Solve the System of Equations 2y=2x^2+4x-8 y=2x^2-28

$2 y = 2 x^{2} + 4 x - 8$ $y = 2 x^{2} - 28$

Replace all occurrences of $y$ with $2 x^{2} - 28$ in each equation.

Replace all occurrences of $y$ in $2 y = 2 x^{2} + 4 x - 8$ with $2 x^{2} - 28$ .

$2 (2 x^{2} - 28) = 2 x^{2} + 4 x - 8$

$y = 2 x^{2} - 28$

Simplify the left side.

Simplify $2 (2 x^{2} - 28)$ .

Apply the distributive property.

$2 (2 x^{2}) + 2 \cdot - 28 = 2 x^{2} + 4 x - 8$

$y = 2 x^{2} - 28$

Multiply.

Multiply $2$ by $2$ .

$4 x^{2} + 2 \cdot - 28 = 2 x^{2} + 4 x - 8$

$y = 2 x^{2} - 28$

Multiply $2$ by $- 28$ .

$4 x^{2} - 56 = 2 x^{2} + 4 x - 8$

$y = 2 x^{2} - 28$

$4 x^{2} - 56 = 2 x^{2} + 4 x - 8$

$y = 2 x^{2} - 28$

$4 x^{2} - 56 = 2 x^{2} + 4 x - 8$

$y = 2 x^{2} - 28$

$4 x^{2} - 56 = 2 x^{2} + 4 x - 8$

$y = 2 x^{2} - 28$

$4 x^{2} - 56 = 2 x^{2} + 4 x - 8$

$y = 2 x^{2} - 28$

Solve for $x$ in $4 x^{2} - 56 = 2 x^{2} + 4 x - 8$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 x^{2} + 4 x - 8 = 4 x^{2} - 56$

$y = 2 x^{2} - 28$

Move all terms containing $x$ to the left side of the equation.

Subtract $4 x^{2}$ from both sides of the equation.

$2 x^{2} + 4 x - 8 - 4 x^{2} = - 56$

$y = 2 x^{2} - 28$

Subtract $4 x^{2}$ from $2 x^{2}$ .

$- 2 x^{2} + 4 x - 8 = - 56$

$y = 2 x^{2} - 28$

$- 2 x^{2} + 4 x - 8 = - 56$

$y = 2 x^{2} - 28$

Add $56$ to both sides of the equation.

$- 2 x^{2} + 4 x - 8 + 56 = 0$

$y = 2 x^{2} - 28$

Add $- 8$ and $56$ .

$- 2 x^{2} + 4 x + 48 = 0$

$y = 2 x^{2} - 28$

Factor the left side of the equation.

Factor $- 2$ out of $- 2 x^{2} + 4 x + 48$ .

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

$- 2 x^{2} + 4 x + 48 = 0$

$y = 2 x^{2} - 28$

Factor $- 2$ out of $4 x$ .

$- 2 x^{2} - 2 (- 2 x) + 48 = 0$

$y = 2 x^{2} - 28$

Factor $- 2$ out of $48$ .

$- 2 x^{2} - 2 (- 2 x) - 2 \cdot - 24 = 0$

$y = 2 x^{2} - 28$

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

$- 2 (x^{2} - 2 x) - 2 \cdot - 24 = 0$

$y = 2 x^{2} - 28$

Factor $- 2$ out of $- 2 (x^{2} - 2 x) - 2 (- 24)$ .

$- 2 (x^{2} - 2 x - 24) = 0$

$y = 2 x^{2} - 28$

$- 2 (x^{2} - 2 x - 24) = 0$

$y = 2 x^{2} - 28$

Factor.

Factor $x^{2} - 2 x - 24$ 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 $- 2$ .

$- 6, 4$

$y = 2 x^{2} - 28$

Write the factored form using these integers.

$- 2 ((x - 6) (x + 4)) = 0$

$y = 2 x^{2} - 28$

$- 2 ((x - 6) (x + 4)) = 0$

$y = 2 x^{2} - 28$

Remove unnecessary parentheses.

$- 2 (x - 6) (x + 4) = 0$

$y = 2 x^{2} - 28$

$- 2 (x - 6) (x + 4) = 0$

$y = 2 x^{2} - 28$

$- 2 (x - 6) (x + 4) = 0$

$y = 2 x^{2} - 28$

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

$x - 6 = 0$

$x + 4 = 0$

$y = 2 x^{2} - 28$

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

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

$x - 6 = 0$

$y = 2 x^{2} - 28$

Add $6$ to both sides of the equation.

$x = 6$

$y = 2 x^{2} - 28$

$x = 6$

$y = 2 x^{2} - 28$

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

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

$x + 4 = 0$

$y = 2 x^{2} - 28$

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = 2 x^{2} - 28$

$x = - 4$

$y = 2 x^{2} - 28$

The final solution is all the values that make $- 2 (x - 6) (x + 4) = 0$ true.

$x = 6, - 4$

$y = 2 x^{2} - 28$

$x = 6, - 4$

$y = 2 x^{2} - 28$

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

Replace all occurrences of $x$ in $y = 2 x^{2} - 28$ with $6$ .

$y = 2 {(6)}^{2} - 28$

$x = 6$

Simplify the right side.

Simplify $2 {(6)}^{2} - 28$ .

Simplify each term.

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

$y = 2 \cdot 36 - 28$

$x = 6$

Multiply $2$ by $36$ .

$y = 72 - 28$

$x = 6$

$y = 72 - 28$

$x = 6$

Subtract $28$ from $72$ .

$y = 44$

$x = 6$

$y = 44$

$x = 6$

$y = 44$

$x = 6$

$y = 44$

$x = 6$

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

Replace all occurrences of $x$ in $y = 2 x^{2} - 28$ with $- 4$ .

$y = 2 {(- 4)}^{2} - 28$

$x = - 4$

Simplify the right side.

Simplify $2 {(- 4)}^{2} - 28$ .

Simplify each term.

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

$y = 2 \cdot 16 - 28$

$x = - 4$

Multiply $2$ by $16$ .

$y = 32 - 28$

$x = - 4$

$y = 32 - 28$

$x = - 4$

Subtract $28$ from $32$ .

$y = 4$

$x = - 4$

$y = 4$

$x = - 4$

$y = 4$

$x = - 4$

$y = 4$

$x = - 4$

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

$(6, 44)$

$(- 4, 4)$

The result can be shown in multiple forms.

Point Form:

$(6, 44), (- 4, 4)$

Equation Form:

$x = 6, y = 44$

$x = - 4, y = 4$

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