Solve the System of @WORD y=x^2-16 , x-y=-4

$y = x^{2} - 16$ , $x - y = - 4$

Step 1

Replace all occurrences of $y$ with $x^{2} - 16$ in each equation.

Replace all occurrences of $y$ in $x - y = - 4$ with $x^{2} - 16$ .

$x - (x^{2} - 16) = - 4$

$y = x^{2} - 16$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x - x^{2} + 16 = - 4$

$y = x^{2} - 16$

Multiply $- 1$ by $- 16$ .

$x - x^{2} + 16 = - 4$

$y = x^{2} - 16$

$x - x^{2} + 16 = - 4$

$y = x^{2} - 16$

$x - x^{2} + 16 = - 4$

$y = x^{2} - 16$

$x - x^{2} + 16 = - 4$

$y = x^{2} - 16$

Step 2

Solve for $x$ in $x - x^{2} + 16 = - 4$ .

Add $4$ to both sides of the equation.

$x - x^{2} + 16 + 4 = 0$

$y = x^{2} - 16$

Add $16$ and $4$ .

$x - x^{2} + 20 = 0$

$y = x^{2} - 16$

Factor the left side of the equation.

Factor $- 1$ out of $x - x^{2} + 20$ .

Reorder $x$ and $- x^{2}$ .

$- x^{2} + x + 20 = 0$

$y = x^{2} - 16$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + x + 20 = 0$

$y = x^{2} - 16$

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 20 = 0$

$y = x^{2} - 16$

Rewrite $20$ as $- 1 (- 20)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 20 = 0$

$y = x^{2} - 16$

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$- (x^{2} - x) - 1 \cdot - 20 = 0$

$y = x^{2} - 16$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 20)$ .

$- (x^{2} - x - 20) = 0$

$y = x^{2} - 16$

$- (x^{2} - x - 20) = 0$

$y = x^{2} - 16$

Factor.

Factor $x^{2} - x - 20$ 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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

$y = x^{2} - 16$

Write the factored form using these integers.

$- ((x - 5) (x + 4)) = 0$

$y = x^{2} - 16$

$- ((x - 5) (x + 4)) = 0$

$y = x^{2} - 16$

Remove unnecessary parentheses.

$- (x - 5) (x + 4) = 0$

$y = x^{2} - 16$

$- (x - 5) (x + 4) = 0$

$y = x^{2} - 16$

$- (x - 5) (x + 4) = 0$

$y = x^{2} - 16$

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

$x - 5 = 0$

$x + 4 = 0$

$y = x^{2} - 16$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

$y = x^{2} - 16$

Add $5$ to both sides of the equation.

$x = 5$

$y = x^{2} - 16$

$x = 5$

$y = x^{2} - 16$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

$y = x^{2} - 16$

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = x^{2} - 16$

$x = - 4$

$y = x^{2} - 16$

The final solution is all the values that make $- (x - 5) (x + 4) = 0$ true.

$x = 5, - 4$

$y = x^{2} - 16$

$x = 5, - 4$

$y = x^{2} - 16$

Step 3

Replace all occurrences of $x$ with $5$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 16$ with $5$ .

$y = {(5)}^{2} - 16$

$x = 5$

Simplify the right side.

Simplify ${(5)}^{2} - 16$ .

Raise $5$ to the power of $2$ .

$y = 25 - 16$

$x = 5$

Subtract $16$ from $25$ .

$y = 9$

$x = 5$

$y = 9$

$x = 5$

$y = 9$

$x = 5$

$y = 9$

$x = 5$

Step 4

Replace all occurrences of $x$ with $- 4$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 16$ with $- 4$ .

$y = {(- 4)}^{2} - 16$

$x = - 4$

Simplify the right side.

Simplify ${(- 4)}^{2} - 16$ .

Raise $- 4$ to the power of $2$ .

$y = 16 - 16$

$x = - 4$

Subtract $16$ from $16$ .

$y = 0$

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

$x = - 4$

Step 5

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

$(5, 9)$

$(- 4, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(5, 9), (- 4, 0)$

Equation Form:

$x = 5, y = 9$

$x = - 4, y = 0$

Step 7

Solve the System of @WORD y=x^2-16 , x-y=-4

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