Solve the System of @WORD y=x^2-9 , x-y=-3

$y = x^{2} - 9$ , $x - y = - 3$

Step 1

Replace all occurrences of $y$ with $x^{2} - 9$ in each equation.

Replace all occurrences of $y$ in $x - y = - 3$ with $x^{2} - 9$ .

$x - (x^{2} - 9) = - 3$

$y = x^{2} - 9$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x - x^{2} + 9 = - 3$

$y = x^{2} - 9$

Multiply $- 1$ by $- 9$ .

$x - x^{2} + 9 = - 3$

$y = x^{2} - 9$

$x - x^{2} + 9 = - 3$

$y = x^{2} - 9$

$x - x^{2} + 9 = - 3$

$y = x^{2} - 9$

$x - x^{2} + 9 = - 3$

$y = x^{2} - 9$

Step 2

Solve for $x$ in $x - x^{2} + 9 = - 3$ .

Add $3$ to both sides of the equation.

$x - x^{2} + 9 + 3 = 0$

$y = x^{2} - 9$

Add $9$ and $3$ .

$x - x^{2} + 12 = 0$

$y = x^{2} - 9$

Factor the left side of the equation.

Factor $- 1$ out of $x - x^{2} + 12$ .

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

$- x^{2} + x + 12 = 0$

$y = x^{2} - 9$

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

$- (x^{2}) + x + 12 = 0$

$y = x^{2} - 9$

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 12 = 0$

$y = x^{2} - 9$

Rewrite $12$ as $- 1 (- 12)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 12 = 0$

$y = x^{2} - 9$

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

$- (x^{2} - x) - 1 \cdot - 12 = 0$

$y = x^{2} - 9$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 12)$ .

$- (x^{2} - x - 12) = 0$

$y = x^{2} - 9$

$- (x^{2} - x - 12) = 0$

$y = x^{2} - 9$

Factor.

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

$y = x^{2} - 9$

Write the factored form using these integers.

$- ((x - 4) (x + 3)) = 0$

$y = x^{2} - 9$

$- ((x - 4) (x + 3)) = 0$

$y = x^{2} - 9$

Remove unnecessary parentheses.

$- (x - 4) (x + 3) = 0$

$y = x^{2} - 9$

$- (x - 4) (x + 3) = 0$

$y = x^{2} - 9$

$- (x - 4) (x + 3) = 0$

$y = x^{2} - 9$

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

$x - 4 = 0$

$x + 3 = 0$

$y = x^{2} - 9$

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

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

$x - 4 = 0$

$y = x^{2} - 9$

Add $4$ to both sides of the equation.

$x = 4$

$y = x^{2} - 9$

$x = 4$

$y = x^{2} - 9$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

$y = x^{2} - 9$

Subtract $3$ from both sides of the equation.

$x = - 3$

$y = x^{2} - 9$

$x = - 3$

$y = x^{2} - 9$

The final solution is all the values that make $- (x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

$y = x^{2} - 9$

$x = 4, - 3$

$y = x^{2} - 9$

Step 3

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

Replace all occurrences of $x$ in $y = x^{2} - 9$ with $4$ .

$y = {(4)}^{2} - 9$

$x = 4$

Simplify the right side.

Simplify ${(4)}^{2} - 9$ .

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

$y = 16 - 9$

$x = 4$

Subtract $9$ from $16$ .

$y = 7$

$x = 4$

$y = 7$

$x = 4$

$y = 7$

$x = 4$

$y = 7$

$x = 4$

Step 4

Replace all occurrences of $x$ with $- 3$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 9$ with $- 3$ .

$y = {(- 3)}^{2} - 9$

$x = - 3$

Simplify the right side.

Simplify ${(- 3)}^{2} - 9$ .

Raise $- 3$ to the power of $2$ .

$y = 9 - 9$

$x = - 3$

Subtract $9$ from $9$ .

$y = 0$

$x = - 3$

$y = 0$

$x = - 3$

$y = 0$

$x = - 3$

$y = 0$

$x = - 3$

Step 5

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

$(4, 7)$

$(- 3, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 7), (- 3, 0)$

Equation Form:

$x = 4, y = 7$

$x = - 3, y = 0$

Step 7

Solve the System of @WORD y=x^2-9 , x-y=-3

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