Grade 10 Academic Math; Linear Systems.. please help?!

I was hoping that someone could help me as my brain is completely failing me at the moment and I can't seem to find any solution in my notes.

Q: Find the point of intersection of the lines:

(1) 3x-2y= 12
(2) 2y-x= -8

.....and I'm supposed to end up with {(x,y)|(2,-3)}

If you are able to use a calculator, you can type each of these equations into your calcluator and find the point of intersection. I'm going to start by telling you how to do this on a TI-84 or TI-83 calcluator. If you CAN'T use a calculator or don't have one, then I will also explain how to do this without a calculator.

With calculator:
1) Solve for "y" (make both equations y=...) for both equations and graph both of them.
2) Press 2nd and then hit "calc" (above TRACE)
3) Button number 5
4) Enter, Enter, Enter (enter 3 times)
5) Your solution will be at the bottom of your screen

How to solve for "y" if your brain is failing you that badly... AKA "isolate y"
3x-2y=12
-2y=-3x+12 ....Subtract 3x
y=(-3x+12)/-2 ....Divide by -2 to completely isolate "y" You can stop here and type it in here like this

Do the same thing with the other one.


Without calculator:

There's multiple different methods on doing this----Such as solving by substitution or elimination. I will be solving it by elimination.

3x-2y=12
-x+2y=-8 (DON'T BE TRICKED! Notice how they put it in reverse order?)
2x=4
x=2

Now that you have one number, you can plug it in to one of the formulas and solve.
2y-2=-8
2y=-6
y=-3

So, x=2 and y=-3

To check your answer, plug in the two values you received.

3(2)-2(-3)=12
6+6=12

2(-3)-2=-8
-6-2=-8


Hope this helped.

Put the equations in the same order ( I wish I had done that before I solved it incorrectly)
3x - 2y = 12
-x + 2y = -8
I will use linear combination, but substitution works OK, too.
Add the equations together:
2x = 4
x = 2

Substitute this value not either of the original equations:
-2 + 2y = -8
Add 2 to both sides
2y = -6
y = -3

(2 , -3)

first solve both equations for y then substitute any x value u want preferably 1 for both equations and the 2 numbers u get should be the intersections

(1) 3x - 2y = 12
(2) 2y - x = - 8
add
2x = 4
x = 2
from
3x - 2y = 12
6 – 2x = 12
2x = – 6
x = – 3
(2, – 3) you are right.
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