Find Where Undefined/Discontinuous (7x)/(x+1)=5-7/(x+1)

$\frac{7 x}{x + 1} = 5 - \frac{7}{x + 1}$

Step 1

Move all the expressions to the left side of the equation.

Subtract $5$ from both sides of the equation.

$\frac{7 x}{x + 1} - 5 = - \frac{7}{x + 1}$

Add $\frac{7}{x + 1}$ to both sides of the equation.

$\frac{7 x}{x + 1} - 5 + \frac{7}{x + 1} = 0$

Step 2

Simplify $\frac{7 x}{x + 1} - 5 + \frac{7}{x + 1}$ .

Find the common denominator.

Write $- 5$ as a fraction with denominator $1$ .

$\frac{7 x}{x + 1} + \frac{- 5}{1} + \frac{7}{x + 1} = 0$

Multiply $\frac{- 5}{1}$ by $\frac{x + 1}{x + 1}$ .

$\frac{7 x}{x + 1} + \frac{- 5}{1} \cdot \frac{x + 1}{x + 1} + \frac{7}{x + 1} = 0$

Multiply $\frac{- 5}{1}$ by $\frac{x + 1}{x + 1}$ .

$\frac{7 x}{x + 1} + \frac{- 5 (x + 1)}{x + 1} + \frac{7}{x + 1} = 0$

Combine the numerators over the common denominator.

$\frac{7 x - 5 (x + 1) + 7}{x + 1} = 0$

Simplify each term.

Apply the distributive property.

$\frac{7 x - 5 x - 5 \cdot 1 + 7}{x + 1} = 0$

Multiply $- 5$ by $1$ .

$\frac{7 x - 5 x - 5 + 7}{x + 1} = 0$

Subtract $5 x$ from $7 x$ .

$\frac{2 x - 5 + 7}{x + 1} = 0$

Add $- 5$ and $7$ .

$\frac{2 x + 2}{x + 1} = 0$

Factor $2$ out of $2 x + 2$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (x) + 2 (1)$ .

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

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Divide $2$ by $1$ .

$2 = 0$

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Find Where Undefined/Discontinuous (7x)/(x+1)=5-7/(x+1)

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