How do you solve 6x-3y equals -3 and 8x-4y equals -4 using the linear combination method?

Answer 1

Set up your equations as an addition, lining up like variables:

6x - 3y = -3

8x - 4y = -4

First we want to eliminate one of the variables. We do this by getting the scalar multiples for one of the variables to match in both equations.

In this example, we can do this by simplifying both equations:

Divide the top equation by 3, divide the bottom equation by 4:

6/3x - 3/3y = -3/3

8/4x - 4/4y = -4/4

~

2x - y = -1

2x - y = -1

Right now, you can already guess that something is wrong, but we will forge on ahead anyways:

In order to get one of the variables eliminated, we have to multiply one of the equations by -1, or we can just simply subtract the equations instead of adding them:

2x - y = -1

+ -2x +y = 1)

------------------

0x +0y = 0

Huh. Isn't that strange - when we eliminated one of the variables, we also eliminated the other variable. What we are left with is 0=0.

What this means is that the two equations are linearly dependent, and they in fact overlap each other.

HOWEVER, because we are left with 0=0, the system of equations is consistent, so there IS a solution, albeit the solution will be infinite.

NOTE: if we had instead been left with a contradiction, such as 0 = 1, or 0=-1 or 0=4, we would say that the system is inconsistent and there is no solution. If you want to be technical, you would say that the solution is the empty set, which you can denote with ø or {}.

Since the two lines overlap each other completely, there is in fact only one line in the system. Simply pick one of the equations and simplify. As we already showed above, both equations simplify to the same linear equation:

2x - y = -1

We may want to isolate one of the variables:

2x = -1 + y

~

2x + 1 = y

So our solution is the line:

y = 2x + 1

Where x is free.

You could also pick y to be your free variable, and write the solution as:

x = -1/2 + 1/2y