# How do you solve the system of equations x^2+y^2=11 and x+y=15 by substitution?

May 28, 2017

See below

#### Explanation:

We set $x = 15 - y$ (from the second equation).

Then,

${x}^{2} + {y}^{2} = 11$

${\left(15 - y\right)}^{2} + {y}^{2} = 11$

$225 - 30 y + 2 {y}^{2} = 11$

$2 {y}^{2} - 30 y + 214 = 0$

${y}^{2} - 15 y + 107 = 0$ (this is very ugly :/)

We know that $x$ is the opposite term in the quadratic formula, as the two equations are symmetrical.

$y = \frac{15 \pm \sqrt{225 - 4 \left(107\right)}}{2} = \frac{15 \pm \sqrt{- 203}}{2}$

So, $y = \frac{15 + i \sqrt{203}}{2}$ and $x = \frac{15 - i \sqrt{203}}{2}$