Find the Holes in the Graph y=(x-1)/((x+2)(x-1))

$y = \frac{x - 1}{(x + 2) (x - 1)}$

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$y = \frac{x - 1}{(x + 2) (x - 1)}$

Rewrite the expression.

$y = \frac{1}{x + 2}$

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

$x - 1$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{x + 2}$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Substitute $1$ for $x$ in $\frac{1}{x + 2}$ and simplify.

Substitute $1$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{1 + 2}$

Add $1$ and $2$ .

$\frac{1}{3}$

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

$(1, \frac{1}{3})$