Solve the System of Equations x^2-9=y y-x=3

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

Step 1

Rewrite the equation as $y = x^{2} - 9$ .

$y = x^{2} - 9$

$y - x = 3$

Step 2

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

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

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

$y = x^{2} - 9$

Simplify the left side.

Remove parentheses.

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

$y = x^{2} - 9$

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

$y = x^{2} - 9$

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

$y = x^{2} - 9$

Step 3

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

Subtract $3$ from both sides of the equation.

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

$y = x^{2} - 9$

Subtract $3$ from $- 9$ .

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

$y = x^{2} - 9$

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.

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

$y = x^{2} - 9$

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

$y = x^{2} - 9$

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 4

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 5

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 6

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

$(4, 7)$

$(- 3, 0)$

Step 7

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 8

Solve the System of Equations x^2-9=y y-x=3

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