Precalculus Examples

$x^{2} + y^{2} = 9$ , $x^{2} + y = 3$

Step 1

Subtract $x^{2}$ from both sides of the equation.

$y^{2} = 9 - x^{2}$

$x^{2} + y = 3$

Step 2

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

$y = \pm \sqrt{9 - x^{2}}$

$x^{2} + y = 3$

Step 3

Simplify $\pm \sqrt{9 - x^{2}}$ .

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

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

$x^{2} + y = 3$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = x$ .

$y = \pm \sqrt{(3 + x) (3 - x)}$

$x^{2} + y = 3$

$y = \pm \sqrt{(3 + x) (3 - x)}$

$x^{2} + y = 3$

Step 4

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

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

$y = \sqrt{(3 + x) (3 - x)}$

$x^{2} + y = 3$

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

$y = - \sqrt{(3 + x) (3 - x)}$

$x^{2} + y = 3$

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

$y = \sqrt{(3 + x) (3 - x)}$

$y = - \sqrt{(3 + x) (3 - x)}$

$x^{2} + y = 3$

$y = \sqrt{(3 + x) (3 - x)}$

$y = - \sqrt{(3 + x) (3 - x)}$

$x^{2} + y = 3$

Step 5

Subtract $x^{2}$ from both sides of the equation.

$y = 3 - x^{2}$

$y = \sqrt{(3 + x) (3 - x)}$

$y = - \sqrt{(3 + x) (3 - x)}$

Step 6

Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution.

$(0, 3)$

$(\sqrt{5}, - 2)$

$(- \sqrt{5}, - 2)$

Step 7