Question #85ee6

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#.

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