Solving Linear Systems: Elimination

The third method for solving systems of linear equations is elimination.

Elimination is probably the method to choose if you cannot easily solve one of the equations in the system for a variable.

The goal of elimination is to eliminate one of the variables so that you can solve for the other one.

Steps to solving by elimination:

  1. Choose the variable you wish to eliminate.
  2. Multiply or divide the equations by constants so that the coefficients of the chosen variable are the same except one is negative and one is positive.
  3. Add the equations together.
  4. Solve the resulting equation for the remaining variable.
  5. Substitute that value into one of the original equations and solve for the remaining variable.

Example 3

Solve the system below using elimination.
2x - 5y = 10
4x + 3y = 7

Step 1: Since you can multiply 2x by -2 to get -4x, choose to eliminate the x.

Step 2: Multiply the first equation by -2:
-4x + 10y = -20
4x + 3y = 7

Step 3: Add the equations:
-4x + 10y = -20
4x + 3y = 7
13y = 13

Step 4: Solve for y:
13y = -13
y = -1

Step 5: Substitute y = -1 into one of the original equations and solve for x.
2x - 5y = 10
2x - 5(-1) = 10
2x + 5 = 10
2x = 5
x = 5

So the solution is (5/2, -1). Check your solution in both of the original equations.

Checking:
2(5/2) - 5(-1) = 10
5 + 5 = 10
10 = 10

4(5/2) + 3(-1) = 7
10 - 3 = 7
7 = 7

Since (5/2, -1) makes both equations true, it is the solution to the system.

 

Example 4

Solve the system below using elimination.
3x + 4y = -11
2x - 5y = 8

Step 1: In this case you will need to multiply both equations by a number to eliminate a variable. Let's choose y.

Think: By what do I need to multiply each equation to get coefficients of y that will add to zero? That is, how do I make the coefficients the same, but have one that is positive and one that is negative?

Step 2: Multiplying equation 1 by and equation 2 by will make the coefficients of y equal to 20 and -20 respectively.

Answer: 5, 4

After multiplying, the new equations are:

x + y =

x - y =

Answer: 15x + 20y = -55, 8x -20y = 32

Step 3: Add the 2 new equations together:

   15x + 20y = -55
+   8x - 20y = 32

x =

Answer: 23x = -23

Step 4: Solve for x.

23x = -23

x =

Answer: x = -1

Step 5: We found that x = -1. Substitute into one of the original equations and solve for y. Let's substitute into equation 2.

2(-1) - 5y = 8
-2 - 5y = 8
-5y = 10
y = -2
Therefore the solution to the system is ( , )

Answer: (-1, -2)

 

Example 5

Step 1: Since the second equation is already solved for y, substitute it into the first equation:

6x + 4(-3/2x + 5) = 10

6x + x + = 10

Answer: 6x+ (-6)x + 20 = 10

Step 2: Continue simplifying:

6x+ (-6)x + 20 = 10

=

Answer: 20 = 10

The final equation has no variables and is FALSE. When this happens, the system is inconsistent and has no solution. The graph of this system would be parallel lines.

Example 6

Solve the system below using elimination.
3y = 2x - 1
4x - 6y = 2

Step 1: First, let's put both equations in the same form. Subtract -2x from both sides of the first equation:

-2x + 3y = -1
4x - 6y = 2

We can eliminate x by multiplying the first equation by .

Answer: 2

We can eliminate x by multiplying the first equation by 2. We get:
-4x + 6y = -2
4x - 6y = 2
0 = 0

After adding we find that the final equation has no variables and is TRUE. When this happens, the system is a dependent system and has infinite solutions. The graphs of the equations in this system would be the same line.



 

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