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(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?
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Solve the 2nd equation for "y":
y = -x + 12
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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
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Substitute that into y = -x +12 to get:
y = -(42/5)+(60/5)
y = 18/5 = 3 3/5
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Cheers,
Stan H.
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