Solve the System of Equations y=2x^2+9x-80 16=x-y

$y = 2 x^{2} + 9 x - 80$ $16 = x - y$

Replace all occurrences of $y$ with $2 x^{2} + 9 x - 80$ in each equation.

Replace all occurrences of $y$ in $16 = x - y$ with $2 x^{2} + 9 x - 80$ .

$16 = x - (2 x^{2} + 9 x - 80)$

$y = 2 x^{2} + 9 x - 80$

Simplify the right side.

Simplify $x - (2 x^{2} + 9 x - 80)$ .

Simplify each term.

Apply the distributive property.

$16 = x - (2 x^{2}) - (9 x) + 80$

$y = 2 x^{2} + 9 x - 80$

Simplify.

Multiply $2$ by $- 1$ .

$16 = x - 2 x^{2} - (9 x) + 80$

$y = 2 x^{2} + 9 x - 80$

Multiply $9$ by $- 1$ .

$16 = x - 2 x^{2} - 9 x + 80$

$y = 2 x^{2} + 9 x - 80$

Multiply $- 1$ by $- 80$ .

$16 = x - 2 x^{2} - 9 x + 80$

$y = 2 x^{2} + 9 x - 80$

$16 = x - 2 x^{2} - 9 x + 80$

$y = 2 x^{2} + 9 x - 80$

$16 = x - 2 x^{2} - 9 x + 80$

$y = 2 x^{2} + 9 x - 80$

Subtract $9 x$ from $x$ .

$16 = - 2 x^{2} - 8 x + 80$

$y = 2 x^{2} + 9 x - 80$

$16 = - 2 x^{2} - 8 x + 80$

$y = 2 x^{2} + 9 x - 80$

$16 = - 2 x^{2} - 8 x + 80$

$y = 2 x^{2} + 9 x - 80$

$16 = - 2 x^{2} - 8 x + 80$

$y = 2 x^{2} + 9 x - 80$

Solve for $x$ in $16 = - 2 x^{2} - 8 x + 80$ .

Rewrite the equation as $- 2 x^{2} - 8 x + 80 = 16$ .

$- 2 x^{2} - 8 x + 80 = 16$

$y = 2 x^{2} + 9 x - 80$

Subtract $16$ from both sides of the equation.

$- 2 x^{2} - 8 x + 80 - 16 = 0$

$y = 2 x^{2} + 9 x - 80$

Subtract $16$ from $80$ .

$- 2 x^{2} - 8 x + 64 = 0$

$y = 2 x^{2} + 9 x - 80$

Factor the left side of the equation.

Factor $- 2$ out of $- 2 x^{2} - 8 x + 64$ .

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

$- 2 x^{2} - 8 x + 64 = 0$

$y = 2 x^{2} + 9 x - 80$

Factor $- 2$ out of $- 8 x$ .

$- 2 x^{2} - 2 (4 x) + 64 = 0$

$y = 2 x^{2} + 9 x - 80$

Factor $- 2$ out of $64$ .

$- 2 x^{2} - 2 (4 x) - 2 \cdot - 32 = 0$

$y = 2 x^{2} + 9 x - 80$

Factor $- 2$ out of $- 2 (x^{2}) - 2 (4 x)$ .

$- 2 (x^{2} + 4 x) - 2 \cdot - 32 = 0$

$y = 2 x^{2} + 9 x - 80$

Factor $- 2$ out of $- 2 (x^{2} + 4 x) - 2 (- 32)$ .

$- 2 (x^{2} + 4 x - 32) = 0$

$y = 2 x^{2} + 9 x - 80$

$- 2 (x^{2} + 4 x - 32) = 0$

$y = 2 x^{2} + 9 x - 80$

Factor.

Factor $x^{2} + 4 x - 32$ 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 $- 32$ and whose sum is $4$ .

$- 4, 8$

$y = 2 x^{2} + 9 x - 80$

Write the factored form using these integers.

$- 2 ((x - 4) (x + 8)) = 0$

$y = 2 x^{2} + 9 x - 80$

$- 2 ((x - 4) (x + 8)) = 0$

$y = 2 x^{2} + 9 x - 80$

Remove unnecessary parentheses.

$- 2 (x - 4) (x + 8) = 0$

$y = 2 x^{2} + 9 x - 80$

$- 2 (x - 4) (x + 8) = 0$

$y = 2 x^{2} + 9 x - 80$

$- 2 (x - 4) (x + 8) = 0$

$y = 2 x^{2} + 9 x - 80$

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

$x - 4 = 0$

$x + 8 = 0$

$y = 2 x^{2} + 9 x - 80$

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

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

$x - 4 = 0$

$y = 2 x^{2} + 9 x - 80$

Add $4$ to both sides of the equation.

$x = 4$

$y = 2 x^{2} + 9 x - 80$

$x = 4$

$y = 2 x^{2} + 9 x - 80$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

$y = 2 x^{2} + 9 x - 80$

Subtract $8$ from both sides of the equation.

$x = - 8$

$y = 2 x^{2} + 9 x - 80$

$x = - 8$

$y = 2 x^{2} + 9 x - 80$

The final solution is all the values that make $- 2 (x - 4) (x + 8) = 0$ true.

$x = 4, - 8$

$y = 2 x^{2} + 9 x - 80$

$x = 4, - 8$

$y = 2 x^{2} + 9 x - 80$

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

Replace all occurrences of $x$ in $y = 2 x^{2} + 9 x - 80$ with $4$ .

$y = 2 {(4)}^{2} + 9 (4) - 80$

$x = 4$

Simplify the right side.

Simplify $2 {(4)}^{2} + 9 (4) - 80$ .

Simplify each term.

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

$y = 2 \cdot 16 + 9 (4) - 80$

$x = 4$

Multiply $2$ by $16$ .

$y = 32 + 9 (4) - 80$

$x = 4$

Multiply $9$ by $4$ .

$y = 32 + 36 - 80$

$x = 4$

$y = 32 + 36 - 80$

$x = 4$

Simplify by adding and subtracting.

Add $32$ and $36$ .

$y = 68 - 80$

$x = 4$

Subtract $80$ from $68$ .

$y = - 12$

$x = 4$

$y = - 12$

$x = 4$

$y = - 12$

$x = 4$

$y = - 12$

$x = 4$

$y = - 12$

$x = 4$

Replace all occurrences of $x$ with $- 8$ in each equation.

Replace all occurrences of $x$ in $y = 2 x^{2} + 9 x - 80$ with $- 8$ .

$y = 2 {(- 8)}^{2} + 9 (- 8) - 80$

$x = - 8$

Simplify the right side.

Simplify $2 {(- 8)}^{2} + 9 (- 8) - 80$ .

Simplify each term.

Raise $- 8$ to the power of $2$ .

$y = 2 \cdot 64 + 9 (- 8) - 80$

$x = - 8$

Multiply $2$ by $64$ .

$y = 128 + 9 (- 8) - 80$

$x = - 8$

Multiply $9$ by $- 8$ .

$y = 128 - 72 - 80$

$x = - 8$

$y = 128 - 72 - 80$

$x = - 8$

Simplify by subtracting numbers.

Subtract $72$ from $128$ .

$y = 56 - 80$

$x = - 8$

Subtract $80$ from $56$ .

$y = - 24$

$x = - 8$

$y = - 24$

$x = - 8$

$y = - 24$

$x = - 8$

$y = - 24$

$x = - 8$

$y = - 24$

$x = - 8$

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

$(4, - 12)$

$(- 8, - 24)$

The result can be shown in multiple forms.

Point Form:

$(4, - 12), (- 8, - 24)$

Equation Form:

$x = 4, y = - 12$

$x = - 8, y = - 24$

Solve the System of Equations y=2x^2+9x-80 16=x-y

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