How to Solve an Algebra Problem by Substitution

On standardized tests, you might be given two algebra equations and asked to solve for variables x and y. One method for solving is substitution. Substitution is particularly effective when one variable is expressed directly in terms of the other. Examples include y = 3x + 6 or x = 4y + 7. The principle behind substitution is to eliminate one variable by substituting a form of it, as expressed by the other equation.

Instructions

1. Pick an example problem. Solve for x and y when y = 3x + 1 and x + y = 13.

2. Here y is directly expressed as 3x + 1. You can therefore substitute y = 3x + 1 into the other equation. So x + (3x + 1) = 13.

3. Solve algebraically for x.x + (3x + 1) = 13x + 3x + 1 = 134x + 1 = 13 -1 = -14x = 124x / 4 = 12 / 4x = 3

4. Solve for y by substituting the value of x (x = 3) into the first equation.y = 3x + 1y = 3(3) + 1y = 9 + 1y = 10. So x = 3 and y = 10.

5. Check your answers by substituting the values of x and y into both algebra equations. x = 3 and y = 10.y = 3x + 110 = 3(3) + 110 = 9 +110 = 10. The first equation checks out.

6. Now check the second equation.x + y = 133 + 10 = 1313 = 13. The second equation checks out.

Tips & Warnings

• Substitution is also effective when one of the variables lacks a coefficient. For example, take the second equation x + y = 13. By subtracting x, you would have y = 13 - x. Alternatively by subtracting y, you would have x = 13 - y, either of which could be substituted into the other equation.

• In the original example, by substituting 3x + 1 into the first equation, you were able to solve x + (3x + 1) by simply removing the parentheses. However, if the sign before the parentheses had been negative, you would have had to change the signs of both terms inside the parentheses. So x – (3x + 1) would have been x – 3x – 1. If the equation had been x - (-3x -1) the result would have been x + 3x + 1.