Precalculus Examples

$x^{2} + y^{2} = 25$ , $x + y = 8$

Step 1

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

$y^{2} = 25 - x^{2}$

$x + y = 8$

Step 2

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

$y = \pm \sqrt{25 - x^{2}}$

$x + y = 8$

Step 3

Simplify $\pm \sqrt{25 - x^{2}}$ .

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Rewrite $25$ as $5^{2}$ .

$y = \pm \sqrt{5^{2} - x^{2}}$

$x + y = 8$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5$ and $b = x$ .

$y = \pm \sqrt{(5 + x) (5 - x)}$

$x + y = 8$

$y = \pm \sqrt{(5 + x) (5 - x)}$

$x + y = 8$

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.

$y = \sqrt{(5 + x) (5 - x)}$

$x + y = 8$

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

$y = - \sqrt{(5 + x) (5 - x)}$

$x + y = 8$

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

$y = \sqrt{(5 + x) (5 - x)}$

$y = - \sqrt{(5 + x) (5 - x)}$

$x + y = 8$

$y = \sqrt{(5 + x) (5 - x)}$

$y = - \sqrt{(5 + x) (5 - x)}$

$x + y = 8$

Step 5

Subtract $x$ from both sides of the equation.

$y = 8 - x$

$y = \sqrt{(5 + x) (5 - x)}$

$y = - \sqrt{(5 + x) (5 - x)}$

Step 6

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

$x = \frac{8 - i \sqrt{14}}{2}, y = \frac{8 + i \sqrt{14}}{2}$

$x = \frac{8 + \sqrt{14} i}{2}, y = \frac{8 - i \sqrt{14}}{2}$

Step 7