# Solving Systems of Equations by Substitution

When we refer to solving a system of equations, we want to solve for a numerical value for one variable. In this section we will describe how to do that using substitution. Below are two conditions that should be met if you want to use substitution as a method for solving a system of equations.

• There must be the same number of equations as variables.   If there are two variables, there must be two equations; three variables, three equations, etc.
• One of the equations can easily be solved for one variable.

The use of substitution does not exclude addition as another logical and efficient method, but if one variable is easily solved for, the substitution method may be quicker to use.

 Substitution Method for Solving Systems of Equations with Two Equations Choose one equation and isolate one variable; this equation will be considered the first equation. (Use the method detailed in earlier units to do this.) Substitute the solution from step 1 into the second equation and solve for the variable in the equation. Using the value found in step 2, substitute it into the first equation and solve for the second variable. Substitute the values for both variables into both equations to show they are correct.

Now let’s take a closer look at using this method for solving systems of two equations. If we are given the two equations:

(a) y – 3x = 5 and
(b) y + x = 3

We want to determine if we can find a numerical value for both x and y. Let’s look at this by working through the steps outlined above.

1. Choose one equation and isolate one variable; this equation will be considered the first equation.

 To isolate the y variable we subtract x from both sides. y + x = 3 y + x – x = 3 – x y = 3 – x

2. Substitute the solution from step 1 into the second equation and solve for the variable in the equation.

 In equation (a) y – 3x = 5, replace the variable y with the value for y obtained in step 1, from equation (b). y – 3x = 5 (3 – x) – 3x = 5 Solve for x by first combining like terms, then isolating the terms with the x variable. 3 – x – 3x = 5 3 – 3 – 4x= 5 – 3 – 4x = 2 Divide both sides by the coefficient of x to isolate x. x = 2/–4 = –½

3. Using the value found in step 2, substitute it into the first equation and solve for the second variable.

 Now we take the value of x = –½ and plug it back into y + x = 3 and solve for the value of y. y + x = 3 y + (– ½) = 3 y – ½ = 3 y – ½ + ½ = 3 + ½ y = 3½

4. Substitute the values for both variables into both equation to show they are correct.

 Now we substitute the value of x = –½ and y = 3½ into both of our original equations. (a) y – 3x = 5 3½ – 3(–½) = 5 3½ + 1½ = 5 5 = 5 (b) y + x = 3 3½ + (–½) = 3 3 = 3

### Example

Solve the system of linear equations given below using substitution.

Suppose there is a piggybank that contains 57 coins, which are only quarters and dimes. The total number of coins in the bank is 57, and the total value of these coins is \$9.45. This information can be represented by the following system of equations:

D + Q = 57
00.10D + 0.25Q = 9.45

Determine how many of the coins are quarters and how many are dimes.

The information for this example is represented by the following system of equations:

D + Q = 57
0.10D + 0.25Q = 9.45

 Determine how many of the coins are quarters and how many are dimes. D = 32 (number of dimes) Q = 25 (number of quarters

To solve this problem using substitution we need to follow the steps outlined in this section.

1. Choose one equation and isolate one variable; this equation will be considered the first equation.

The equation D + Q = 57 is one that can be easily solved for D.

 We want to isolate D, so we subtract Q from both sides of the equation. D + Q = 57 D + Q – Q = 57 – Q D = 57 – Q

2. Substitute the solution from step 1 into the second equation and solve for the variable in the equation.

 Now we substitute the value for D, which is 57 – Q, into the other equation, 0.10D + 0.25Q = 9.45. This leaves us with one equation with only one variable, Q. We find a numeric value for Q by isolating Q. 0.10D + 0.25Q = 9.45, and D = 57 – Q, so 0.10 (57 – Q) + 0.25Q = 9.45 First we must get rid of the parenthesis using the distributive property. Then we combine like terms. 5.7 – 0.10Q + 0.25Q = 9.45 5.7 + 0.15Q = 9.45 Now subtract 5.7 from both sides of these equation to isolate the term containing the variable Q. 5.7 + 0.15Q – 5.7 = 9.45 –5.7 0.15Q = 3.75 Divide both sides by 0.15. (0.15/0.15)Q = 3.75/0.15 Q = 25

3. Using the value found in step 2, substitute it into the first equation and solve for the second variable.

 We found that Q = 25, so we substitute that into the equation D + Q = 57. When we do this we find D = 32. D + Q = 57 D + 25 = 57 D + 25 – 25 = 57 – 25 D = 32

4. Substitute the values for both variables into both equation to show they are correct.

 Now we should substitute the value of D = 32 and Q = 25 into both of our original equations. D + Q = 57 32 + 25 = 57 57 = 57 0.10D + 0.25Q = 9.45 .10(32) + .25(25) = 9.45 3.20 + 6.25 = 9.45 9.45 = 9.45

These values work in both equations so we have the correct answer.

If you wish to try a practice problem now, return to the introduction and select the practice or additional practice button.