Find the Excluded Values 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 $1$ .

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

Step 3

Set the denominator in $\frac{3 (x - 1)}{x + 1}$ equal to $0$ to find where the expression is undefined.

$x + 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5