SOLUTION: this is a "solving systems of equations using substitution" question. I tried to solve it using a negative and a positive number but i cant. i've been stuck on this question for 30

Question 237533: this is a "solving systems of equations using substitution" question. I tried to solve it using a negative and a positive number but i cant. i've been stuck on this question for 30 minutes. Please help me as soon as possible.
2x-3y=6
x+y=12
*when you solve x+y=12, does it become y=x-12 or y=x+12?
*also, when you plug in for y ---> 2x-3(x-12)=6, does it become 2x-3x+12=6 or 2x-3x-12=6 or 2x+3x+12=6 or 2x+3x-12=6?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
2x-3y=6
x+y=12
*when you solve x+y=12, does it become y=x-12 or y=x+12?
*also, when you plug in for y ---> 2x-3(x-12)=6, does it become 2x-3x+12=6 or 2x-3x-12=6 or 2x+3x+12=6 or 2x+3x-12=6?
----------------------------
Solve the 2nd equation for "y":
y = -x + 12
----
Substitute that into the 1st equation and solve for "x":
2x -3(-x+12) = 6
2x + 3x - 36 = 6
5x = 42
x = 42/5 = 8 2/5
-----------------------
Substitute that into y = -x +12 to get:
y = -(42/5)+(60/5)
y = 18/5 = 3 3/5
---------------------
Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Add to both sides

NOW make the substitution:

You should be able to take it from there. Write back if you need more help. Hint: The values of x and y are both fairly ugly fractions.

John

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