Solve by Addition/Elimination x^2-y^2=4 x^2+y^2=36

$x^{2} - y^{2} = 4$ $x^{2} + y^{2} = 36$

Simplify the left side.

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

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

$x^{2} + y^{2} = 36$

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

$x^{2} + y^{2} = 36$

Simplify the left side.

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

$y^{2} + x^{2} = 36$

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

$y^{2} + x^{2} = 36$

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

Add the two equations together to eliminate $y^{2}$ from the system.

	$-$	$y^{2}$	$+$	$x^{2}$	$=$		$4$
$+$		$y^{2}$	$+$	$x^{2}$	$=$	$3$	$6$
			$2$	$x^{2}$	$=$	$4$	$0$

Divide each term in $2 x^{2} = 40$ by $2$ and simplify.

Divide each term in $2 x^{2} = 40$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{40}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{40}{2}$

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

$x^{2} = \frac{40}{2}$

Simplify the right side.

Divide $40$ by $2$ .

$x^{2} = 20$

Substitute the value found for $x^{2}$ into one of the original equations, then solve for $y^{2}$ .

Substitute the value found for $x^{2}$ into one of the original equations to solve for $y^{2}$ .

$- y^{2} + 20 = 4$

Move all terms not containing $y^{2}$ to the right side of the equation.

Subtract $20$ from both sides of the equation.

$- y^{2} = 4 - 20$

Subtract $20$ from $4$ .

$- y^{2} = - 16$

Divide each term in $- y^{2} = - 16$ by $- 1$ and simplify.

Divide each term in $- y^{2} = - 16$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{- 16}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{- 16}{- 1}$

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

$y^{2} = \frac{- 16}{- 1}$

Simplify the right side.

Divide $- 16$ by $- 1$ .

$y^{2} = 16$

This is the final solution to the independent system of equations.

$x^{2} = 20$

$y^{2} = 16$

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

$x = \pm \sqrt{20}$

$y^{2} = 16$

Simplify $\pm \sqrt{20}$ .

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \pm \sqrt{4 (5)}$

$y^{2} = 16$

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

$x = \pm \sqrt{2^{2} \cdot 5}$

$y^{2} = 16$

$x = \pm \sqrt{2^{2} \cdot 5}$

$y^{2} = 16$

Pull terms out from under the radical.

$x = \pm 2 \sqrt{5}$

$y^{2} = 16$

$x = \pm 2 \sqrt{5}$

$y^{2} = 16$

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.

$x = 2 \sqrt{5}$

$y^{2} = 16$

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

$x = - 2 \sqrt{5}$

$y^{2} = 16$

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