## Algebra on the GRE

# Using substitution to solve systems of linear equations on the GRE

You may be asked on the GRE to solve simultaneous linear equations. Our previous article on the equations of lines showed the method of elimination to do this. In this post, we'll show you another method: substitution. It is always good to have a variety of tools up your sleeve.

Consider the following two linear equations from the __Official Guide__:

*1) 4x + 3y = 132) x + 2y = 2*

Equation 2 looks a little simpler, so let’s start with that.

1. Get an expression for x

x + 2y = 2

→ x = 2 – 2y

2. Substitute the expression you have for x into equation 1

4x + 3y = 13

→ 4 *(2 - 2y)* + 3y = 13

→ 8 – 8y + 3y = 13

→ 8 – 5y = 13

→ 8 – 13 = 5y

→ 5y = -5

→ y = -1

3. Replace your value for y into one of the equations to get a value for x

x + 2y = 2

→ x + 2*(-1)* = 2

→ x -2 = 2

→ x = 4

4. Put the values for x and y into the other equation to make sure you are right

4x + 3y = 13

→ 4(4) + 3(-1) = 13

→ 16 – 3 = 13

The equation is true, so you have done everything correctly.

You may be thinking that that is all very nice, but *what is the actual use of doing such a series of calculations?*

One potential application occurs when you consider plotting the lines on the xy-plane. Remember at the start we said the two equations are linear equations. This means that if you draw them on the xy-plane, they are lines. By solving the two simultaneous linear equations you have worked out where these two lines intersect. They intersect at the point with the coordinates (4, -1).

You now have a second method for solving linear equations and a typical application of what you have done. You can calculate the point at which two lines meet!