Solve linear equations #2 (2 variables - Algebra 1)

In this article, I will discuss how you can solve linear equations but this time I will be showing you how to solve linear equations using the linear combination method.  No matter what method you are using, whether it be the substitution method or linear combination method (or matrix as the students in Algebra 2 will be learning), the end result should ALWAYS be the same. 

The idea behind linear combination method is to manipulate one of the equations by multiplying or dividing by some constant so that you can add/subtract the 2 equations.  Once you have added/subtracted the equations together, you will end up with one equation with one variable.  And solving one equation with one variable should become second nature to you at this point (if not, please reach out to your teachers, or even call us for additional help).

Let us use the same equations that we have used for the substitution method just to prove that the solution to the linear combination method HAS TO yield the same answer.  Let us recall that the equations are:

equation 1:  3x - 5y = 3
equation 2:  9x - 20y = 6

If I can multiply the first equation by 3, I will have the following:

equation 1:  3(3x - 5y = 3)
new equation 1':  9x - 15y = 9

equation 1':  9x - 15y = 9
equation 2:  9x - 20y = 6

Let us subtract equation 1' from equation 2 (1' - 2)

9x - 9x - 15y - (-20y) = 9 - 6
5y = 3
y = 3/5

Substitute y = 3/5 into either equation 1 or equation 2

equation 1:  3x - 5y = 3
3x - 5(3/5) = 3
3x - 3 = 3
3x = 6
x = 2

Solution is (2, 3/5).  As you can see from this answer, linear combination and substitution method yields the identical solution as it should.  Click here to see the substitution method solution.

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