Solve 3 linear equations (Algebra 2 using linear combination method)

In the previous article, I showed you how to solve a system of 3 linear equations by using the substitution method.  You can also solve this same system of linear equations using the linear combination method.  Just to recap, the idea behind linear combination method is to find ways to cancel out one of the variable at time, so you will end up only trying to solve one variable at a time.

We will use the same set of equations to prove that no matter what method you choose to use, you should and MUST get the same answer.  The system of linear equations that we are trying to solve is:

equation #1:  4x + y + 6z = 7
equation #2:  3x + 3y + 2z = 17
equation #3:  -x -y +z = -9

We will try to get rid of the variable "y" first. 

equation 4 = equation 1 + equation 3
equation 4:  3x + 7z = -2

Multiply equation 1' by 3.
equation 1': 
3[4x + y + 6z = 7]
12x + 3y + 18z = 21

equation 5 = equation 1' - equation 2
equation 5: 
12x + 3y + 18z - (3x + 3y + 2z) = 21 - 17
9x + 16z = 4

Solve equation 4 and equation 5

equation 4:  3x + 7z = -2
equation 5:  9x + 16z = 4

Multiply equation 4 by 3:
3 x equation 4:  3(3x + 7z = -2)
equation 4':  9x + 21z = -6

Subtract equation 5 - equation 4':

9x + 16z - (9x + 21z) = 4 - (-6)
-5z = 10
z = -2

Substitute z = -2 into equation 4 or equation 5

equation 4:  3x + 7z = -2
3x + 7(-2) = -2
3x = 12
x = 4

Substitute z = -2 and x = 4 into equation 1 or equation 2 or equation 3:

equation #3:  -x -y +z = -9
-4 -y -2 = -9
-y = -3
y = 3

Solution is the same as the substitution method (4, 3, -2).

If you have any question regarding this type of problems, please feel free to reach out to me or any of the instructors in my center.

Michael Huang
Center Director
Mathnasium of Glen Rock/Ridgewood
T:     201-444-8020   
E:  glenrock@mathnasium.com
www.mathnasium.com/glenrock