Systems of equations
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The first part has the solutions x/y=-2 or -3. So x=-2y or -3y.

Substitute for x in x+y=8: -2y+y=8, -y=8 so y=-8 and x=16; or

-3y+y=8, -2y=8, y=-4 and x=12.

Therefore the solutions are (16,-8) and (12,-4).

If you plug these into the original quadratic you will find that this pair of solutions makes the quadratic = 0.

by Top Rated User (1.1m points)

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