Lesson Objectives
  • Demonstrate an understanding of Systems of Linear Equations in Two Variables
  • Learn how to solve a System of Linear Equations in Two Variables using Elimination
  • Learn how to solve a System of Linear Equations in Two Variables with No Solution
  • Learn how to solve a System of Linear Equations in Two Variables with Infinitely Many Solutions

How to Solve a System of Linear Equations using Elimination


So far, we have learned how to solve a system of linear equations in two variables using graphing and substitution. Our graphing method involves graphing each equation and looking for the point of intersection. This point lies on both lines and is, therefore, a solution for our system. We found that the graphing method was very inefficient and didn’t work well for non-integer values. We also learned how to use the substitution method. This method involves solving one of the equations of the system for one of the variables. We can then plug in for that variable in the other equation. This allows us to obtain a linear equation in one variable and find one of our unknowns. We can finish up the process by plugging in our known value into one of the equations and solving for the final unknown.
Another Algebraic method used to solve a system of linear equations in two variables is known as elimination. The elimination method gets its name from a process where one of the variables gets eliminated, and we are left with a linear equation in one variable. Let's take a look at the steps used in the elimination method.

Elimination Method

  • Write both equations in standard form
  • Transform the coefficients of one pair of variable terms into opposites
  • Add the two left sides of the equations together, set this equal to the sum of the two right sides of the equations
  • Solve the linear equation in one variable
  • Find the other unknown using substitution
  • Check
Let's take a look at an example.
Example 1: Solve each system using the elimination method.
-7x + 9y = -15
2x - 9y = -15
Let's label our equations as:
1) -7x + 9y = -15
2) 2x - 9y = -15
Step 1) Write both equations in standard form. Standard form varies a lot between textbooks. For our purposes, we just want the x term and y term on one side and the constant on the other. It helps to have integers. In our case, each equation is already in standard form.
Step 2) Transform the coefficients of one pair of variable terms into opposites. For this example, we can see that we have 9y and (-9y). The coefficients for y are opposites. This means if we add (9y + (-9y)), we will get 0. In other words, the y variable will be eliminated.
Step 3) Add the two left sides and set this equal to the sum of the two right sides. $$-7x + 9y = -15$$ $$\underline{+ 2x - 9y = -15}$$ The 9y + (-9y) will cancel: $$\require{cancel}-7x + \cancel{9y} = -15$$ $$\underline{+ 2x - \cancel{9y} = -15}$$ On the left, we add (-7x + 2x), this gives us -5x. On the right, we add -15 + (-15), this gives us (-30). $$-5x = -30$$ Step 4) Solve the linear equation in one variable
-5x = -30
x = 6
Step 5) Find the other unknown using substitution. We can plug in a 6 for x in either original equation. Let's plug into equation 2:
2(6) - 9y = -15
12 - 9y = -15
-9y = -15 - 12
-9y = -27
y = 3
Our solution for the system is (6,3)
Step 6) Check
-7(6) + 9(3) = -15
-42 + 27 = -15
-15 = -15
2(6) - 9(3) = -15
12 - 27 = -15
-15 = -15
Example 2: Solve each system using the elimination method.
-4x - 7y = 16
x + y = 2
Let's label our equations as:
1) -4x - 7y = 16
2) x + y = 2
Step 1) Write both equations in standard form. In our case, each equation is already in standard form.
Step 2) Transform the coefficients of one pair of variable terms into opposites. If opposites are not present, we can pick whichever variable looks easier to eliminate. In this case, we will choose to eliminate the x variable. One coefficient is 1 and the other is (-4). We can change equation 2 by multiplying both sides by 4:
2) 4x + 4y = 8
Now we have (-4x) and 4x. When we add our equations, the x variable will be eliminated.
Step 3) Add the two left sides and set this equal to the sum of the two right sides. $$-4x - 7y = 16$$ $$\underline{+ 4x + 4y = 8}$$ The 4x + (-4x) will cancel: $$\cancel{-4x} - 7y = 16$$ $$\underline{\cancel{+ 4x} + 4y = 8}$$ On the left, we add (-7y + 4y), this gives us -3y. On the right, we add 16 + 8, this gives us (24). $$-3y = 24$$ Step 4) Solve the linear equation in one variable
-3y = 24
y = -8
Step 5) Find the other unknown using substitution. We can plug in a (-8) for y in either original equation. Let's plug into equation 2:
x + (-8) = 2
x = 10
Our solution for the system is (10,-8)
Step 6) Check
-4(10) - 7(-8) = 16
-40 + 56 = 16
16 = 16
10 + (-8) = 2
2 = 2
Example 3: Solve each system using the elimination method.
9 = 3y - 5x
-2x = -16 - 4y
Let's label our equations as:
1) 9 = 3y - 5x
2) -2x = -16 - 4y
Step 1) Write both equations in standard form:
1) -5x + 3y = 9
2) -2x + 4y = -16
Step 2) Transform the coefficients of one pair of variable terms into opposites. If opposites are not present, we can pick whichever variable looks easier to eliminate. In this case, we will have to modify both equations in order to create opposite coefficients. We have a coefficient on y in equation 1, which is 3 and a coefficient on y in equation 2, which is 4. We can multiply both sides of equation 1 by 4:
1) -20x + 12y = 36
Now we can multiply both sides of equation 2 by (-3):
2) 6x - 12y = 48
Our y variable will now be eliminated when we add equations. We have opposite coefficients of (12 and -12).
Step 3) Add the two left sides and set this equal to the sum of the two right sides. $$-20x + 12y = 36$$ $$\underline{+ 6x - 12y = 48}$$ The 12y + (-12y) will cancel: $$-20x + \cancel{12y} = 36$$ $$\underline{\hspace{.25em}+6x - \cancel{12y} = 48}$$ On the left, we add (-20x + 6x), this gives us -14x. On the right, we add 36 + 48, this gives us (84). $$-14x = 84$$ Step 4) Solve the linear equation in one variable
-14x = 84
x = -6
Step 5) Find the other unknown using substitution. We can plug in a (-6) for x in either original equation. Let's plug into equation 1:
9 = 3y - 5(-6)
9 = 3y + 30
-21 = 3y
y = -7
Our solution for the system is (-6,-7)
Step 6) Check
9 = 3(-7) - 5(-6)
9 = -21 + 30
9 = 9
-2(-6) = -16 - 4(-7)
12 = -16 + 28
12 = 12

Inconsistent Systems - Elimination Method

At this point, we should know that there are linear systems which will not have a solution. This occurs when we have two parallel lines. Parallel lines have the same slope and will never intersect. This means there will never be a point which lies on both lines and there can't be a solution for the system. We saw with the substitution method, that solving an inconsistent system or a system with no solution leads to a false statement. We will see the same result when we use the elimination method. A system with no solution will lead to the variables dropping out and a false statement. Let's look at an example.
Example 4: Solve each system using the elimination method.
-33x = 36y + 25
0 = -16 - 24y - 22x
Let's place our equations in standard form and label each:
1) -33x - 36y = 25
2) -22x - 24y = 16
Now we can look to eliminate one of our variables. We do this by transforming one or both equations such that one pair of coefficients will be opposites. Let's choose to eliminate x. We see one coefficient as -33 and the other as -22. We can multiply -33 by 2 and obtain -66. We can multiply -22 by -3 and obtain +66. Let's transform each equation:
1) -66x - 72y = 50
2) 66x + 72y = -48
We can already see there will be problems. Notice how we have opposite coefficients on both x and y. Let's perform our addition. We want to add the left sides and set this equal to the sum of the right sides.
$$-66x - 72y = 50$$ $$\underline{+66x - 72y = -48}$$ We will see both variables cancel out: $$\cancel{-66x - 72y} = 50$$ $$\underline{\cancel{+66x - 72y} = -48}$$ We will end up with 0 on the left and 2 (50 - 48 = 2) on the right:
0 = 2 (false)
When we see a false statement, we know we have an inconsistent system. We can say there is no solution for the system.

Dependent Equations - Elimination Method

We have also seen a system with dependent equations. This means we have a system with the same two equations involved. The equations have been algebraically manipulated to look different, but they are the same. Since a linear equation in two variables has an infinite number of solutions, we can say there is an infinite number of solutions for this type of system. We will know we have dependent equations when our variables drop out and we are left with a true statement. Let's look at an example.
Example 5: Solve each system using the elimination method.
30y = 60x + 30
2x = -1 + y
Let's place our equations in standard form and label each:
1) -60x + 30y = 30
2) 2x - y = -1
Before we move further, notice that we can divide both sides of equation 1 by -30 and obtain equation 2:
For the sake of demonstration, let's eliminate one of the variables. We can eliminate both x and y by multiplying both sides of equation 2 by 30.
2) 60x - 30y = -30
Now let's add the left sides of the equations and set this equal to the sum of the right sides.
$$-60x + 30y = +30$$ $$\underline{+60x - 30y = -30}$$ We can see that everything will be eliminated and we will be left with 0 = 0. $$\cancel{-60x + 30y} = \cancel{+30}$$ $$\underline{\cancel{+60x - 30y} = \cancel{-30}}$$ $$0 = 0$$ Since 0 = 0 is a true statement, we know we have dependent equations. We state that this system has an infinite number of solutions.