Write as a Set of Linear Factors 1/(x-1)+3=7/(x-1)

$\frac{1}{x - 1} + 3 = \frac{7}{x - 1}$

Step 1

Subtract $\frac{7}{x - 1}$ from both sides of the equation.

$\frac{1}{x - 1} + 3 - \frac{7}{x - 1} = 0$

Step 2

Simplify $\frac{1}{x - 1} + 3 - \frac{7}{x - 1}$ .

Combine the numerators over the common denominator.

$3 + \frac{1 - 7}{x - 1} = 0$

Subtract $7$ from $1$ .

$3 + \frac{- 6}{x - 1} = 0$

Move the negative in front of the fraction.

$3 - \frac{6}{x - 1} = 0$

To write $3$ as a fraction with a common denominator, multiply by $\frac{x - 1}{x - 1}$ .

$\frac{3 (x - 1)}{x - 1} - \frac{6}{x - 1} = 0$

Combine the numerators over the common denominator.

$\frac{3 (x - 1) - 6}{x - 1} = 0$

Simplify the numerator.

Factor $3$ out of $3 (x - 1) - 6$ .

Factor $3$ out of $- 6$ .

$\frac{3 (x - 1) + 3 \cdot - 2}{x - 1} = 0$

Factor $3$ out of $3 (x - 1) + 3 \cdot - 2$ .

$\frac{3 (x - 1 - 2)}{x - 1} = 0$

Subtract $2$ from $- 1$ .

$\frac{3 (x - 3)}{x - 1} = 0$