How do you solve $x^2+y=8$ and $x-2y=-6$ ?

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How do you solve $x^2+y=8$ and $x-2y=-6$ ?

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Answer 1 · 2017-01-28T10:02:11

$(-5/2, 7/2), (2, 4)$

Explanation:

$x^2+y=8$
$x-2y=-6$
In the first equation solve for y in terms of x:
$x^2+y=8$ => $y=8-x^2$
Substitute for $y$ in the second equation:
$x-2(8-x^2)=-6$ => simplify:
$x-16+2x^2=-6$ => add 6 to both sides and rewrite as:
$2x^2+x-10=0$
Solve by quadratic formula or by factoring,
$(2x+5)(x-2)=0$
$x=-5/2 or x=2$
Solve for y:
$y=8-(-5/2)^2 or y=8-2^2$
$y=8-25/4 or y=8-4$
$y=7/4 or y=4$
Hence the solutions are:
$(-5/2, 7/2), (2, 4)$

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How do you solve $x^2+y=8$ and $x-2y=-6$ ?

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Explanation:

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How do you solve x^2+y=8x^2+y=8 and x-2y=-6 x-2y=-6 ?

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How do you solve $x^2+y=8$ and $x-2y=-6$ ?