Solve the System of Equations y=x^2-9 -2x+y=15

$y = x^{2} - 9$ $- 2 x + y = 15$

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

Replace all occurrences of $y$ in $- 2 x + y = 15$ with $x^{2} - 9$ .

$- 2 x + x^{2} - 9 = 15$

$y = x^{2} - 9$

Simplify the left side.

Remove parentheses.

$- 2 x + x^{2} - 9 = 15$

$y = x^{2} - 9$

$- 2 x + x^{2} - 9 = 15$

$y = x^{2} - 9$

$- 2 x + x^{2} - 9 = 15$

$y = x^{2} - 9$

Solve for $x$ in $- 2 x + x^{2} - 9 = 15$ .

Subtract $15$ from both sides of the equation.

$- 2 x + x^{2} - 9 - 15 = 0$

$y = x^{2} - 9$

Subtract $15$ from $- 9$ .

$- 2 x + x^{2} - 24 = 0$

$y = x^{2} - 9$

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$- 2 u + u^{2} - 24 = 0$

$y = x^{2} - 9$

Factor $- 2 u + u^{2} - 24$ 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 $- 24$ and whose sum is $- 2$ .

$- 6, 4$

$y = x^{2} - 9$

Write the factored form using these integers.

$(u - 6) (u + 4) = 0$

$y = x^{2} - 9$

$(u - 6) (u + 4) = 0$

$y = x^{2} - 9$

Replace all occurrences of $u$ with $x$ .

$(x - 6) (x + 4) = 0$

$y = x^{2} - 9$

$(x - 6) (x + 4) = 0$

$y = x^{2} - 9$

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

$x - 6 = 0$

$x + 4 = 0$

$y = x^{2} - 9$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

$y = x^{2} - 9$

Add $6$ to both sides of the equation.

$x = 6$

$y = x^{2} - 9$

$x = 6$

$y = x^{2} - 9$

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

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

$x + 4 = 0$

$y = x^{2} - 9$

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = x^{2} - 9$

$x = - 4$

$y = x^{2} - 9$

The final solution is all the values that make $(x - 6) (x + 4) = 0$ true.

$x = 6, - 4$

$y = x^{2} - 9$

$x = 6, - 4$

$y = x^{2} - 9$

Replace all occurrences of $x$ with $6$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 9$ with $6$ .

$y = {(6)}^{2} - 9$

$x = 6$

Simplify the right side.

Simplify ${(6)}^{2} - 9$ .

Raise $6$ to the power of $2$ .

$y = 36 - 9$

$x = 6$

Subtract $9$ from $36$ .

$y = 27$

$x = 6$

$y = 27$

$x = 6$

$y = 27$

$x = 6$

$y = 27$

$x = 6$

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 $16$ .

$y = 7$

$x = - 4$

$y = 7$

$x = - 4$

$y = 7$

$x = - 4$

$y = 7$

$x = - 4$

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

$(6, 27)$

$(- 4, 7)$

The result can be shown in multiple forms.

Point Form:

$(6, 27), (- 4, 7)$

Equation Form:

$x = 6, y = 27$

$x = - 4, y = 7$

Solve the System of Equations y=x^2-9 -2x+y=15

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