# How do you solve x² + y² = 16 and x + y = 4?

Jun 4, 2018

$\left(x , y\right) = \left(0 , 4\right)$ or $\left(4 , 0\right)$

#### Explanation:

From the second equation we get
$y = 4 - x$
pluggin this in the first equation

${x}^{2} + {\left(4 - x\right)}^{2} = 16$
${x}^{2} + 16 - 8 x + {x}^{2} = 16$
$2 {x}^{2} - 8 x = 0$
$2 x \left(x - 4\right) = 0$
so
$x = 0$
or
$x = 4$