Precalculus Examples

$y = x + 5$ , $y = \frac{x^{2} - 25}{x - 5}$

Step 1

Simplify $\frac{x^{2} - 25}{x - 5}$ .

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Step 1.1

Simplify the numerator.

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Step 1.1.1

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

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

$y = x + 5$

Step 1.1.2

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

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

$y = x + 5$

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

$y = x + 5$

Step 1.2

Cancel the common factor of $x - 5$ .

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Step 1.2.1

Cancel the common factor.

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

$y = x + 5$

Step 1.2.2

Divide $x + 5$ by $1$ .

$y = x + 5$

Step 2

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

Step 3

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$y = x + 5, y = \frac{x^{2} - 25}{x - 5}$

Step 4