4(x+y)=x-16

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4(x+y) = x-16 4x+4y = x-16 3x+4y = -16
the rest of the equation I forgot to place in: -3(x-y)=y+10
So you have this system of equations\[4(x+y)=x-16\]\[-3(x-y)=y+10\]First solve the first equation for one of the variables\[4x+4y=x-16\]\[4y=-3x-16\]\[y=-\frac{3}{4}x-4\]Now plug that value into the second equation and solve for the second variable\[-3x+3y=y+10\]\[-3x+2y=10\]\[-3x+2\left(-\frac{3}{4}x-4\right)=10\]\[-3x-\frac{3}{2}x-8=10\]\[\frac{-6-3}{2}x=18\]\[-\frac{9}{2}x=18\]\[x=-4\]Now plug that value of x into either equation to solve for y.\[4(x+y)=x-16\]\[4(-4+y)=-4-16\]\[-16+4y=-20\]\[4y=4\]\[y=1\]So the solution is x=-4, y=1

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oops, should be -1/
the second last line I dropped the negative it should say \[4y=-4\]\[y=-1\]So the solution is x=-4, y=-1
just to let you know, you dont only have to do the substitution process, you can also do elimination. 4(x+y)=x−16 −3(x−y)=y+10 distribute to get 4x+4y=x−16 -3x+3y=y+10 the subtract the common terms from each other to get final answer

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