# 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

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 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(57940)   (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(17059)   (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