Substitution Method vs. Eliminations Methond

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!

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