**Q: Solve by using the substitution and elimination method:**

**(1) 3x – 4y = -1
(2) -10 + 8y = 6x**

A: Ok, this is 2 questions in 1. We need to solve it 2 different ways. Let’s first start with the substitution method:

I am going to solve for x in equation (2). No reason that I picked equation (2) and no reason that I decided to solve for x (it could have been y). I just had to pick one…. So, solving for x in equation (2):

-10 + 8y = 6x

[divide everything by 6]

-10/6 + 8/6y = 6x/6

[reduce the fractions]

-5/3 + 4/3y = x

OKAY, now look at equation (1):

3x – 4y = -1

We are going to put the blue equation that we solved for (from equation 2), into the x in equation (1) like so:

3(-5/3 + 4/3y) – 4y = -1

Now, multiply through:

-15/3 + 12/3y -4y = -1

Reduce:

-5 + 4y – 4y = -1

Simplify:

-5 = -1

WHAT?!?! -5 = -1? Can’t be. We know that isn’t true. Therefore, this means that there are **no solutions** to this system of equations.

OKAY, now let’s try solving using the elimination method. We should get the same answer, we just go about solving for it a different way:

(1) 3x – 4y = -1

(2) -10 + 8y = 6x

Re-arrange equation (2) so that the x’s and y’s are under those of equation (1)… I am going to subtract the 6x to the left side, and add the 10 to the right side to get:

-6x + 8y = 10

So:

(1) 3x – 4y = -1

(2) -6x + 8y = 10

Now, I am going to multiply equation (1) by 2 to each number:

(1) 2[3x – 4y = -1]

(2) -6x + 8y = 10

Simplify to get:

(1) 6x – 8y = -2

(2) -6x + 8y = 10

Add the two equations together… notice the x’s and y’s cancel out on the left side? So you get:

0 = 8

What?? 0 does not equal 8. Therefore, again, we find the answer is **no solutions!**