# How do you solve the system x² + y² = 20 and x + y = 6?

Jul 7, 2016

$y = 6 - x$

By substitution

$\therefore {x}^{2} + {\left(6 - x\right)}^{2} = 20$

${x}^{2} + 36 - 12 x + {x}^{2} = 20$

$2 {x}^{2} - 12 x + 16 = 0$

$2 \left({x}^{2} - 6 x + 8\right) = 0$

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

$x = 4 \mathmr{and} 2$

$\therefore x + y = 6$

$4 + y = 6 \text{ and } 2 + y = 6$

$y = 2 \text{ and } y = 4$

Hence, the solution set is $\left\{4 , 2\right\}$ and $\left\{2 , 4\right\}$.

Hopefully this helps!