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

Jan 28, 2017

$\left(- \frac{5}{2} , \frac{7}{2}\right) , \left(2 , 4\right)$

#### Explanation:

${x}^{2} + y = 8$
$x - 2 y = - 6$
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:
$x - 2 \left(8 - {x}^{2}\right) = - 6$ => simplify:
$x - 16 + 2 {x}^{2} = - 6$ => add 6 to both sides and rewrite as:
$2 {x}^{2} + x - 10 = 0$
Solve by quadratic formula or by factoring,
$\left(2 x + 5\right) \left(x - 2\right) = 0$
$x = - \frac{5}{2} \mathmr{and} x = 2$
Solve for y:
$y = 8 - {\left(- \frac{5}{2}\right)}^{2} \mathmr{and} y = 8 - {2}^{2}$
$y = 8 - \frac{25}{4} \mathmr{and} y = 8 - 4$
$y = \frac{7}{4} \mathmr{and} y = 4$
Hence the solutions are:
$\left(- \frac{5}{2} , \frac{7}{2}\right) , \left(2 , 4\right)$