Solve the System of Equations y=-2x^2+2 and y=-6

$y = - 2 x^{2} + 2$ and $y = - 6$

Step 1

Eliminate the equal sides of each equation and combine.

$- 2 x^{2} + 2 = - 6$

Step 2

Solve $- 2 x^{2} + 2 = - 6$ for $x$ .

Move all terms not containing $x$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$- 2 x^{2} = - 6 - 2$

Subtract $2$ from $- 6$ .

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

Divide each term in $- 2 x^{2} = - 8$ by $- 2$ and simplify.

Divide each term in $- 2 x^{2} = - 8$ by $- 2$ .

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $x^{2}$ by $1$ .

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

Simplify the right side.

Divide $- 8$ by $- 2$ .

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

Step 3

Substitute $2$ for $x$ .

$y = - 6$

Step 4

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

$(2, - 6)$

$(- 2, - 6)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(2, - 6), (- 2, - 6)$

Equation Form:

$x = 2, y = - 6$

$x = - 2, y = - 6$

Step 6

Solve the System of Equations y=-2x^2+2 and y=-6

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