Solve the System of Equations y=x^2-4 y=-2x-5

$y = x^{2} - 4$ $y = - 2 x - 5$

Eliminate the equal sides of each equation and combine.

$x^{2} - 4 = - 2 x - 5$

Solve $x^{2} - 4 = - 2 x - 5$ for $x$ .

Add $2 x$ to both sides of the equation.

$x^{2} - 4 + 2 x = - 5$

Add $5$ to both sides of the equation.

$x^{2} - 4 + 2 x + 5 = 0$

Add $- 4$ and $5$ .

$x^{2} + 2 x + 1 = 0$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$x^{2} + 2 x + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x + 1)}^{2} = 0$

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Evaluate $y$ when $x = - 1$ .

Substitute $- 1$ for $x$ .

$y = - 2 \cdot - 1 - 5$

Substitute $- 1$ for $x$ in $y = - 2 \cdot - 1 - 5$ and solve for $y$ .

Remove parentheses.

$y = - 2 \cdot - 1 - 5$

Simplify $- 2 \cdot - 1 - 5$ .

Multiply $- 2$ by $- 1$ .

$y = 2 - 5$

Subtract $5$ from $2$ .

$y = - 3$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- 1, - 3)$

The result can be shown in multiple forms.

Point Form:

$(- 1, - 3)$

Equation Form:

$x = - 1, y = - 3$