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What are the solutions to the following system of equations? Select the correct answer. y=x^2-4x+8 4x+y=12

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Reduce the second equation to y= -4x + 12, then subtract the two equations. (Quadratic on top), to get y=x^2 -4. Then, plug in an x,y coordinate if you have one.
User Nick Weisser
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y=x^2-4x+8 and 4x+y=12 so y=12-4x, insert y in the quadratic to get
12-4x=x^2-4x+8
12=x^2+8
0=x^2-4=(x-2)(x+2)
So the solutions are x=2 and x=-2.
User Jylee
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7.2k points

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