Question

Question #85ee6

Answer 1

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):
`2x - 3y = -1` substitute 4 with `x` and it gives you `2 xx 4 - 3y = -1`.

Now you have:
`8 - 3y = -1`
`-3y = -1 - 8`
`-3y = -9`
`y = (-9)/(-3)`
Therefore, `y = 3`.

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