# Solving Systems of Equations by Addition/Subtraction

A second method that can be used to solve systems of equations is addition and subtraction. Below are two conditions that should be met if you want to use addition/subtraction for solving a system of equations.

• There must be the same number of equations as variables.
• One variable from both equations has the same coefficient in both equations, or the coefficients are multiplies of one another.

In many cases this technique may be more efficient than using substitution. The method you choose to use is a matter of preference.

 Addition/Subtraction Method for Solving Systems of Two Equations Algebraically manipulate both equations so that all the variables are on one side of the equal sign and in the same order. (Line the equations up, one on top of the other.) If needed, multiply one of the equations by a constant so that there is one variable in each equation that has the same coefficient. Subtract one equation from the other. Solve the resulting equation for the one variable. (Use the technique described in previous units.) Using the value found in the step 4, substitute it into either equation and solve for the remaining variable. Substitute the values for both variables into the equation not used in step 5 to be sure our solution is correct.

While there seems to be a lot of steps to this process, it really is rather simple. Let’s work through one example to see how the steps work. We have two equations below:

P = 2 + 2Q
P
= 10 – 6Q

To solve for the numeric value of each variable, we should follow the steps outlined above.

1. Algebraically manipulate both equations so that all the variables are on one side of the equal sign and in the same order. (Line the equations up, one on top of the other.)

 When we do this we get the equations shown at the right. Notice how the terms containing P are lined up, as are the terms containing Q. P – 2Q = 2 P + 6Q = 10

1. If needed, multiply one of the equations by a constant so that there is one variable in each equation that has the same coefficient.

The variable P in both equations has a coefficient of one, so we move on to the next step.

1. Subtract one equation from the other.

1. Solve the resulting equation for the one variable.

1. Using the value found in the step 4, substitute it into either equation and solve for the remaining variable.

 In the previous step we found that Q = 1. We plug this value back into either of our initial equations and solve for P. P = 2 + 2Q P = 2 + 2(1) P = 4

1. Substitute the values for both variables into the equation not used in step 5 to be sure our solution is correct.

 Now we substitute the value of Q = 1 and P = 4 into the equation not used in step 5. P = 10 – 6Q 4 = 10 – 6(1) 4 = 4.

These values work in this equation so we have the correct answer.

Keep in mind that you can use either method for solving systems of equations. Let’s revisit the example we used in the last section, but this time solve it using addition/subtraction.

Example
Solve the system of linear equations given below using addition/subtraction.

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
0.10D + 0.25Q = 9.45

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

The information in this example can be 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)

W
e were given the two equations:

D + Q = 57     (1)
0.10D + 0.25Q = 9.45     (2)

1. Algebraically manipulate both equations so that all the variables are on one side of the equal sign and in the same order.

 When we do this we get the equations shown at the right. Notice how the terms containing D are lined up, as are the terms containing Q. D + Q = 57 0.10D + 0.25Q = 9.45

2. If needed, multiply one of the equations by a constant so that there is one variable in each equation that has the same coefficient.

Because neither of the variables has the same coefficient in both equations, we will need to multiply one equation by a number that will result in the variable in one equation having the same coefficient as in the other. In this example we can see the coefficient of Q in equation (1) is 4 times larger than the coefficient of Q in equation (2).

 If we multiply the second equation by 4, then the variable Q will have the same coefficient in both equations. This means that both sides of equation (2) must be multiplied by 4. 0.10D + 0.25Q = 9.45 4(0.10D + 0.25Q = 9.45) 4(0.10)D + 4(0.25)Q = 4(9.45) 0.40D + 1.00Q = 37.8

3. Subtract one equation from the other.
 Now we subtract one equation from the other. D + Q = 57 - (0.40D + Q = 37.8) 0.60D = 19.2

4. Solve the resulting equation for the one variable.
 We now solve the equation from step 3 to find a value for the D variable.

5. Using the value found in the step 4, substitute it into either equation and solve for the remaining variable.

 Now we substitute the value of D = 32 into one of the equations to solve for Q. D + Q = 57 32 + Q – 32= 57 – 32 Q = 25

6. Substitute the values for both variables into the equation not used in step 5 to be sure our solution is correct.

 Now we substitute the value of D = 32 and Q = 25 into the equation not used in step 5. 0.10D + 0.25Q = 9.45 .10(32) + .25(25) = 9.45 3.20 + 6.25 = 9.45 9.45 = 9.45

So we find that Q = 25 and D = 32. If you look back at the last section you will find this is the same result we found then.

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