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

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Answer 1 · 2015-05-10T23:58:09

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

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

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

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

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

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

How do you find the solution of the system of equations y=4−x2y = 4-x^2 and 3x+y=03x+y = 0?