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I'm given a problem on my exam review that I don't remember how to solve? I have to find the values for x and y that make this true: http://puu.sh/6AmQU.png Can anyone help me solve this?
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solve the system \[2y+5=x-1\\ y-2=3x\]
2y + 6 = x, (y-2)/3 = x?
first equation is the same as \(2y+6=x\) substitute it in to the second equation and get \[y-2=3(2y+6)\]
solve that one for \(y\) then substitute back and solve for \(x\)
Ahhhh that's right. Now it's starting to look familiar. Gimmie a second to solve that
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\[y-2=3(2y+6)\\ y-2=6y+18\\ -20=5y\\ -5=y\]
I got y = -4 for some reason
You added 18 to positive 2 instead of negative, your -20 should have been 16
y = -4 checks out because when it's plugged in, x = -2 in both equations
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