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

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Answer 1 · 2015-05-10T23:58:09

Given
$"[1] "y=4-x^2$
$"[2] "3x+y=0$

Rearranging $"[2]"$
$"[3] "y=-3x$

Substituting $"[3]"$ into $"[1]"$
$"[4] "-3x = 4 - x^2$

Rearranging $"[4]"$
$"[5] " x^2-3x-4 = 0$

Factoring the left-hand side of $"[5]"$
$"[6] "(x-4)(x+1) = 0$

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

How do you find the solution of the system of equations y = 4-x^2y = 4-x^2 and 3x+y = 03x+y = 0?