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

May 3, 2016

$y = 4 - x \to {x}^{2} + {\left(4 - x\right)}^{2} = 16$

${x}^{2} + 16 - 8 x + {x}^{2} = 16$

$2 {x}^{2} - 8 x = 0$

$2 x \left(x - 4\right) = 0$

$x = 0 \mathmr{and} 4$

$0 + y = 4 \mathmr{and} 4 + y = 4$

$y = 4 \mathmr{and} 0$

The solution sets are $\left\{0 , 4\right\} \mathmr{and} \left\{4 , 0\right\}$

Hopefully this helps!