# How do you find the solution of the system of equations y = 4-x^2 and 3x+y = 0?

May 10, 2015

Given
$\text{[1] } y = 4 - {x}^{2}$
$\text{[2] } 3 x + y = 0$

Rearranging $\text{[2]}$
$\text{[3] } y = - 3 x$

Substituting $\text{[3]}$ into $\text{[1]}$
$\text{[4] } - 3 x = 4 - {x}^{2}$

Rearranging $\text{[4]}$
$\text{[5] } {x}^{2} - 3 x - 4 = 0$

Factoring the left-hand side of $\text{[5]}$
$\text{[6] } \left(x - 4\right) \left(x + 1\right) = 0$

Solution:
$x = 4$ or $x = - 1$