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

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

Step 1

Eliminate the equal sides of each equation and combine.

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

Step 2

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

Add $x$ to both sides of the equation.

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

Add $2$ to both sides of the equation.

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

Add $- 4$ and $2$ .

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

Factor $x^{2} + x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $1$ .

$- 1, 2 + y = - x - 2$

Write the factored form using these integers.

$(x - 1) (x + 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x + 2 = 0 + y = - x - 2$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 1) (x + 2) = 0$ true.

$x = 1, - 2$

Step 3

Evaluate $y$ when $x = 1$ .

Substitute $1$ for $x$ .

$y = - (1) - 2$

Simplify $- (1) - 2$ .

Multiply $- 1$ by $1$ .

$y = - 1 - 2$

Subtract $2$ from $- 1$ .

$y = - 3$

Step 4

Evaluate $y$ when $x = - 2$ .

Substitute $- 2$ for $x$ .

$y = - (- 2) - 2$

Simplify $- (- 2) - 2$ .

Multiply $- 1$ by $- 2$ .

$y = 2 - 2$

Subtract $2$ from $2$ .

$y = 0$

Step 5

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

$(1, - 3)$

$(- 2, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(1, - 3), (- 2, 0)$

Equation Form:

$x = 1, y = - 3$

$x = - 2, y = 0$

Step 7

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

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