Systems of Linear Equations

A system of equations is a set of two or more equations that contain multiple variables.  

For example:


When given a problem with variables, it will require n equations to solve. The majority of systems of equations questions on the SAT will ask you to solve for two variables using two equations.

For two-variable systems, there are three types:

One Solution

These are the traditional systems. They won’t always tell you there’s one solution, but if you’re simply asked to solve for one of the variables, then it’s this.



For the system of equations above, what is the value of x + y?

A) 2
B) 3
C) 4
D) 5


  1. Underline what you are being asked to solve for (the x + y in the example).
  2. If the answer choices are all coordinates ( 3, 5), then one at a time plug them into the two equations and find which one creates two working equations.
  3. If the answers aren’t coordinates, choose to use elimination or substitution to solve (elimination and substitution are covered on the next page).

What one solution looks like on a graph:


Use substitution when it is ridiculously easy to get one equation to look like \(y=\) or \(x=\). Ridiculously easy means using addition or subtraction.

To use substitution, you must isolate a variable and get the coefficient of the variable in one of the equations to be 1 (i.e. \(x=3+2y\)).  


We use substitution to rewrite the second equation as:


Then we are simply solving for the lone variable.

Elimination / Combination / Subtraction

Combination involves adding, or subtracting, one equation to/from the other equation and in the process eliminating one of the variables. This method only works when we make the coefficients of one of the variables (both x’s or both y’s) equal.  


By multiplying our first equation by two, we can rewrite our equations as:


Now the coefficients for \(y\) are equal, so if we subtract one equation from the other, we will eliminate all of the \(y\)s.


To finish our solution completely, we plug \(x=6\) in to either equation and solve for \(y\).