There are two ways to solve a system of linear equations. This post focuses on the elimination method. There are two examples of **using the elimination method to solve a system of equations** in this post.

When **solving a system of two equations with elimination**, you want look at the x and y terms. Be on the lookout for variables that have the same coefficient with the same or opposite signs. This method’s name describes it accurately. When you use elimination, you “eliminate” one of the variables by adding or subtracting the given equations together. The variable terms with the same coefficient is eliminated. The signs of the variable terms to be eliminated determine whether the equations are added or subtracted. If the signs are the same, you subtract the equations. You add the equations when the signs of the variable terms are different.

After eliminating one of the variables, you have a simple one step equation to solve. The resulting equation should involve multiplication of a real number. The method to solve the equation will depend on the type of coefficient. Once the value of the first variable is determined, you will need to substitute the value you just found into one of the original equations and solve for the other variable.

Now you have the values of the ordered pair and should be a solution to both of the original equations. You really should test the values to ensure an accurate answer.

### Example 1 Solve the System of Equations with the Elimination Method

x + 2y = 3 (1)

4x – 2y = 7 (2)

Looking at the equations, it is clear that the y-terms have the same coefficients with opposite signs. Thus, the equations 1 and 2 should be added term by term to get:

5x = 10 (3).

This simple equation only takes one step to solve. To isolate x, use the division property of the equality to give

x = 2 (4).

Use the result in equation 4 and substitute it into either equation 1 or 2. Using equation 1,

2 + 2y = 3 (5).

Subtracting 2 from both sides of equation 5 gives,

2y = 3 (6).

To finish up take equation 6 and divide it by 2 and

y = ½ or 0.5 (7).

The ordered pair that will satisfy both of the original equations is (2, 0.5).

Remember how this related to the graph of a system of equations. These equations have one unique solution, which means (2,0.5) is the point where the graphs will cross. To check the solution, choose one of the original equations and do the math. Using equation 1 and (2,0.5).

4(2) – 2(0.5) = 7 (8) –> substitute the x- and y-values into equation 1

8 – 1 = 7 (9) –> multiply

7 = 7 (10) –> subtract

In equation 10, 7 is equal to 7 and thus the ordered pair is the solution to the system of equations:

x + 2y = 3 (1)

4x – 2y = 7 (2)

### Example 2 Solve the System of Equations with the Elimination Method

5x + 3y = -19 (1)

8x + 3y = 25 (2)

Filed under: Algebra 2, Equations and Graphs, Solving Systems of Equations Tagged: | 2 variable system of equations, Algebra 2, solve a system of equations with 2 variables, solve a system of equations with elimination method, Solving a system of equations, solving a system with 2 equations

M1LK3Y, on February 21, 2012 at 8:09 pm said:AWSOME!!!

bogdan, on November 1, 2012 at 12:25 am said:thnx but i need help with one step and two step equations

bogdan, on November 1, 2012 at 12:27 am said:but still ty 🙂

Mr. Pi, on November 1, 2012 at 10:03 pm said:You are welcome. I hope to add more algebra 1 to my site, but I just don’t have time right now.