Find the Holes in the Graph (-5x+15)/(3x-9)

$\frac{- 5 x + 15}{3 x - 9}$

Factor $5$ out of $- 5 x + 15$ .

Factor $5$ out of $- 5 x$ .

$\frac{5 (- x) + 15}{3 x - 9}$

Factor $5$ out of $15$ .

$\frac{5 (- x) + 5 (3)}{3 x - 9}$

Factor $5$ out of $5 (- x) + 5 (3)$ .

$\frac{5 (- x + 3)}{3 x - 9}$

Factor $3$ out of $3 x - 9$ .

Factor $3$ out of $3 x$ .

$\frac{5 (- x + 3)}{3 (x) - 9}$

Factor $3$ out of $- 9$ .

$\frac{5 (- x + 3)}{3 x + 3 \cdot - 3}$

Factor $3$ out of $3 x + 3 \cdot - 3$ .

$\frac{5 (- x + 3)}{3 (x - 3)}$

Cancel the common factor of $- x + 3$ and $x - 3$ .

Factor $- 1$ out of $- x$ .

$\frac{5 (- (x) + 3)}{3 (x - 3)}$

Rewrite $3$ as $- 1 (- 3)$ .

$\frac{5 (- (x) - 1 (- 3))}{3 (x - 3)}$

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

$\frac{5 (- (x - 3))}{3 (x - 3)}$

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

$\frac{5 (- 1 (x - 3))}{3 (x - 3)}$

Cancel the common factor.

$\frac{5 (- 1 (x - 3))}{3 (x - 3)}$

Rewrite the expression.

$\frac{5 \cdot (- 1)}{3}$

Simplify the numerator.

Multiply $5$ by $- 1$ .

$\frac{5 (- 1)}{3}$

Multiply $5$ by $- 1$ .

$\frac{- 5}{3}$

Move the negative in front of the fraction.

$- \frac{5}{3}$

To find the holes in the graph, look at the denominator factors that were cancelled.

$x - 3$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{5}{3}$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Because the cancelled expression is constant with respect to $x$ , it is the $y$ coordinate for every hole.

$- \frac{5}{3}$

The holes in the graph are the points where any of the cancelled factors are equal to $0$ .

$(3, - \frac{5}{3})$

Find the Holes in the Graph (-5x+15)/(3x-9)

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