Lesson Objectives
  • Demonstrate an understanding of how to solve a system of linear equations in two variables using graphing
  • Demonstrate an understanding of how to solve a system of linear equations in two variables using substitution
  • Demonstrate an understanding of the addition property of equality
  • Learn how to solve a system of linear equations in two variables using elimination
  • Learn how to identify a system of linear equations in two variables with no solution
  • Learn how to identify a system of linear equations in two variables with infinitely many solutions

How to Solve a Linear System using Elimination


So far, we have learned how to solve a system of linear equations in two variables using graphing and substitution. In this lesson, we will learn another method known as "elimination". The goal of the elimination method is to eliminate one of the variables and obtain a linear equation in one variable. This allows us to get a solution for one of the variables. We can then use substitution to find the other unknown. The elimination method works best when one pair of variable terms are opposites, this means they will have opposite coefficients.

Solving a Linear System using the Elimination Method

  • Place both equations in standard form
    • ax + by = c
  • Transform one or both equations in such a way that one pair of variable terms are opposites
    • This means the variables will have opposite coefficients
  • Add the left sides of the equations and set this equal to the sum of the right sides, the result is a linear equation in one variable
    • This is legal since we are adding equal quantities to both sides of an equation
    • If a = b and c = d, then a + c = b + d
  • Solve the equation and find the other unknown using substitution
  • Check
      Plug in for x and y in both original equations
Let's look at a few examples.
Example 1: Solve each linear system using elimination
-5x + 3y = 18
5x - 2y = -12
Step 1) Place both equations in standard form.
In this case, both equations are in standard form. We will label our equations as 1 and 2.
1) -5x + 3y = 18
2) 5x - 2y = -12
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
In this case, we have -5x and 5x, which are opposites.
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$-5x + 3y = -12$$ $$\underline{+5x - 2y = -12}$$ The -5x and 5x will cancel: $$\require{cancel}\cancel{-5x} + 3y = 18$$ $$\underline{\cancel{+5x} - 2y = -12}$$ $$\hspace{4em}y = -24$$ On the left, we are left with (3y) + (-2y), which is just (y). On the right, we have (18) + (-12), which is (6).
Step 4) Solve the equation and find the other unknown.
In this case, we already know that y is 6. Let's plug this in for y in one of the original equations. Let's choose to plug in for y in equation 1.
-5x + 3y = 18
-5x + 3(6) = 18
-5x + 18 = 18
-5x = 0
x = 0
Our solution is the ordered pair (0,6).
Step 5) Check.
-5x + 3y = 18
-5(0) + 3(6) = 18
18 = 18
5x - 2y = -12
5(0) - 2(6) = -12
-12 = -12
Example 2: Solve each linear system using elimination
4x + 10y = -22
5y = 5x - 25
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) 4x + 10y = -22
2) -5x + 5y = -25
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
In this case, we can multiply both sides of equation 2 by (-2). This will give us (10y) in equation 1 and (-10y) in equation 2.
2) 10x - 10y = 50
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$\hspace{.5em}4x + 10y = -22$$ $$\underline{10x - 10y = 50}$$ The 10y and (-10y) will cancel: $$\hspace{.5em}4x + \cancel{10y} = -22$$ $$\underline{10x - \cancel{10y} = 50}$$ $$14x \hspace{3em} = 28$$ On the left, we are left with (10x) + (6x), which gives us (14x). On the right, we have (-22) + (50), which is (28).
This leaves us with the equation:
14x = 28
Step 4) Solve the equation and find the other unknown.
14x = 28
x = 2
Let's plug a (2) in for x in equation 1.
4x + 10y = -22
4(2) + 10y = -22
(8) + 10y = -22
10y = -22 - 8
10y = -30
y = -3
Our solution is the ordered pair (2,-3).
Step 5) Check.
4x + 10y = -22
4(2) + 10(-3) = -22
8 + (-30) = -22
-22 = -22
5y = 5x - 25
5(-3) = 5(2) - 25
-15 = 10 - 25
-15 = -15
Example 3: Solve each linear system using elimination
-9y = 2x - 16
-7x = 22 + 12y
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) -2x - 9y = -16
2) -7x - 12y = 22
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
We can multiply equation 1 by (7) and equation 2 by (-2). This will give us (-14x) in equation 1 and (14x) in equation 2.
1) -14x - 63y = -112
2) 14x + 24y = -44
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$-14x - 63y = -112$$ $$\underline{+14x + 24y = -44}$$ The -14x and 14x will cancel: $$\cancel{-14x} - 63y = -112$$ $$\underline{\cancel{+14x} + 24y = -44}$$ $$\hspace{3.2em}-39y = -156$$ On the left, we are left with (-63y) + (24y), which is (-39y). On the right, we have (-112) + (-44), which is (-156).
This leaves us with the equation:
-39y = -156
Step 4) Solve the equation and find the other unknown.
-39y = -156
y = 4
Let's plug a 4 in for y in equation 1.
-2x - 9y = -16
-2x - 9(4) = -16
-2x - 36 = -16
-2x = 20
x = -10
Our solution is the ordered pair (-10,4).
Step 5) Check.
-9y = 2x - 16
-9(4) = 2(-10) - 16
-36 = -20 - 16
-36 = -36
-7x = 22 + 12y
-7(-10) = 22 + 12(4)
70 = 22 + 48
70 = 70

Special Case Linear Systems

At this point, we have run across special case scenarios while using graphing and substitution methods. The majority of our problems with linear systems will have exactly one solution. This type of system is said to be "consistent", with equations that are "independent". When we have a system which contains two parallel lines, there will not be a solution. This type of system is said to be "inconsistent". Additionally, we will see systems that contain the same equation. These equations are said to be "dependent". For this type of system, there are infinitely many solutions.

Linear Systems with No Solution

When using the elimination method, if both are variables are eliminated and we are left with a false statement, we know we have an inconsistent system. Our answer will be stated as "no solution" or ∅. Let's look at an example.
Example 4: Solve each linear system using elimination
16 = 55x + 10y
-22x - 4y = -4
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) 55x + 10y = 16
2) -22x - 4y = -4
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
We can multiply equation 1 by (2) and equation 2 by (5). This will give us (20y) in equation 1 and (-20y) in equation 2.
1) 110x + 20y = 32
2) -110x - 20y = -20
We should be able to see a problem. Both variables are set up such that they have opposite coefficients. If we add the left sides together and set this equal to the sum of the right sides, we will get a false statement.
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$+110x + 20y = 32$$ $$\underline{-110x - 20y = -20}$$ The 110x and (-110x) will cancel and the 20y and (-20y) will also cancel: $$\cancel{+110x} + \cancel{+20y} = 32$$ $$\cancel{-110x} + \cancel{-20y} = -20$$ $$\hspace{6.65em}0 = 12$$ On the left, we are left with (0). On the right, we have (32) + (-20), which is (12).
0 = 12 (false)
We can stop and report our answer as "no solution".

Solving Linear Systems with Infinitely Many Solutions

When using the elimination method, if both are variables are eliminated and we are left with a true statement, we know we have dependent equations. Our answer will be stated as "infinitely many solutions". Let's look at an example.
Example 5: Solve each linear system using elimination
15y + 3 = -24x
-150y - 30 = 240x
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) 24x + 15y = -3
2) -240x - 150y = 30
At this point, we should be able to tell that multiplying equation 1 by (-10) will give us equation 2. Therefore, both equations of the system are the same and we will have infinitely many solutions. To show the full process, let's continue with the steps.
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
We can multiply equation 1 by (10). This will give us (240x) and (-240x).
1) 240x + 150y = -30
2) -240x - 150y = 30
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$+240x + 150y = -30$$ $$\underline{-240x - 150y = 30}$$ The 240x and (-240x) will cancel and the 150y and (-150y) will also cancel: $$\cancel{+240x} + \cancel{+150y} = -30$$ $$\cancel{-240x} + \cancel{-150y} = 30$$ $$\hspace{7.25em}0 = 0$$ On the left, we are left with (0). On the right, we have (30) + (-30), which is (0).
0 = 0 (true)
We can stop and report our answer as "infinitely many solutions".