Find the Holes in the Graph y=((3x+1)(x-1))/(x-1)

$y = \frac{(3 x + 1) (x - 1)}{x - 1}$

Step 1

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$y = \frac{(3 x + 1) (x - 1)}{x - 1}$

Divide $3 x + 1$ by $1$ .

$y = 3 x + 1$

Step 2

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

$x - 1$

Step 3

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $3 x + 1$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Substitute $1$ for $x$ in $3 x + 1$ and simplify.

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

$3 \cdot 1 + 1$

Simplify.

Multiply $3$ by $1$ .

$3 + 1$

Add $3$ and $1$ .

$4$

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

$(1, 4)$

Step 4