# Question 85ee6

Dec 12, 2016

Oh, hey! I learned this last year!

A) SUBSTITUTION EQUATIONS
So, you will be given two equations.
Make 2x−3y=−1 as equation 1.
Make y=x−1 as equation 2.

Think carefully whether to substitute Equation 1 into Equation 2 OR Equation 2 into Equation 1.
Always look at two equations for like terms - you can see that both of these equations have a y.

Yes! You would rather substitute Equation 2 into Equation 1 because it's easier.

So, substitute the y in Equation 1 and it will give you 2x−3(x−1)=−1.

To solve 2x−3(x−1)=−1, you cannot touch any other numbers if you do not expand the bracket FIRST.

To expand the −3(x−1) bracket, this is how you do it: −3×x+−3×−1.
Therefore, −3x+3 is your expanded bracket. [BE CAUTIOUS ABOUT NEGATIVES & POSITIVES!!]

So then it leaves you with 2x−3x+3=−1.
Send the +3 to the right hand side of equal sign and it will become -3.
2x−3x=−1−3
−x=−4#
$x = 4$

[THAT'S NOT DONE YET! UNLESS IF YOU DON'T WANT FULL MARKS! (LOL)]

So now you've know that $x = 4$, you then gotta find what is $y$.
So substitute 4 into $x$ either in Equation 1 or 2 (you choose).
Option 1 (Substitute into Equation 1):
$2 x - 3 y = - 1$ substitute 4 with $x$ and it gives you $2 \times 4 - 3 y = - 1$.

Now you have:
$8 - 3 y = - 1$
$- 3 y = - 1 - 8$
$- 3 y = - 9$
$y = \frac{- 9}{- 3}$
Therefore, $y = 3$.

I'll show you how to do Elimination some other time.