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

$y = 2 x$ $y = x^{2} - 8$

Eliminate the equal sides of each equation and combine.

$2 x = x^{2} - 8$

Solve $2 x = x^{2} - 8$ for $x$ .

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

$2 x - x^{2} = - 8$

Add $8$ to both sides of the equation.

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

Factor the left side of the equation.

Factor $- 1$ out of $2 x - x^{2} + 8$ .

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

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

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

$- (x^{2}) + 2 x + 8 = 0$

Factor $- 1$ out of $2 x$ .

$- (x^{2}) - (- 2 x) + 8 = 0$

Rewrite $8$ as $- 1 (- 8)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot - 8 = 0$

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

$- (x^{2} - 2 x) - 1 \cdot - 8 = 0$

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (- 8)$ .

$- (x^{2} - 2 x - 8) = 0$

Factor.

Factor $x^{2} - 2 x - 8$ 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 $- 8$ and whose sum is $- 2$ .

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

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x - 4 = 0$

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

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$x = 4, - 2$

Evaluate $y$ when $x = 4$ .

Substitute $4$ for $x$ .

$y = {(4)}^{2} - 8$

Substitute $4$ for $x$ in $y = {(4)}^{2} - 8$ and solve for $y$ .

Remove parentheses.

$y = 4^{2} - 8$

Simplify $4^{2} - 8$ .

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

$y = 16 - 8$

Subtract $8$ from $16$ .

$y = 8$

Evaluate $y$ when $x = - 2$ .

Substitute $- 2$ for $x$ .

$y = {(- 2)}^{2} - 8$

Simplify ${(- 2)}^{2} - 8$ .

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

$y = 4 - 8$

Subtract $8$ from $4$ .

$y = - 4$

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

$(4, 8)$

$(- 2, - 4)$

The result can be shown in multiple forms.

Point Form:

$(4, 8), (- 2, - 4)$

Equation Form:

$x = 4, y = 8$

$x = - 2, y = - 4$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.