Math
  • 2nd Grade Math
  • 3rd Grade Math
  • 4th Grade Math
  • 5th Grade Math
  • 6th Grade Math
  • 7th Grade Math
  • 8th Grade Math
  • 9th Grade Math
  • 10th Grade Math
  • 11 & 12 Grade Math
  • Concepts of Sets
  • Probability
  • Boolean Algebra
  • Math Coloring Pages
  • Multiplication Table
  • Cool Maths Games
  • Math Flash Cards
  • Online Math Quiz
  • Math Puzzles
  • Binary System
  • Math Dictionary
  • Conversion Chart
  • Homework Sheets
  • Math Problem Ans
  • Free Math Answers
  • Printable Math Sheet
  • Funny Math Answers
  • Employment Test
  • Math Patterns
  • Link Partners
  • Privacy Policy

XML RSS

Recent Articles

RSS

Successor and Predecessor | Successor of a Whole Number | Predecessor

Feb 14, 24 11:52 PM

Successor and Predecessor of a Whole Number

Arranging Numbers | Ascending Order | Descending Order |Compare Digits

Feb 14, 24 11:27 PM

Arranging Numbers

Comparison of Numbers | Compare Numbers Rules | Examples of Comparison

Feb 14, 24 10:02 PM

Rules for Comparison of Numbers

Worksheet on Expanded form of a Number | Expanded Form of a Number

Feb 12, 24 11:38 PM

Worksheet on Place Value and Face Value | Place Value of the Digit

Feb 12, 24 11:16 PM

© and ™ math-only-math.com. All Rights Reserved. 2010 - 2024.

Corbettmaths

Solving Equations Practice Questions

Click here for questions, click here for answers.

equation, solve

GCSE Revision Cards

solving linear equations practice questions

5-a-day Workbooks

solving linear equations practice questions

Primary Study Cards

solving linear equations practice questions

Privacy Policy

Terms and Conditions

Corbettmaths © 2012 – 2024

mathportal.org

  • Calculators

Math Calculators, Lessons and Formulas

It is time to solve your math problem

  • HW Help (paid service)
  • Math Lessons
  • Math Formulas
  • Linear Equations
  • Solving linear equations and inequalities

Test on Solving elementary linear equations

Linear equations.

  • Elementary equations
  • Equations involving brackets
  • Equations with fractions and decimals

Linear inequalities

  • Elementary inequalities
  • Inequalities involving brackets
  • Inequalities with fractions and decimals

Graphing functions

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

Welcome to MathPortal. This website's owner is mathematician Miloš Petrović. I designed this website and wrote all the calculators, lessons, and formulas .

If you want to contact me, probably have some questions, write me using the contact form or email me on [email protected]

Email (optional)

[FREE] Fun Math Games & Activities Packs

Always on the lookout for fun math games and activities in the classroom? Try our ready-to-go printable packs for students to complete independently or with a partner!

In order to access this I need to be confident with:

Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – 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] Math Equations Check for Understanding Quiz (Grade 6 to 8)

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

Use this quiz to check your grade 6 to 8 students’ understanding of math equations. 10+ questions with answers covering a range of 6th, 7th and 8th grade math equations topics to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

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

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

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

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

Still stuck?

At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.

Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence.

One on one math tuition

Find out how we can help your students achieve success with our math tutoring programs .

[FREE] Common Core Practice Tests (Grades 3 to 6)

Prepare for math tests in your state with these Grade 3 to Grade 6 practice assessments for Common Core and state equivalents.

40 multiple choice questions and detailed answers to support test prep, created by US math experts covering a range of topics!

Privacy Overview

Linear Equations in Two Variables Questions

Linear equations in two variables questions presented here cover a variety of questions asked regarding linear equations in two variables with solutions and proper explanations. By practising these questions students will develop problem-solving skills.

Linear equations in two variables are linear polynomials with two unknowns. They are of the general form ax + by + c = 0, where x and y are the two variables, a and b are non-zero real numbers and c is a constant. The graphical representation of a linear equation in two variables is a straight line.

Linear Equations in Two Variables Questions with Solutions

Below are some practice questions on linear equations in two variables with detailed solutions.

Question 1: Solve for x and y:

\(\begin{array}{l}\frac{1}{2x}-\frac{1}{y}=-1,\:\:\frac{1}{x}+\frac{1}{2y}=8\:\:\:(x\neq0,\;y\neq0)\end{array} \)

Put 1/x = u and 1/y = v. The given equations become

u/2 – v = –1 ⇒ u – 2v = – 2 ….(i)

u + v/2 = 8 ⇒ 2u + v = 16 ….(ii)

Multiplying equation (ii) by 2 on both sides and adding (i) and (ii), we get

(u + 4u) + ( –2v + 2v) = –2 + 32

⇒ u = 6 ⇒ x = ⅙ and y = ¼

Question 2: Solve the system of linear equations 2x + 3y = 17 and 3x – 2y = 6 by the cross multiplication method.

By cross multiplication

\(\begin{array}{l}\frac{x}{\left\{ 3\times (-6)-(-2)\times(-17)\right\}}=\frac{y}{\left\{ -17\times 3-(-6)\times 2 \right\}}=\frac{1}{\left\{2\times (-2)-3 \times3\right\}}\end{array} \)

⇒ x/( –52) = y/( –39) = 1/( – 13)

⇒ x = 52/13 = 4 and y = 39/13 = 3

Hence x = 4 and y = 3 is the solution of given equations.

Question 3: Solve the following system of equations by substitution method:

2x + 3y = 0 and 3x + 4y = 5

Given equations,

2x + 3y = 0 ….(i)

3x + 4y = 5 …..(ii)

From (i) we get, y = – 2x/3, substituting value of y in (ii), we get

3x + 4(–2x/3) = 5

⇒ 9x – 8x = 15

Then y = (–2 × 15)/3 = – 10

Therefore, x = 15 and y = – 10 is solution of given system of equations.

Video Lesson on Consistent and Inconsistent Equations

solving linear equations practice questions

Question 4: Find the value of k for which the given system of equations has infinitely many solutions: x + (k + 1)y = 5 and (k + 1)x + 9y + (1 – 8k) = 0.

The given equations will have infinitely many solutions if a 1 /a 2 = b 1 /b 2 = c 1 /c 2

Hence, 1/(k + 1) = (k + 1)/9 = – 5/(1 – 8k)

Solving the equations we get k = 2.

Question 5: If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel. Find the value of k.

If the lines are parallel, then they are inconsistent system of equations and a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2

Now, it should be 3/2 = 2k/5

Then we have 2k/5 = 30/20 = 3/2, which satisfies the condition of inconsistency.

Question 6: Find the value of k for which the system of equations has a non-zero solution

5x + 3y = 0 and 10x + ky = 0

The given equations are homogenous equations, they will have a non-zero solution if a 1 /a 2 = b 1 /b 2

Then, 5/10 = 3/k

⇒ 1/2 = 3/k

Question7: The monthly incomes of A and B are in the ratio 8:7 and their expenditures are in the ratio 19:16. If each saves ₹ 5000 per month, find the monthly income of each.

Let the monthly incomes of A and B be 8x and 7x rupees respectively, and let their monthly expenditure be 19y and 16y rupees respectively.

Monthly savings of A = 8x – 19y = 5000 ….(i)

Monthly savings of B = 7x – 16y = 5000 ….(ii)

Multiplying (i) 16 and (ii) by 19 and subtracting (ii) from (i) we get

(16 × 8x – 19 × 7x) = 5000 (16 – 19)

⇒ 5x = 15000 ⇒ x = 3000

Monthly income of A is (8 × 3000) = ₹24,000

Monthly income of B is (7 × 3000) = ₹21,000

Question 8: The sum of a two-digit number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the original number.

Let the original number be (10x + y)

According to the question,

(10x + y) + (10y + x) = 99

⇒ 11(x + y) = 99

⇒ x + y = 9 ….(i)

And x – y = 3 ….(ii)

Adding equations (i) and (ii), we get,

Hence the required number is 63.

Question 9: A man’s age is three times the sum of the ages of his two sons. After 5 years, his age will be twice the sum of his two son’s age. Find the age of the man.

Let the age of the man be x and the sum of the ages of his two sons be y.

x = 3y ⇒ x – 3y = 0 ….(i)

And (x + 5) = 2(y + 5 + 5)

⇒ x – 2y = 15 ….(ii)

Subtracting equation (i) from (ii) we get

Y = 15 and from (i) x = 45.

The present age of the man is 45 years.

Question 10: A man can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours, Find his speed of rowing in still water. Also, find the speed of the stream.

Let the speed of the man in still water be x km/hr and let the speed of the current be y kn/hr.

Speed in downstream = (x + y) km/hr

Speed in upstream = (x – y) km/hr

But speed in downstream = 20/2 km/hr = 10 km/hr

And speed in upstream = 4/2 km/hr = 2 km/hr

∴ x + y = 10 and x – y = 2

Solving both the equations we get;

x = 6 and y = 4.

Hence, the speed of the man in still water is 6 km/hr and the speed of the current is 4 km/hr.

Question 11: Find the four angles of a cyclic Quadrilateral ABCD in which ∠A = (2x – 1) o , ∠B = (y + 5) o , ∠C = (2y + 15) o , and ∠D = (4x – 7) o .

We know that sum of opposite angles of a cyclic quadrilateral is 180 o

∴ ∠A + ∠C = 180 o and ∠B + ∠C = 180 o

∠A + ∠C = 180 o ⇒ (2x – 1) + (2y + 15) = 180 o

⇒ x + y = 83 ….(i)

∠B + ∠C = 180 o ⇒ (y + 5) + (4x – 7) = 180 o

⇒ 4x + y = 182 …..(ii)

Subtracting (i) from (ii) we get

3x = 182 – 83 ⇒ x = 33

Substituting in (i), we get y = 50

∴ ∠A = 2 × 33 – 1 = 65 o

∠B = 50 + 5 = 55 o

∠C = 2 × 50 + 15 = 115 o

∠D = 4 × 33 – 7 = 125 o .

Question 12: 8 men and 12 boys can finish a piece of work in 5 days, while 6 men and 8 boys can finish it in 7 days. Find time taken by a man and a boy alone to finish the same work.

Let 1 man can finish the work in x days and let 1 boy can finish the work in y days.

1 man’s one day work = 1/x

1 boy’s one day work = 1/y

8 men’s 1 day’s work + 12 boy’s one day’s work = ⅕

⇒ 8/x + 12/y = ⅕

⇒ 8u + 12v = ⅕ ….(i) where u = 1/x and v = 1/y

Similarly, 6u + 8v = 1/7 ….(ii)

On solving (i) and (ii) we get, x = 70 and y = 140

∴ One man alone can finish the work in 70 days and one boy alone can finish the work in 140 days.

Related Articles:

Practice questions:.

1. Five years ago Anna was three times older than Mira and ten years later Anna will be two times older than Mira. What are the present ages of Anna and Mira?

2. The difference of two numbers is 4 and the difference of their reciprocals is 4/21. Find the numbers.

3. Find the value of k for which the system of equations 5x – 3y = 0, and 2x + ky = 0 has a non-zero solution.

4. Find the value of a and b for which each of the following systems of linear equations

(a – 1)x + 3y = 2 and 6x + (1 – 2b)y = 6 has infinite number of solutions.

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

solving linear equations practice questions

  • Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

close

Complete Test Preparation Inc.

How to Solve Linear Equations in one variable – Tutorial and Practice Questions

solving linear equations practice questions

  • Posted by Brian Stocker MA
  • Date November 27, 2014
  • Comments 3 comments

Quick Review and Tutorial – How to Solve Linear Equations in One Variable

Linear equations in one variable x is an equation with the following form:

where a and b are some real numbers. If a = 0 and b is different from 0, then the equation has no solution.

Let’s solve one simple example of a linear equation with one variable:

4x – 2 = 2x + 6

When we are given this type of equations, we are always moving variables to the one side of the equation, and real numbers to the other side of the equals sign. Always remember: if you are changing sides, you are changing signs. Let’s move all variables to the left, and real number to the right side:

4x  – 2  =  2x  + 6

4x – 2x = 6 + 2

When 2x goes to the left it becomes now -2x, and -2 goes to the right and becomes +2. After calculations, we find that x is 4, which is a solution of our linear equation.

Let’s solve a little more complex linear equation:

(2x – 6)/4 + 4 = x

2x – 6 + 16 = 4x

2x – 4x = -16 + 6

We multiply whole equation by 4, to lose the fractional line. Now we have a simple linear equation. If we change sides, we change the signs. At the end we do the final calculations.

Linear Equation in One Variable Practice Questions

1. Solve the linear equation:

-x – 7 = -3x – 9

a. -1 b. 0 c.  1 d . 2

2. Solve the linear equation:

3(x + 2) – 2(1 – x) = 4x + 5

3. Find the solution for the following linear equation:

 5x/2 = (3x + 24)/6

4. If a and b are real numbers, solve the following equation:

(a+2)x-b = -2+(a+b)x

5. Find the solution for the following linear equation:

1/(4x – 2) = 5/6

a. 0.2 b. 0.4 c. 0.6 d. 0.8

-x – 7 = -3x – 9 -x + 3x = -9 + 7 2x = -2 x = (-2):2 x = -1

3(x + 2) – 2(1 – x) = 4x + 5 3x + 6 – 2 + 2x = 4x + 5 5x + 4 = 4x + 5 5x – 4x = 5 – 4 x = 1

12

(a + 2)x – b = -2 + (a + b)x ax + 2x – b = -2 + ax + bx ax + 2x – ax – bx = -2 + b 2x – bx = -2 + b (2 – b)x = -(2 – b) x=-(2-b):(2-b) x = -1

1/(4x – 2) = 5/6 … We can cross multiply: 5(4x – 2) = 1 * 6 … Now, we distribute 5 to the parenthesis: 20x – 10 = 6 … We need x term alone on one side: 20x = 6 + 10 20x = 16 … Dividing both sides by 20: x = 16/20 … Simplifying by 2 and having 10 in the denominator provides us finding the decimal equivalent of x: x = 8/10 = 0.8

author avatar

Previous post

Congruent Triangles - Review, Short Tutorial and Practice Questions

How to solve linear equations with 2 variables - tutorial and practice, you may also like.

MathMC

Basic Math Video Tutorials

How to answer basic math multiple choice.

Inequality

How to Solve Linear Inequalities – Quick Review and Practice

' src=

thanks!! found this easy to answer

' src=

thanks alot for this preparation

' src=

Wow good work

Leave A Reply Cancel reply

Your email address will not be published. Required fields are marked *

HIGH SCHOOL

  • ACT Tutoring
  • SAT Tutoring
  • PSAT Tutoring
  • ASPIRE Tutoring
  • SHSAT Tutoring
  • STAAR Tutoring

GRADUATE SCHOOL

  • MCAT Tutoring
  • GRE Tutoring
  • LSAT Tutoring
  • GMAT Tutoring
  • AIMS Tutoring
  • HSPT Tutoring
  • ISAT Tutoring
  • SSAT Tutoring

Search 50+ Tests

Loading Page

math tutoring

  • Elementary Math
  • Pre-Calculus
  • Trigonometry

science tutoring

Foreign languages.

  • Mandarin Chinese

elementary tutoring

  • Computer Science

Search 350+ Subjects

  • Video Overview
  • Tutor Selection Process
  • Online Tutoring
  • Mobile Tutoring
  • Instant Tutoring
  • How We Operate
  • Our Guarantee
  • Impact of Tutoring
  • Reviews & Testimonials
  • Media Coverage
  • About Varsity Tutors

ACT Math Test : Solving Linear Equations

Study concepts, example questions & explanations for act math test, all act math test resources, example questions, example question #1 : solving linear equations.

solving linear equations practice questions

Example Question #2 : Solving Linear Equations

solving linear equations practice questions

You can then just flip each fraction (in doing so, you're doing the same thing to both sides, namely taking the reciprocal), to get:

solving linear equations practice questions

Example Question #5 : Solving Linear Equations

solving linear equations practice questions

Whenever the ACT asks you to solve for the value of a variable, you have two options: you can either use your answers as assets and backsolve, or you can solve for the variable directly. In this case, both are equally good options, and you should use whichever method is most comfortable. Generally, however, solving directly is usually the best option on the ACT Math section.

solving linear equations practice questions

Example Question #6 : Solving Linear Equations

solving linear equations practice questions

Example Question #7 : Solving Linear Equations

solving linear equations practice questions

Example Question #4 : Solving Linear Equations

solving linear equations practice questions

To solve this problem algebraically, distribute the multiplication across each set of parentheses, remembering to multiply each term within the parentheses by its coefficient. That gives you:

solving linear equations practice questions

Report an issue with this question

If you've found an issue with this question, please let us know. With the help of the community we can continue to improve our educational resources.

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC 101 S. Hanley Rd, Suite 300 St. Louis, MO 63105

Or fill out the form below:

Contact Information

Complaint details.

Learning Tools by Varsity Tutors

IMAGES

  1. Algebra 1 Worksheet: Solving Systems of Linear Equations Using Elimination

    solving linear equations practice questions

  2. Solving Linear Equations Worksheet Pdf

    solving linear equations practice questions

  3. 1 1 Solving Linear Equations

    solving linear equations practice questions

  4. Solving Linear Equations Worksheets by The STEM Master

    solving linear equations practice questions

  5. Equations Worksheets

    solving linear equations practice questions

  6. Solving Linear Equations In One Variable Worksheet

    solving linear equations practice questions

VIDEO

  1. Solving Linear Equations

  2. Solving Linear Equations

  3. Solving Linear Equations

  4. Solving Linear Equations

  5. Solving Linear Equations

  6. Solving Linear Equations

COMMENTS

  1. Algebra

    Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.

  2. Linear equations, functions, & graphs

    Algebra (all content) 20 units · 412 skills. Unit 1 Introduction to algebra. Unit 2 Solving basic equations & inequalities (one variable, linear) Unit 3 Linear equations, functions, & graphs. Unit 4 Sequences. Unit 5 System of equations. Unit 6 Two-variable inequalities. Unit 7 Functions. Unit 8 Absolute value equations, functions, & inequalities.

  3. Linear equations & graphs

    Algebra 1 Unit 4: Linear equations & graphs 1,600 possible mastery points Mastered Proficient Familiar Attempted Not started Quiz Unit test About this unit Let's explore different ways to find and visualize slopes and intercepts, and how these concepts can help us solve real-world problems. Two-variable linear equations intro Learn

  4. Solve Linear Equations Practice

    Solve Linear Equations Practice - MathBitsNotebook (A1) Directions: Solve the following equations, for the indicated variable. Be careful! The students' choices may, or may not, be correct. NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators.

  5. 8.E: Solving Linear Equations (Exercises)

    8.3 - Solve Equations with Variables and Constants on Both Sides. In the following exercises, solve the equations with constants on both sides. 8p + 7 = 47. 10w − 5 = 65. 3x + 19 = −47. 32 = −4 − 9n. In the following exercises, solve the equations with variables on both sides. 7y = 6y − 13. 5a + 21 = 2a.

  6. Equations with variables on both sides (practice)

    Lesson 1: Linear equations with variables on both sides Math > Algebra 1 > Solving equations & inequalities > Linear equations with variables on both sides Equations with variables on both sides Google Classroom Solve for f . − f + 2 + 4 f = 8 − 3 f f = Stuck? Review related articles/videos or use a hint. Report a problem Do 4 problems

  7. Solving Linear Equations

    Get Started Solving Linear Equations Solving linear equations means finding the value of the variable (s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation.

  8. Algebra

    Absolute Value Inequalities - In this final section of the Solving chapter we will solve inequalities that involve absolute value. As we will see the process for solving inequalities with a < < (i.e. a less than) is very different from solving an inequality with a > > (i.e. greater than). Here is a set of practice problems to accompany the ...

  9. IXL

    These can help: Add, subtract, multiply, and divide integers. Model and solve equations using algebra tiles. Write and solve equations that represent diagrams. Lesson: Solving equations.

  10. Solving Linear Equations

    Pre-Calculus Reviews & Testimonials About Varsity Tutors Call Now to Set Up Tutoring: » Solving Linear Equations , what is the value of The first step on this problem is to take the equation you're given, . To do that, divide both sides by .

  11. Solving Linear Equations: Practice Problems

    A linear equation is simply an algebraic expression that represents a line. These equations commonly contain one or two variables, usually x or y. These are called first-degree equations because ...

  12. Practice Test on Linear Equations

    In math equations, students can practice test on linear equations in one variable showing step-by-step solution using addition, subtraction, multiplication and division. While practicing this sheet keep in mind the basic concept on linear equations. Notes to practice test on linear equations:

  13. Linear Equations

    Work out the unknown variable by doing the opposite of what it says. Example with two steps. Show step. To solve. we need to: 1 Rearrange the equation so the unknown variable (x) is on its own on one side. Here the opposite of +6 is −6. 2 Work out what the unknown variable (x) is by doing the opposite of what it says.

  14. Linear Equations Questions with Solutions

    Solution: Given equation: x + 3 = -2. Now, keep the variables on one side and constants on the other side. Hence, the equation becomes, x = -2 -3 x = -5 Hence, the value of x is -5. 4. Verify that x = 4 is the root of the equation 3x/2 = 6.

  15. Linear equations in any form

    Course: Algebra 1 > Unit 5. Lesson 6: Summary: Forms of two-variable linear equations. Slope from equation. Slope from equation. Writing linear equations in all forms. Linear equations in any form. Forms of linear equations review. Forms of linear equations: FAQ. Math >.

  16. Solving Equations Practice Questions

    Click here for Answers. equation, solve. Practice Questions. Previous: Ray Method Practice Questions. Next: Equations involving Fractions Practice Questions. The Corbettmaths Practice Questions on Solving Equations.

  17. Math tests and quizzes on solving linear equations.

    9. 10. The online math tests and quizzes about solving linear equations, involving brackets, fractions and decimals.

  18. Linear Equations

    Point slope form The point slope form of a linear equation can be written as katex is not defined where katex is not defined is the slope of the line and katex is not defined 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.

  19. Solving Equations

    Example 1: solve equations involving like terms. Solve for x. x. 5q-4q=9 5q −4q = 9. Combine like terms. Combine the q q terms on the left side of the equation. To do this, subtract 4q 4q from both sides. (5 q-4 q)=9-4 q (5q −4q) = 9− 4q. The goal is to simplify the equation by combining like terms.

  20. Linear Equations in Two Variables

    By practising these questions students will develop problem-solving skills. Linear equations in two variables are linear polynomials with two unknowns. They are of the general form ax + by + c = 0, where x and y are the two variables, a and b are non-zero real numbers and c is a constant. The graphical representation of a linear equation in two ...

  21. Reasoning with linear equations (practice)

    Choose 1 answer: Multiply/divide both sides by the same non-zero constant A Multiply/divide both sides by the same non-zero constant Multiply/divide only one side by a non-zero constant B Multiply/divide only one side by a non-zero constant Rewrite one side (or both) by combining like terms C Rewrite one side (or both) by combining like terms

  22. How to Solve Linear Equations in one variable

    Linear equations in one variable x is an equation with the following form: ax = b. where a and b are some real numbers. If a = 0 and b is different from 0, then the equation has no solution.

  23. Solving Equations Worksheets, Questions and Revision

    Solving Linear Equations. Linear equations are a type of equation that appear all over the place in maths. They can look quite simple, like x+2=5 or they can look a little more complicated. There are 5 key types of linear equation you will need to solve. Exam questions can contain multiple types to make it even harder.

  24. Solving Linear Equations

    Test Prep Academic Tutoring Call Now to Set Up Tutoring: » Solving Linear Equations . You can then reduce the fraction . That leaves you with , but rather the value of . That simplifies to Calculus Tutors in San Diego