Substitution Method

One of the methods to solve a system of linear equations in two variables algebraically is the "substitution method". In this method, we find the value of any one of the variables by isolating it on one side and taking every other term on the other side of the equation. Then we substitute that value in the second equation.

The substitution method is preferable when one of the variables in one of the equations has a coefficient of 1. It involves simple steps to find the values of variables of a system of linear equations by substitution method. Let's learn about it in detail in this article.

What is Substitution Method?

The substitution method is a simple way to solve a system of  linear equations algebraically and find the solutions of the variables. As the name suggests, it involves finding the value of the x-variable in terms of the y-variable from the first equation and then substituting or replacing the value of the x-variable in the second equation. In this way, we can solve and find the value of the y-variable. And at last, we can put the value of y in any of the given equations to find x This process can be interchanged as well where we first solve for x and then solve for y.

In simple words, the substitution method involves substituting the value of any one of the variables from one equation into the other equation. Let us take an example of solving two equations x-2y=8 and x+y=5 using the substitution method.

solving system of linear equations by substitution method gives the result as x equals 6 and y equals minus 1

☛ Note:  The other three algebraic methods of solving linear equations . To learn each of these methods, click on the respective links given below.

  • Elimination method
  • Cross multiplication method
  • Graphical method

Solving Systems of Equations by Substitution Method

The steps to apply or use the substitution method to solve a system of equations are given below:

  • Step 1:  Simplify the given equation by expanding the parenthesis if needed.
  • Step 2: Solve any one of the equations for any one of the variables. You can use any variable based on the ease of calculation.
  • Step 3: Substitute the obtained value of x or y in the other equation.
  • Step 4: Now, simplify the new equation obtained using arithmetic operations  and solve the equation for one variable.
  • Step 5: Now, substitute the value of the variable from  Step 4  in any of the given equations to solve for the other variable.

Here is an example of solving system of equations by using substitution method: 2x+3(y+5)=0 and x+4y+2=0.

Step 1:  Simplify the first equation to get 2x + 3y + 15 = 0. Now we have two equations as,

2x + 3y + 15 = 0 _____ (1)

x + 4y + 2 = 0 ______ (2)

Step 2: We are solving equation (2) for x. So, we get x = -4y - 2.

Step 3: Substitute the obtained value of x in the equation (1). i.e., we are substituting x = -4y-2 in the equation 2x + 3y + 15 = 0, we get, 2(-4y-2) + 3y + 15 = 0.

Step 4: Now, simplify the new equation. We get, -8y-4+3y+15=0

-5y + 11 = 0

Step 5: Now, substitute the value of y in any of the given equations. Let us substitute the value of y in equation (2).

x + 4y + 2 = 0

x + 4 × (11/5) + 2 = 0

x + 44/5 + 2 = 0

x + 54/5 = 0

Therefore, after solving the given system of equations by substitution method, we get x = -54/5 and y= 11/5.

Difference Between Elimination and Substitution Method

Both elimination and substitution methods are ways to solve linear equations algebraically. When the substitution method becomes a little difficult to apply in equations involving large numbers or fractions, we can use the elimination method to ease our calculations. Let us understand the difference between these two methods through the table given below:

Important Notes on Substitution Method:

  • To start with the substitution method, first, select the equation that has coefficient 1 for at least one of the variables and solve for the same variable (with coefficient 1). This makes the process easier.
  • Before starting with the substitution method, combine all like terms (if any).
  • After solving for one variable, we can select any of the given equations or any equation in the whole process to find the other variable.
  • If we get any true statement like 3 = 3, 0 = 0, etc while solving using the substitution method, then it means that the system has infinitely many solutions.
  • If we get any false statement like 3 = 2, 0 = 1, etc then the system has no solution.

☛  Related Topics:

  • Substitution Method Calculator
  • Substitution Method Class 10
  • System of Equations Solver

Substitution Method Examples

Example 1:  Solve the system of linear equations by substitution method: 5m−2n=17 and 3m+n=8.

The given two equations are:

5m−2n=17 ____ (1)

3m+n=8 _____ (2)

The solution of the given two equations can be found by the following steps:

  • From equation 2 we can find the value of n in terms of m, where n = 8 - 3m
  • Substitute the value of n in equation 1. We get, 5m - 2(8-3m)=17

5m - 2(8-3m)=17

5m - 16 + 6m =17

11m = 17 + 16

  • Substitute the value of m in equation 2, we get, 3×3+n=8

Answer:  ∴ The solution is m=3 and n=-1.

Example 2: Jacky has two numbers such that the sum of two numbers is 20 and the difference between them is 10. Find the numbers.

Let the two numbers be x and y such that x>y. It is given that,

x+y=20 ___ (1)

and x−y=10 ___ (2).

We will now solve by substitution.

From equation 1, we get x = 20-y. Substitute this value in equation 2 to find the value of y.

Now, substitute the value of y in equation 1, we get x+5=20, which gives us x=15.

Answer:  Therefore, the two numbers are 15 and 5.

Example 3: Solve the given system of linear equations by substitution method:

- 2x - 5 + 3x + y = 0 ___ (1)

3x + y = 11 ___ (2)

As we can see that the first equation can be further simplified by combining like terms . After simplifying it, we get x+y-5=0. From this equation, let us find the value of x in terms of y, which is x = 5-y. Now substitute this value in equation 2, we get 3(5-y)+y=11.

Now, let us substitute the value of y in equation 1. We get x+2-5=0, which can be simplified to x = 3.

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how to solve 2 equations by substitution

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Practice Questions

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FAQs on Substitution Method

What is the substitution method in algebra.

In algebra, the substitution method is one of the ways to solve linear equations in two variables . In this method, we substitute the value of a variable found by one equation in the second equation. It is very easy to use when we have smaller numbers, but in the case of large numbers or fractional coefficients, it becomes tedious to apply the substitution method.

What are the Steps for the Substitution Method?

The three simple steps for the substitution method are given below:

  • Find the value of any one variable from any of the equations in terms of the other variable.
  • Substitute it in the other equation and solve.
  • Substitute the value of the second variable again in any of the equations.

When would you Use the Substitution Method?

The substitution method can be applied to any pair of linear equations with two variables . It is advisable to use the substitution method when we have smaller coefficients in terms or when the equations are given in form x = ay+c and/or y=bx+p.

What do we Substitute in the Substitution Method?

In the substitution method, we substitute the value of one variable found by simplifying an equation in the other equation. For example, if there are two variables in the equations, say m and n, then we can first find the value of m in terms of n from any one of the equations, and then we substitute that value in the second equation to get an answer of n. Then, we again substitute the value of n in any of the given equations to find m.

What do the Substitution Method and the Elimination Method have in Common?

Both methods involve the process of substitution. In both methods, we find the value of one variable first and then substitute it in any of the given equations. 

What is the First Step in the Substitution Method?

The first step in the substitution method is to find the value of any one of the variables from one equation in terms of the other variable. For example, if there are two equations x+y=7 and x-y=8, then from the first equation we can find that x=7-y. The further steps involve substituting this in the other equation and then solving.

What is the Process of Solving Systems by Substitution?

With a system of equations with variables x and y, we first find the value of x in terms of y from any one of the equations given. Then, we substitute that value in the other equation to find the value of y. At last, we again substitute the value of y in any given equation to find x.

Is the Substitution Method Only for Linear Equations?

No, substitution method can be applied for any type of equations . For example, the equations y = x 2  and y = 3x + 4 can be solved by using the substitution method.

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The Substitution Method

A way to solve systems of linear equations in 2 variables

Video on Solving by Substitution

First, let's review how the substitution property works in general.

Review Example 1

substitution property example 1

Review Example 2

substitution property example 2

Substitution Example 1

picture of algebraic method solution

Let's re-examine system pictured up above.

$ \red{y} = 2x + 1 \text{ and } \red{y} = 4x -1 $

We are going to use substitution like we did in review example 2 above.

example 3

Now we have 1 equation and 1 unknown, we can solve this problem as the work below shows.

The last step is to again use substitution, in this case we know that x = 1, but in order to find the y value of the solution, we just substitute x = 1 into either equation.

$$ y = 2x + 1 \\ y = 2\cdot \red{1} + 1 = 2 + 1 =3 \\ \\ \boxed{ \text{ or you use the other equation}} \\ y = 4x -1 \\ y = 4\cdot \red{1}- 1 \\ y = 4 - 1 = 3 \\ \boxed { ( 1,3) } $$

Substitution Example 2

What is the solution of the system of equations below:

$ y = 2x + 1 \\ 2y = 3x - 2 $

Identify the best equation for substitution and then substitute into other equation.

example 3

Solve for x

Substitute the value of x (-4 in this case) into either equation.

$$ y = 2x + 1 \\ y = 2\cdot \red{-4} + 1 = -8 + 1 = -7 \\ 2y = 3x - 2\\ 2y = 3\cdot-4 -2 \\ \boxed{ \text{ or you use the other equation}} \\ 2y = 3x -2 \\ 2y = 3 ( \red{-4}) -2 \\ 2y = -12 -2 \\ 2y = -14 \frac{1}{2}\cdot2y =\frac{1}{2}\cdot-14 \\ y = -7 $$

$$ \boxed { ( -4, -7 ) } $$

You can also solve the system by graphing and see a picture of the solution below:

Double Check Substitution Method

Substitution Practice Problems

Solve the system below using substitution

$$ y = x+1 \\ y = 2x +2 $$

The solution of this system is the point of intersection: (-1, 0).

$$ y = x + 1 \quad y = 2x + 2 \\ \hspace{1.2cm} \downarrow \hspace{1.4cm} \downarrow \\ \hspace{6mm} x + 1 = 2x + 2 \\ \hspace{7mm} \text{-}x \hspace{1.4cm} \text{-}x \\ \hspace{7mm} \rule{3.2cm}{0.25mm} \\ \hspace{1.7cm} 1 = x + 2 \\ \hspace{1.6cm} \text{-}2 \hspace{1.4cm} \text{-}2 \\ \hspace{7mm} \rule{3.2cm}{0.25mm} \\ \hspace{1.2cm} -1 = x \\ \hspace{1.6cm} \downarrow \\ \hspace{5mm} y = 2x + 2 \\ \hspace{7mm} y = 2 * (-1) + 2 = 0 \\[5mm] \text{Solution:} \hspace{3mm} (-1, 0) $$

Use substitution to solve the following system of linear equations:

  • Line 1: y = 3x – 1
  • Line 2: y = x – 5

Set the Two Equations equal to each other then solve for x

Substitute the x value, -2, into the value for 'x' for either equation to determine y coordinate of solution

$$ y = \red{x} -5 \\ y = \red{-2} -5 = -7 $$

The solution is the point (-2, -7)

Use the substitution method to solve the system:

  • Line 1: y = 5x – 1
  • Line 2: 2y= 3x + 12

Solution of system of equations by substitution method

This system of lines has a solution at the point (2, 9).

Use substitution to solve the system:

  • Line 1: y = 3x + 1
  • Line 2: 4y = 12x + 4

This system has an infinite number of solutions. Because 12x + 4 = 12x is always true for all values of x.

Solve the system of linear equations by substitution

  • Line 1: y= x + 2
  • Line 2: y= x + 8

This system of linear equation has no solution .

These lines have the same slope (slope = 1) so they never intersect.

  • Line 1: y = x + 1
  • Line 2: 2y = 3x

Solution answer

The solution of this system is (1, 3).

  • Line 2: 4y = 12x + 3

No Solutions of system of equations by substitution method

Whenever you arrive at a contradiction such as 3 = 4, your system of linear equations has no solutions. When you use these methods (substitution, graphing , or elimination ) to find the solution what you're really asking is at what

Solve the system using substitution.

  • Line 1: y = x + 5
  • Line 2: y = 2x + 2

Practice Problem seven  solution of system of equations

The solution of this system is the point of intersection: (3, 8).

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How to Solve Simultaneous Equations Using Substitution Method

Last Updated: January 13, 2024 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. 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 213,348 times.

Simultaneous equations are two linear equations with two unknown variables that have the same solution. Solving equations with one unknown variable is a simple matter of isolating the variable; however, this isn’t possible when the equations have two unknown variables. By using the substitution method, you must find the value of one variable in the first equation, and then substitute that variable into the second equation. [1] X Research source While it involves several steps, the substitution method for solving simultaneous equations requires only basic algebra skills.

Finding the Value of y

Step 1 Choose the equation you want to work with first.

Finding the Value of x

Step 1 Plug in the                     y              {\displaystyle y}   value into either equation.

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  • ↑ https://www.mathsteacher.com.au/year9/ch05_simult/01_sub/method.htm
  • ↑ https://www.purplemath.com/modules/systlin4.htm
  • ↑ https://flexbooks.ck12.org/cbook/ck-12-cbse-math-class-10/section/3.5/primary/lesson/solving-simultaneous-linear-equations-by-substitution/
  • ↑ https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-systems-topic/cc-8th-systems-with-substitution/a/systems-of-equations-with-substitution

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4.2: Solve Systems of Equations by Substitution

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Learning Objectives

By the end of this section, you will be able to:

  • Solve a system of equations by substitution
  • Solve applications of systems of equations by substitution

Before you get started, take this readiness quiz.

  • Simplify −5(3−x). If you missed this problem, review Exercise 1.10.43 .
  • Simplify 4−2(n+5). If you missed this problem, review Exercise 1.10.41 .
  • Solve for y. 8y−8=32−2y If you missed this problem, review Exercise 2.3.22 .
  • Solve for x. 3x−9y=−3 If you missed this problem, review Exercise 2.6.22 .

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a System of Equations by Substitution

We will use the same system we used first for graphing.

\(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

We’ll fill in all these steps now in Exercise \(\PageIndex{1}\).

Exercise \(\PageIndex{1}\): How to Solve a System of Equations by Substitution

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

This figure has three columns and six rows. The first row says, “Step 1. Solve one of the equations for either variable.” To the right of this, the middl row reads, “We’ll solve the first equation for y.” The third column shows the two equations: 2x + y = 7 and x – 2y = 6. It shows that 2x + y = 7 becomes y = 7 – 2x.

Exercise \(\PageIndex{2}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{-2 x+y=-11} \\ {x+3 y=9}\end{array}\right.\)

Exercise \(\PageIndex{3}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+3 y=10} \\ {4 x+y=18}\end{array}\right.\)

SOLVE A SYSTEM OF EQUATIONS BY SUBSTITUTION.

  • Solve one of the equations for either variable.
  • Substitute the expression from Step 1 into the other equation.
  • Solve the resulting equation.
  • Substitute the solution in Step 3 into one of the original equations to find the other variable.
  • Write the solution as an ordered pair.
  • Check that the ordered pair is a solution to both original equations.

If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in Exercise \(\PageIndex{4}\).

Exercise \(\PageIndex{4}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+y=-1} \\ {y=x+5}\end{array}\right.\)

The second equation is already solved for y . We will substitute the expression in place of y in the first equation.

Exercise \(\PageIndex{5}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+y=6} \\ {y=3 x-2}\end{array}\right.\)

Exercise \(\PageIndex{6}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x-y=1} \\ {y=-3 x-6}\end{array}\right.\)

(−1,−3)

If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for y .

Exercise \(\PageIndex{7}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{3 x+y=5} \\ {2 x+4 y=-10}\end{array}\right.\)

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

Exercise \(\PageIndex{8}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x+y=2} \\ {3 x+2 y=-1}\end{array}\right.\)

(1,−2)

Exercise \(\PageIndex{9}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{-x+y=4} \\ {4 x-y=2}\end{array}\right.\)

In Exercise \(\PageIndex{7}\) it was easiest to solve for y in the first equation because it had a coefficient of 1. In Exercise \(\PageIndex{10}\) it will be easier to solve for x .

Exercise \(\PageIndex{10}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-2 y=-2} \\ {3 x+2 y=34}\end{array}\right.\)

We will solve the first equation for xx and then substitute the expression into the second equation.

Exercise \(\PageIndex{11}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-5 y=13} \\ {4 x-3 y=1}\end{array}\right.\)

(−2,−3)

Exercise \(\PageIndex{12}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-6 y=-6} \\ {2 x-4 y=4}\end{array}\right.\)

When both equations are already solved for the same variable, it is easy to substitute!

Exercise \(\PageIndex{13}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=-2 x+5} \\ {y=\frac{1}{2} x}\end{array}\right.\)

Since both equations are solved for y , we can substitute one into the other.

Exercise \(\PageIndex{14}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=3 x-16} \\ {y=\frac{1}{3} x}\end{array}\right.\)

Exercise \(\PageIndex{15}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=-x+10} \\ {y=\frac{1}{4} x}\end{array}\right.\)

Be very careful with the signs in the next example.

Exercise \(\PageIndex{16}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x+2 y=4} \\ {6 x-y=8}\end{array}\right.\)

We need to solve one equation for one variable. We will solve the first equation for y .

Exercise \(\PageIndex{17}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-4 y=-4} \\ {-3 x+4 y=0}\end{array}\right.\)

\((2,\frac{3}{2})\)

Exercise \(\PageIndex{18}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x-y=0} \\ {2 x-3 y=5}\end{array}\right.\)

\((−\frac{1}{2},−2)\)

In Example , it will take a little more work to solve one equation for x or y .

Exercise \(\PageIndex{19}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x-3 y=6} \\ {15 y-20 x=-30}\end{array}\right.\)

We need to solve one equation for one variable. We will solve the first equation for x .

Exercise \(\PageIndex{20}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x-3 y=12} \\ {-12 y+8 x=48}\end{array}\right.\)

infinitely many solutions

Exercise \(\PageIndex{21}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x+2 y=12} \\ {-4 y-10 x=-24}\end{array}\right.\)

Look back at the equations in Exercise \(\PageIndex{22}\). Is there any way to recognize that they are the same line?

Let’s see what happens in the next example.

Exercise \(\PageIndex{22}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x-2 y=-10} \\ {y=\frac{5}{2} x}\end{array}\right.\)

The second equation is already solved for y , so we can substitute for y in the first equation.

Exercise \(\PageIndex{23}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{3 x+2 y=9} \\ {y=-\frac{3}{2} x+1}\end{array}\right.\)

no solution

Exercise \(\PageIndex{24}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x-3 y=2} \\ {y=\frac{5}{3} x-4}\end{array}\right.\)

Solve Applications of Systems of Equations by Substitution

We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5.

HOW TO USE A PROBLEM SOLVING STRATEGY FOR SYSTEMS OF LINEAR EQUATIONS.

  • Read the problem. Make sure all the words and ideas are understood.
  • Identify what we are looking for.
  • Name what we are looking for. Choose variables to represent those quantities.
  • Translate into a system of equations.
  • Solve the system of equations using good algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable?

Exercise \(\PageIndex{25}\)

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Exercise \(\PageIndex{26}\)

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

The numbers are 3 and 7.

Exercise \(\PageIndex{27}\)

The sum of two number is −6. One number is 10 less than the other. Find the numbers.

The numbers are 2 and −8.

In the Exercise \(\PageIndex{28}\), we’ll use the formula for the perimeter of a rectangle, P = 2 L + 2 W .

Exercise \(\PageIndex{28}\)

Add exercises text here.

Exercise \(\PageIndex{29}\)

The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle.

The length is 12 and the width is 8.

Exercise \(\PageIndex{30}\)

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width of the rectangle.

The length is 23 and the width is 6.

For Exercise \(\PageIndex{31}\) we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle.

Exercise \(\PageIndex{31}\)

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Exercise \(\PageIndex{32}\)

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of the angles are 22 degrees and 68 degrees.

Exercise \(\PageIndex{33}\)

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of the angles are 36 degrees and 54 degrees.

Exercise \(\PageIndex{34}\)

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 + $40 for each training session. How many training sessions would make the salary options equal?

Exercise \(\PageIndex{35}\)

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

There would need to be 160 policies sold to make the total pay the same.

Exercise \(\PageIndex{36}\)

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Kenneth would need to sell 1,000 suits.

Access these online resources for additional instruction and practice with solving systems of equations by substitution.

  • Instructional Video-Solve Linear Systems by Substitution
  • Instructional Video-Solve by Substitution

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  • Solving System of Linear Equations

Substitution Method

The substitution method is most useful for systems of 2 equations in 2 unknowns. The main idea here is that we solve one of the equations for one of the unknowns, and then substitute the result into the other equation.

Substitution method can be applied in four steps

Solve one of the equations for either x = or y = .

Substitute the solution from step 1 into the other equation.

Solve this new equation.

Solve for the second variable.

Example 1: Solve the following system by substitution

Step 1: Solve one of the equations for either x = or y = . We will solve second equation for y.

Step 2: Substitute the solution from step 1 into the second equation.

Step 3: Solve this new equation.

Step 4: Solve for the second variable

The solution is: (x, y) = (10, -5)

Note: It does not matter which equation we choose first and which second. Just choose the most convenient one first!

Example 2: Solve by substitution

Step 1: Solve one of the equations for either x = or y =. Since the coefficient of y in equation 2 is -1, it is easiest to solve for y in equation 2.

Step 3: Solve this new equation ( for x ).

The solution is: $(x, y) = (1, 2)$

Exercise: Solve the following systems by substitution

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Systems of Linear Equations: Solving by Substitution

Definitions Graphing Special Cases Substitution Elimination/Addition Gaussian Elimination More Examples

The method of solving "by substitution" works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable.

Here is how it works. (I'll use one of the systems from the " solving by graphing " page.)

Content Continues Below

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Solving Systems of Equations by Substitution on MathHelp.com

Solving Systems by Substitution

  • Solve the following system by substitution.

2 x − 3 y = −2 4 x + y = 24

The instructions tell me to solve "by substitution". This means that I need to solve one of the equations for one of the variables, and plug the result into the other equation in place of the variable I've solved for. It does not matter which equation or which variable I pick. There is no right or wrong choice of equation or variable; the answer will be the same, regardless. But — some choices may be better than others.

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For instance, in this case, can you see that it would probably be simplest to solve the second equation for " y = ", since there is already a y floating around loose in the middle of that equation? I could solve the first equation for either variable, but I'd get fractions, and solving the second equation for x would also give me fractions. It wouldn't be "wrong" to make a different choice, but it would probably be more difficult.

Being lazy, I'll solve the second equation for y :

4 x + y = 24 y = −4 x + 24

Now I'll plug this in (that is, I'll "substitute " this) for y in the first equation, and solve the resulting one-variable equation for the value of x :

2 x − 3(−4 x + 24) = −2 2 x + 12 x − 72 = −2 14 x = 70 x = 5

Now I can plug this x -value back into either equation, and solve for the corresponding value of y . But since I already have an expression for " y = ", it will be simplest to just plug into this:

y = −4(5) + 24 = −20 + 24 = 4

Then my solution is the following point:

solution: ( x , y ) = (5, 4)

How do you pick the equation to substitute into?

There is no particular rule about which equation to plug into. You picked one of the equations and solved it for one of the variables. You can now pick any of the *other* equations that contains that solved-for variable, and plug in for that variable. The idea is to extract some information from the first equation, and then plug its info into one of the other equations, and see where that takes you.

Both the first equation you solve, and the different equation you plug into, are entirely your choice. There can be many correct ways to solve a system by substitution, but the most important thing to remember is to plug into a different equation rather than the one you'd started with.

For instance, in the above exercise, if I had substituted my " −4 x + 24 " expression into the same equation as I'd used to solve for " y = ", I would have gotten a true, but useless, statement:

4 x + (−4 x + 24) = 24 4 x − 4 x + 24 = 24 24 = 24

Yes, twenty-four does equal twenty-four, but who cares? How does this help? It doesn't, other than to suggest that something might be wrong with your process or assumptions.

So, when using substitution, make sure you substitute from one equation into the other equation, or you'll just be wasting your time.

y = 36 − 9 x 3 x + y /3 = 12

We already know (from the previous page ) that these equations are actually both the same line; that is, that this is a dependent system. We know what this looks like graphically: we get two identical line equations, and we get a graph that displays just the one line. But what does this look like algebraically?

The first equation is already solved for y , so I'll substitute that into the second equation:

3 x + (36 − 9 x )/3 = 12 3 x + 12 − 3 x = 12 12 = 12

Well, um... yes, twelve does equal twelve, but so what?

I did substitute the result from the first equation into the second equation, so this unhelpful result is not because of some screw-up on my part. It's just that this is what a dependent system looks like when you try to find a solution.

Remember that, when you're trying to solve a system, you're trying to use the second equation to narrow down the choices of points on the first equation. You're trying to find the one single point that works in both equations. But in a dependent system, the "second" equation is really just another copy of the first equation, and all the points on the one line will work in the other line.

In other words, I got an unhelpful result because the second line equation didn't tell me anything new. This tells me that the system is actually dependent, and that the solution is the whole line:

solution: y = 36 − 9 x

This is always true, by the way. When you try to solve a system and you get a statement like " 12 = 12 " or " 0 = 0 " — something that's true, but unhelpful (I mean, duh!, of course twelve equals twelve!) — then you have a dependent system. We already knew (from the previous page) that this system was dependent, but now you know what the algebra looks like.

(Keep in mind that your text may format the answer to look something like " ( t , 36 − 9 t ) ", or something similar, using some variable, some "parameter", other than x . But this "parametrized" form of the solution means the exact same thing as "the solution is the line y = 36 − 9 x ".)

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7 x + 2 y = 16 −21 x − 6 y = 24

Neither of these equations is particularly easier than the other for solving. I'll get fractions, no matter which equation and which variable I choose. So, um... I guess I'll take the first equation, and I'll solve it for, um, y , because at least the 2 (from the " 2 y ") will divide evenly into the 16 .

7 x + 2 y = 16 2 y = −7 x + 16 y = −(7/2) x + 8

Now I'll plug this into the other equation:

−21 x − 6(−(7/2) x + 8) = 24 −21 x + 21 x − 48 = 24 −48 = 24

Um... I don't think that's right....

In this case, I got a nonsense result. All my math was right, but I got an obviously wrong answer. So what happened?

Keep in mind that, when solving, you're trying to find where the lines intersect. But what if they don't intersect?

Then you're going to get some kind of wrong answer when you assume that there is a solution (as I did when I tried to find that solution). We knew, from the previous page, that this system represents two parallel lines. But I tried, by substitution, to find the intersection point anyway. And I got a "garbage" result. Since there wasn't any intersection point, my attempt led to utter nonsense.

solution: no solution (inconsistent system)

This is always true, by the way. When you get a nonsense result, this is the algebraic indication that the system of equations is inconsistent.

Note that this is quite different from the previous example:

A true-but-useless result (like " 12 = 12 ") is quite different from a nonsense "garbage" result (like " −48 = 24 "), just as two identical lines are quite different from two distinct parallel lines.

A useless result means a dependent system which has a solution (every point on the whole line, rather than just one point); a nonsense result means an inconsistent system which has no solution of any kind. Don't confuse the two!

https://www.purplemath.com/modules/systlin4.htm

You can use the Mathway widget below to practice solving systems of equations by using the method of substitution (or skip the widget, and continue to the next page ). Try the entered exercise, or type in your own exercise. Then click the button, select "Solve by Substitution" from the box, and compare your answer to Mathway's.

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Solving Systems of Equations by Substitution: Explanation, Review, and Examples

  • The Albert Team
  • Last Updated On: February 16, 2023

how to solve 2 equations by substitution

To understand solving systems of equations by substitution, let’s first think about what substitution means.

We use substitution in many situations. First, when the pitcher on a softball team hurts her shoulder, another player can take her place as a substitute. Next, we can substitute vegan alternatives to animal products in a recipe. Finally, when we run into construction on our drive home and take a different road, we’re substituting one route for another. In each of these examples of substitution, we are replacing one entity with another equivalent one to solve a problem or reach a goal.

In the study of Algebra, we learn how to substitute variables for mathematical values in expressions. There is an algebraic property of equality called the Substitution Property , which states:

For example, we can substitute 7 for x in the following equation.

We can do this because 5+2=7 .

The equation above has only one variable. What about situations where we have two or more variables and two or more equations? These  systems of equations  can seem more challenging, but solving systems of linear equations by substitution is often the easiest way to find solutions. 

Solving Systems of Equations Algebraically

There are multiple methods for solving systems of equations, including solving systems algebraically. Solving systems algebraically involves manipulating the equations we are given to uncover the values of each of the variables. 

But when must a system of linear equations be solved algebraically? When solving systems of equations, we should generally choose the method that takes the least effort and leaves the least room for error.

Let’s look at the system of linear equations below:

First, notice we are given the value of one of the variables, x . So, we can easily substitute that value for x into the other equation and solve for y . 

To review how to solve equations, check out our post: Solving One-Step Equations .

When we practice solving systems of equations, students are often told which method to use to find the solution. The directions might say: solve by graphing, solve by elimination, or solve by substitution. We will cover the first two methods in other posts. Here, we will focus on how to solve a system of equations algebraically using substitution.

How to Solve a System of Equations by Substitution

To solve a system of equations by substitution, we can rewrite a two-variable equation as a single variable equation by substituting the value of a variable from one equation into the other.

Let’s start by solving the system of equations that we looked at above:

As we decide how to solve systems of equations with substitution, we almost always have options. We have to decide which variable to substitute and which equation to substitute it into. 

In this example, the choice is clear. Since the first equation says that x=4 , we will substitute x with 4 in the second equation so that the second equation becomes:

Next, we can use the subtraction property of equality to subtract 4 from each side of the equation: 

We just solved this system of linear equations with substitution! The solution to this system is (4,8) .

Knowing that the solution to a system of linear equations is the point of intersection, we can confirm graphically that the coordinate pair (4,8) is the solution to this system of equations.

how to solve 2 equations by substitution

Pro Tip: Online graphing calculators like Desmos can help you check your work quickly and easily.

Solving Systems of Equations by Substitution Steps

So, the steps for using the substitution method to solve a system of linear equations are:

  • Rewrite one of the equations to isolate one variable.
  • In the other equation, substitute the value of your isolated variable in for that variable.
  • Solve this second equation for the other variable. You should have a numerical value.
  • Substitute your numerical value into one of the two original equations and solve for the other variable.
  • Check your work. You can do this by graphing or by substituting the solutions into the original equations.

Substitution Method Examples

It’s helpful to use these steps when we consider how to solve systems of equations by substitution. Now, we can apply these steps to various systems to see if they work.

Solving Systems of Equations by Substitution Examples (One Solution)

Let’s see if these steps work for another system of equations:

1. Rewrite one of the equations to isolate one variable. Let’s solve the first equation for y :

2. In the other equation, substitute the value of your isolated variable in for that variable. So, we will substitute 10-x in for y into the second equation so that it becomes:

3. Solve this second equation for the other variable. In this case, we are solving for x :

4. Substitute your numerical value into one of the two original equations and solve for the other variable. We’ll substitute 6 for x into the first equation and solve for y .

5. Check your work either. This time, we’ll confirm algebraically that the coordinate pair ({\color{red}{6}},{\color{blue}{4}}) works by substituting those values into the other equation, x-y=2 :

Solving Systems of Equations by Substitution Examples (No Solution)

The systems of equations we have solved so far had one solution, but systems of equations may also have zero, multiple, or an infinite number of solutions. Let’s solve a no solution system of equations by substitution:

Notice that y is isolated in the second equation. So, we can substitute (-x+1) for y into the first equation so that it becomes: 

At this point, we have a statement that is not true. A false statement tells us that there is no solution to the system of equations.

If we graph this system, we will see that these are equations of parallel lines, and parallel lines never intersect.

how to solve 2 equations by substitution

Solving Systems of Equations by Substitution Examples (Infinite Solutions)

Let’s solve another system of linear equations by substitution:

In this system, the first equation almost has y isolated, so let’s rewrite that one:

Now we can substitute -x+4 for y in the second equation and solve algebraically:

Our equation is a true statement. However, it doesn’t tell us the values of our variables. Therefore, there are an infinite number of solutions to this system of equations. How can that be? Both equations graph as the same line. We can verify this by rewriting each equation into slope-intercept form.

Earlier, we found that the first equation can be rewritten as:

Next, the second equation becomes:

Since the equations are the same, the lines fully overlap. So, the system will have an infinite number of solutions because there are an infinite number of coordinate pairs that lie on both lines.

The Substitution Method: Keys to Remember

  • Substitution is a helpful strategy in both life and math.
  • Solving systems of equations algebraically involves using the Properties of Algebra.
  • Substitution may be the obvious way to approach a system of equations, or question directions may require using substitution to solve systems of linear equations.
  • Substitution allows us to eliminate one variable in a two-variable equation and solve for the other. 
  • Once the value of one variable is determined, we can use substitution again to solve for the other variable(s).
  • Systems of equations may have zero, one, multiple, or infinite solutions which can be determined and checked algebraically.

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Module 4: Systems of Linear Equations and Inequalities

4.2: solving a 2×2 system of linear equations by substitution, section 4.2 learning objectives.

4.2:  Solving a 2×2 System of Linear Equations by Substitution 

  • Solve systems of linear equations using substitution
  • Recognize when systems of linear equations have no solution or an infinite number of solutions

Solve a system of equations using the substitution method

In the previous section, we verified that ordered pairs were potential solutions to systems, and we used graphs to classify how many solutions a system of two linear equations had. What if we are not given a point of intersection, or it is not obvious from a graph? Can we still find a solution to the system? Of course you can, using algebra!

In this section we will learn the substitution method for finding a solution to a system of linear equations in two variables. We have used substitution in different ways throughout this course.  For example, when we were using formulas, we substituted values that we knew into the formula to solve for values that we did not know.  The idea is similar when applied to solving systems, there are just a few different steps in the process. You will first solve for one variable, and then substitute that expression into the other equation. Let’s start with an example to see what this means.

Find the value of [latex]x[/latex] for this system.

[latex]\begin{array}{r}4x+3y=−14\\y=2\end{array}[/latex]

[latex]\begin{array}{r}4x+3y=−14 \\ 4x+3\left(2\right)=−14\end{array}[/latex]

Simplify and solve the equation for [latex]x[/latex] .

[latex]\begin{array}{r}4x+6=−14\\ \underline{\hspace{.25in}-\hspace{.03in}6 \hspace{.23in}-6}\\ 4x\hspace{.31in}=−20\end{array}[/latex]

[latex] {\hspace{.31in}\displaystyle \frac{4x}{4}=\frac{-20}{4}}[/latex]

[latex]\hspace{.3in}x=−5[/latex]

[latex]x=−5[/latex]

You can substitute a value for a variable even if it is an expression. Here’s an example.

Solve the system of equations for [latex]x[/latex]   and [latex]y[/latex].

[latex]\begin{array}{l}y+x=3\\x=y+5\end{array}[/latex]

[latex]\begin{array}{r}y+x=3\\y+\left(y+5\right)=3\end{array}[/latex]

Simplify and solve the equation for [latex]y[/latex] .

[latex]\begin{array}{r}2y+5=3\,\,\,\,\,\\{\underline{\hspace{.25in}−\hspace{.03in}5 \hspace{.07in}−5}}\,\,\,\\2y\hspace{.33in}=−2\end{array}[/latex]

[latex]\hspace{.34in} \displaystyle \frac{2y}{2}=\frac{-2}{2}[/latex]

[latex]\hspace{.4in}y=−1[/latex]

Now find [latex]x[/latex]   by substituting this value for [latex]y[/latex] into either equation. We will use the first equation here.

[latex]y+x=3[/latex]

[latex]\begin{array}{r} -1+x=3\hspace{.03in}\\ \underline{+1\hspace{.36in}+1}\\x=4\hspace{.03in}\end{array}[/latex]

Finally, we can check the solution [latex]x=4[/latex], [latex]y=−1[/latex] by substituting these values into each of the original equations.

[latex]\begin{array}{r}y+x=3\\−1+4=3\\3=3\\\text{TRUE}\end{array}[/latex]

[latex]\begin{array}{r}x=y+5\\4=−1+5\\4=4\\\text{TRUE}\end{array}[/latex]

[latex]x=4[/latex] and [latex]y=−1[/latex]

Equivalently, the solution is [latex](4,−1)[/latex].

Remember, a solution to a system of equations must be a solution to each of the equations within the system. The ordered pair [latex](4,−1)[/latex] does work for both equations, so you know that it is a solution to the system.

Let’s look at another example whose substitution involves the distributive property.

Solve the system for [latex]x[/latex]   and [latex]y[/latex].

[latex]\begin{array}{r}y = 3x + 6\\−2x + 4y = 4\end{array}[/latex]

The first equation tells you how to express [latex]y[/latex] in terms of [latex]x[/latex], so it makes sense to substitute [latex]3x+6[/latex] into the second equation for [latex]y[/latex].

[latex]\begin{array}{r}−2x+4y=4\\−2x+4\left(3x+6\right)=4\end{array}[/latex]

[latex]−2x+12x+24=4\hspace{.8in}[/latex]

[latex]\begin{array}{r}10x+24=4\,\,\,\,\,\,\,\\\underline{\hspace{.38in}−\hspace{.03in}24\hspace{.08in}−24\,}\\10x\hspace{.38in}=−20\end{array}[/latex]

[latex]\hspace{.93in}\displaystyle \frac{10x}{10}=\frac{-20}{10}\hspace{.5in}[/latex]

[latex]\hspace{1.01in}x=-2\hspace{.47in}[/latex]

To find [latex]y[/latex], substitute this value for [latex]x[/latex]back into one of the original equations.

[latex]\begin{array}{l}y=3x+6\\y=3\left(−2\right)+6\\y =−6+6\\y=0\end{array}[/latex]

Check the solution [latex]x=−2[/latex], [latex]y=0[/latex] by substituting them into each of the original equations.

[latex]\begin{array}{l}y=3x+6\\0=3\left(−2\right)+6\\0=−6+6\\0=0\\\text{TRUE}\end{array}[/latex]

[latex]\begin{array}{r}−2x+4y=4\\−2\left(−2\right)+4\left(0\right)=4\\4+0=4\\4=4\\\text{TRUE}\end{array}[/latex]

[latex]x=−2[/latex] and [latex]y=0[/latex]

Equivalently, the solution is (−2, 0).

In the examples above, one of the equations was already given to us in terms of the variable x or y . This allowed us to quickly substitute into the other equation and solve for one of the unknowns.

Sometimes you may have to rewrite one of the equations in terms of one of the variables first before you can substitute. In the example below, you will first need to isolate one of the variables before you can substitute it into the other equation.

Solve for [latex]x[/latex] and [latex]y[/latex].

[latex]\begin{array}{r}2x+3y=22\\3x+y=19\end{array}[/latex]

[latex]3x+y=19[/latex], can easily be solved for [latex]y[/latex], so it makes sense to start there.

Solve [latex]3x+y=19[/latex] for [latex]y[/latex].

[latex]\begin{array}{r}3x+y=19\hspace{.42in}\\ \underline{-3x \hspace{.41in}-3x}\hspace{.31in}\\ \hspace{.78in}y=19-3x\end{array}[/latex]

Substitute [latex]19–3x[/latex] for [latex]y[/latex] in the first equation.

[latex]\begin{array}{r}2x+3y=22\\2x+3(19–3x)=22\end{array}[/latex]

[latex]2x+57-9x=22\hspace{.36in}[/latex]

[latex]\begin{array}{r}-7x+57=22\,\,\,\,\,\\{\underline{\hspace{.39in}-\hspace{.04in}57\hspace{.09in}-57}}\,\,\\-7x\hspace{.41in}=-35\end{array}[/latex]

[latex]\displaystyle \hspace{.7in}\frac{-7x}{-7}=\frac{-35}{-7}\hspace{.28in}[/latex]

[latex]\hspace{.7in}x=5\hspace{.3in}[/latex]

Substitute [latex]x=5[/latex] back into one of the original equations to solve for y.

[latex]\begin{array}{r}3x+y=19\,\,\,\,\,\,\,\,\,\,\,\,\\3\left(5\right)+y=19\,\,\,\,\,\,\,\,\,\,\,\,\\15+y=19\,\,\,\,\,\,\,\,\,\,\,\,\\ \underline{−15\hspace{.3in}-15}\,\,\,\,\,\hspace{.2in}\\y=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Check both solutions by substituting them into each of the original equations.

[latex]\begin{array}{r}2x+3y=22\\2(5)+3\left(4\right)=22\\10+12=22\\22=22\\\text{TRUE}\\\\3x+y=19\\3\left(5\right)+4= 19\\19=19\\\text{TRUE}\end{array}[/latex]

[latex]x=5[/latex] and [latex]y=4[/latex]

Equivalently, the solution is (5, 4).

In the following video, you will be given an example of solving a systems of two equations using the substitution method.

If you had chosen the other equation to start with in the previous example, you would still be able to find the same solution.  It is really a matter of preference because sometimes solving for a variable will result in having to work with fractions.  As you become more experienced with algebra, you will be able to anticipate what choices will lead to more desirable outcomes.

Recognize systems of equations that have no solution or an infinite number of solutions

When we learned methods for solving linear equations in one variable, we found that some equations didn’t have any solutions, and others had an infinite number of solutions.  We saw this behavior again when we started describing solutions to systems of equations in two variables.

Recall this example from Module 1 for solving linear equations in one variable:

Solve for x . [latex]12+2x–8=7x+5–5x[/latex]

[latex] \displaystyle \begin{array}{l}12+2x-8=7x+5-5x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x+4=2x+5\end{array}[/latex]

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,2x+4=2x+5\\\,\,\,\,\,\,\,\,\underline{-2x\,\,\,\,\,\,\,\,\,\,-2x\,\,\,\,\,\,\,\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4= \,5\end{array}[/latex]

This false statement implies there are no solutions to this equation.  In the same way, you may see an outcome like this when you use the substitution method to find a solution to a system of linear equations in two variables. In the next example, you will see an example of a system of two equations that does not have a solution.

Solve for x and y .

[latex]\begin{array}{l}y=5x+4\\10x−2y=4\end{array}[/latex]

[latex]\begin{array}{r}10x−2y=4\,\,\,\,\,\,\,\,\,\,\,\,\\10x–2\left(5x+4\right)=4\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Expand the expression on the left.

[latex]10x–10x–8=4[/latex]

Combine like terms on the left side of equation. But since [latex]10x-10x=0[/latex], we are left with

[latex]−8=4[/latex].

Since this is a false statement, we conclude that there is no solution.

No Solution

You get the false statement [latex]−8=4[/latex]. What does this mean? The graph of this system sheds some light on what is happening.

Two parallel lines. One line is y=5x+4, and the other line is 10x-2y=4.

The lines are parallel, they never intersect and there is no solution to this system of linear equations. Note that the result [latex]−8=4[/latex] is not a solution. It is simply a false statement and it indicates that there is no solution.

We have also seen linear equations in one variable and systems of equations in two variables that have an infinite number of solutions.  In the next example, you will see what happens when you apply the substitution method to a system with an infinite number of solutions.

Solve for x and y.

[latex]\begin{array}{r}\,\,\,y=−\frac{1}{2}x+\frac{1}{3}\\3x+6y=2\end{array}[/latex]

Show Solution

We can substitute [latex]y=−\frac{1}{2}x+\frac{1}{3}[/latex] for [latex]y[/latex] in the second equation. This gives you find the following:

[latex]\begin{array}{r}3x+6y=2\\3x+6\left(-\frac{1}{2}x+\frac{1}{3}\right)=2\\3x-3x+2=2\\2=2\end{array}[/latex]

This true statement implies that there are an infinite number of solutions, since the statement is true for all values of [latex]x[/latex]. The discussion below describes why this is the case and what exactly it means.

This time you get a true statement: [latex]2=2[/latex]. But what does this type of answer mean? Again, graphing can help you make sense of this system.

how to solve 2 equations by substitution

This system consists of two equations that both represent the same line . Every point along the line will be a solution to the system, and that’s why the substitution method yields a true statement. In this case, there are an infinite number of solutions.

In the previous section, we learned that we can write the solution set in this case as the following:

[latex]\{(x,y)\hspace{.01in}|\hspace{.01in}y=-\frac{1}{2}x+\frac{1}{3}\}[/latex]

In the following video you will see an example of solving a system that has an infinite number of solutions.

In the next video you will see an example of solving a system of equations that has no solutions.

The substitution method is one way of solving systems of equations. To use the substitution method, use one equation to find an expression for one of the variables in terms of the other variable. Then substitute that expression in place of that variable in the second equation. You can then solve this equation as it will now have only one variable. Solving using the substitution method will yield one of three results: a single value for each variable within the system (indicating one solution), an untrue statement (indicating no solutions), or a true statement (indicating an infinite number of solutions).

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Step by Step Solution

Reformatting the input :.

Changes made to your input should not affect the solution:  (1): "a8"   was replaced by   "a^8". 

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :                       2/(a^8)-(9/(a-2))=0 

Step  1  :

Equation at the end of step  1  :, step  2  :, equation at the end of step  2  :, step  3  :, calculating the least common multiple :.

  3.1      Find the Least Common Multiple       The left denominator is :         a 8         The right denominator is :         a-2  

      Least Common Multiple:        a 8  •  (a-2)  

Calculating Multipliers :

  3.2      Calculate multipliers for the two fractions     Denote the Least Common Multiple by  L.C.M      Denote the Left Multiplier by  Left_M      Denote the Right Multiplier by  Right_M      Denote the Left Deniminator by  L_Deno      Denote the Right Multiplier by  R_Deno      Left_M = L.C.M / L_Deno = a-2     Right_M = L.C.M / R_Deno = a 8

Making Equivalent Fractions :

  3.3        Rewrite the two fractions into equivalent fractions Two fractions are called equivalent if they have the same numeric value. For example :   1/2   and  2/4  are equivalent,  y/(y+1) 2   and  (y 2 +y)/(y+1) 3   are equivalent as well. To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

Adding fractions that have a common denominator :

  3.4         Adding up the two equivalent fractions Add the two equivalent fractions which now have a common denominator Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

Step  4  :

Pulling out like terms :.

  4.1       Pull out like factors :     -9a 8 + 2a - 4   =    -1 • (9a 8 - 2a + 4)  

Polynomial Roots Calculator :

  4.2      Find roots (zeroes) of :        F(a) = 9a 8 - 2a + 4 Polynomial Roots Calculator is a set of methods aimed at finding values of   a   for which   F(a)=0   Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  a  which can be expressed as the quotient of two integers The Rational Root Theorem states that if a polynomial zeroes for a rational number   P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient In this case, the Leading Coefficient is  9  and the Trailing Constant is  4.   The factor(s) are: of the Leading Coefficient :  1,3 ,9   of the Trailing Constant :  1 ,2 ,4   Let us test ....

Note - For tidiness, printing of 13 checks which found no root was suppressed Polynomial Roots Calculator found no rational roots

Equation at the end of step  4  :

Step  5  :, when a fraction equals zero :.

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero. Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator. Here's how:

Now, on the left hand side, the   a 8  •  a-2   cancels out the denominator, while, on the right hand side, zero times anything is still zero. The equation now takes the shape :     -9a 8 +2a-4   = 0

Equations of order 5 or higher :

  5.2       Solve    -9a 8 +2a-4 = 0 Handling of functions of an even degree greater than 6 is not implemented yet

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IMAGES

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  2. Solving a System of 2 Equations with Substitution

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VIDEO

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COMMENTS

  1. Substitution method review (systems of equations)

    The second equation is solved for x , so we can substitute the expression − y + 3 in for x in the first equation: 3 x + y = − 3 3 ( − y + 3) + y = − 3 − 3 y + 9 + y = − 3 − 2 y = − 12 y = 6 Plugging this value back into one of our original equations, say x = − y + 3 , we solve for the other variable: x = − y + 3 x = − ( 6) + 3 x = − 3

  2. Substitution Method

    Solution: Step 1: Simplify the first equation to get 2x + 3y + 15 = 0. Now we have two equations as, 2x + 3y + 15 = 0 _____ (1) x + 4y + 2 = 0 ______ (2) Step 2: We are solving equation (2) for x. So, we get x = -4y - 2.

  3. How to solve systems of linear equations by substitution, examples

    Step 1 We are going to use substitution like we did in review example 2 above. Now we have 1 equation and 1 unknown, we can solve this problem as the work below shows. The last step is to again use substitution, in this case we know that x = 1, but in order to find the y value of the solution, we just substitute x = 1 into either equation.

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  7. 4.2: Solving Linear Systems by Substitution

    Solve by substitution: Solution: Step 1: Solve for either variable in either equation. If you choose the first equation, you can isolate y in one step. 2x + y = 7 2x + y− 2x = 7− 2x y = − 2x + 7. Step 2: Substitute the expression − 2x + 7 for the y variable in the other equation. Figure 4.2.1.

  8. How to Solve Simultaneous Equations Using Substitution Method

    1 Choose the equation you want to work with first. It doesn't matter which equation you choose, but you might want to look for one that will give you numbers that are easier to work with. [2] For example, if your simultaneous equations are 1) and 2) , you will probably want to begin with the first equation, because the is already by itself. 2

  9. 4.2: Solve Systems of Linear Equations with Two Variables

    Solve a System of Equations by Substitution. We will now solve systems of linear equations by the substitution method. We will use the same system we used first for graphing. {2 x + y = 7 x − 2 y = 6 {2 x + y = 7 x − 2 y = 6. We will first solve one of the equations for either x or y. We can choose either equation and solve for either ...

  10. 4.2: Solve Systems of Equations by Substitution

    When both equations are already solved for the same variable, it is easy to substitute! Exercise 4.2.13 4.2. 13. Solve the system by substitution. {y = −2x + 5 y = 12x { y = − 2 x + 5 y = 1 2 x. Answer. Substitute 1 2 x 1 2 x for y in the first equation. Replace the y with 1 2 x 1 2 x. Solve the resulting equation.

  11. Systems of equations with substitution: -3x-4y=-2 & y=2x-5

    So this first blue equation would then become -3x-4 but instead of putting a y there the second constraint tells us that y needs to be equal to 2x-5. So it's 4 (2x-5) and all of that is going to be equal to -2. So now we get just one equation with one unknown. and now we just have to solve for x. So, let's see if we can do that.

  12. Substitution method

    The substitution method is most useful for systems of 2 equations in 2 unknowns. The main idea here is that we solve one of the equations for one of the unknowns, and then substitute the result into the other equation. Substitution method can be applied in four steps Step 1: Solve one of the equations for either x = or y = . Step 2:

  13. Solving a System by Substitution -- Explained!

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