Solve the System of Equations y=x^2 x^2+2y=12

$y = x^{2}$ $x^{2} + 2 y = 12$

Step 1

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

Replace all occurrences of $y$ in $x^{2} + 2 y = 12$ with $x^{2}$ .

$x^{2} + 2 (x^{2}) = 12$

$y = x^{2}$

Simplify the left side.

Add $x^{2}$ and $2 x^{2}$ .

$3 x^{2} = 12$

$y = x^{2}$

$3 x^{2} = 12$

$y = x^{2}$

$3 x^{2} = 12$

$y = x^{2}$

Step 2

Solve for $x$ in $3 x^{2} = 12$ .

Divide each term in $3 x^{2} = 12$ by $3$ and simplify.

Divide each term in $3 x^{2} = 12$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{12}{3}$

$y = x^{2}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{12}{3}$

$y = x^{2}$

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

$x^{2} = \frac{12}{3}$

$y = x^{2}$

$x^{2} = \frac{12}{3}$

$y = x^{2}$

$x^{2} = \frac{12}{3}$

$y = x^{2}$

Simplify the right side.

Divide $12$ by $3$ .

$x^{2} = 4$

$y = x^{2}$

$x^{2} = 4$

$y = x^{2}$

$x^{2} = 4$

$y = x^{2}$

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

$x = \pm \sqrt{4}$

$y = x^{2}$

Simplify $\pm \sqrt{4}$ .

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

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

$y = x^{2}$

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

$x = \pm 2$

$y = x^{2}$

$x = \pm 2$

$y = x^{2}$

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.

$x = 2$

$y = x^{2}$

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

$x = - 2$

$y = x^{2}$

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

$x = 2, - 2$

$y = x^{2}$

$x = 2, - 2$

$y = x^{2}$

$x = 2, - 2$

$y = x^{2}$

Step 3

Replace all occurrences of $x$ with $2$ in each equation.

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

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

$x = 2$

Simplify the right side.

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

$y = 4$

$x = 2$

$y = 4$

$x = 2$

$y = 4$

$x = 2$

Step 4

Replace all occurrences of $x$ with $- 2$ in each equation.

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

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

$x = - 2$

Simplify the right side.

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

$y = 4$

$x = - 2$

$y = 4$

$x = - 2$

$y = 4$

$x = - 2$

Step 5

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

$(2, 4)$

$(- 2, 4)$

Step 6

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 2, y = 4$

$x = - 2, y = 4$

Step 7

Solve the System of Equations y=x^2 x^2+2y=12

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