Precalculus Examples

$x^{2} + y^{2} = 13$ , $x^{2} - y^{2} = 5$

Step 1

Simplify the left side.

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Step 1.1

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

$y^{2} + x^{2} = 13$

$x^{2} - y^{2} = 5$

$y^{2} + x^{2} = 13$

$x^{2} - y^{2} = 5$

Step 2

Simplify the left side.

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Step 2.1

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

$- y^{2} + x^{2} = 5$

$y^{2} + x^{2} = 13$

$- y^{2} + x^{2} = 5$

$y^{2} + x^{2} = 13$

Step 3

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

		$y^{2}$	$+$	$x^{2}$	$=$	$1$	$3$
$+$	$-$	$y^{2}$	$+$	$x^{2}$	$=$		$5$
			$2$	$x^{2}$	$=$	$1$	$8$

Step 4

Divide each term in $2 x^{2} = 18$ by $2$ and simplify.

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Step 4.1

Divide each term in $2 x^{2} = 18$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{18}{2}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{18}{2}$

Step 4.2.1.2

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

$x^{2} = \frac{18}{2}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $18$ by $2$ .

$x^{2} = 9$

Step 5

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

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Step 5.1

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

$y^{2} + 9 = 13$

Step 5.2

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

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Step 5.2.1

Subtract $9$ from both sides of the equation.

$y^{2} = 13 - 9$

Step 5.2.2

Subtract $9$ from $13$ .

$y^{2} = 4$

Step 6

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

$x^{2} = 9$

$y^{2} = 4$

Step 7

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

$x = \pm \sqrt{9}$

$y^{2} = 4$

Step 8

Simplify $\pm \sqrt{9}$ .

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Step 8.1

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

$y^{2} = 4$

Step 8.2

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

$x = \pm 3$

$y^{2} = 4$

$x = \pm 3$

$y^{2} = 4$

Step 9

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

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Step 9.1

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

$x = 3$

$y^{2} = 4$

Step 9.2

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

$x = - 3$

$y^{2} = 4$

Step 9.3

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

$x = 3, - 3$

$y^{2} = 4$

$x = 3, - 3$

$y^{2} = 4$

Step 10

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

$y = \pm \sqrt{4}$

$x = 3, - 3$

Step 11

Simplify $\pm \sqrt{4}$ .

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Step 11.1

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

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

$x = 3, - 3$

Step 11.2

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

$y = \pm 2$

$x = 3, - 3$

$y = \pm 2$

$x = 3, - 3$

Step 12

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

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Step 12.1

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

$y = 2$

$x = 3, - 3$

Step 12.2

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

$y = - 2$

$x = 3, - 3$

Step 12.3

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

$y = 2, - 2$

$x = 3, - 3$

$y = 2, - 2$

$x = 3, - 3$

Step 13

The final result is the combination of all values of $x$ with all values of $y$ .

$(3, 2), (3, - 2), (- 3, 2), (- 3, - 2)$

Step 14

The result can be shown in multiple forms.

Point Form:

$(3, 2), (3, - 2), (- 3, 2), (- 3, - 2)$

Equation Form:

$x = 3, y = 2$

$x = 3, y = - 2$

$x = - 3, y = 2$

$x = - 3, y = - 2$

Step 15