Solve the System of Equations y=x^2+3 and y=-2x^2+3

$y = x^{2} + 3$ and $y = - 2 x^{2} + 3$

Eliminate the equal sides of each equation and combine.

$x^{2} + 3 = - 2 x^{2} + 3$

Solve $x^{2} + 3 = - 2 x^{2} + 3$ for $x$ .

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

Add $2 x^{2}$ to both sides of the equation.

$x^{2} + 3 + 2 x^{2} = 3$

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

$3 x^{2} + 3 = 3$

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

Subtract $3$ from both sides of the equation.

$3 x^{2} = 3 - 3$

Subtract $3$ from $3$ .

$3 x^{2} = 0$

Divide each term in $3 x^{2} = 0$ by $3$ and simplify.

Divide each term in $3 x^{2} = 0$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{0}{3}$

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

$x^{2} = \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$x^{2} = 0$

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Evaluate $y$ when $x = 0$ .

Substitute $0$ for $x$ .

$y = - 2 {(0)}^{2} + 3$

Substitute $0$ for $x$ in $y = - 2 {(0)}^{2} + 3$ and solve for $y$ .

Remove parentheses.

$y = - 2 \cdot 0^{2} + 3$

Simplify $- 2 \cdot 0^{2} + 3$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = - 2 \cdot 0 + 3$

Multiply $- 2$ by $0$ .

$y = 0 + 3$

Add $0$ and $3$ .

$y = 3$

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

$(0, 3)$

The result can be shown in multiple forms.

Point Form:

$(0, 3)$

Equation Form:

$x = 0, y = 3$

Solve the System of Equations y=x^2+3 and y=-2x^2+3

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