Find the Excluded Values -3/(x+3)=x/(x+3)-x/5

$\frac{- 3}{x + 3} = \frac{x}{x + 3} - \frac{x}{5}$

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

Subtract $\frac{x}{x + 3}$ from both sides of the equation.

$\frac{- 3}{x + 3} - \frac{x}{x + 3} = - \frac{x}{5}$

Add $\frac{x}{5}$ to both sides of the equation.

$\frac{- 3}{x + 3} - \frac{x}{x + 3} + \frac{x}{5} = 0$

Simplify $\frac{- 3}{x + 3} - \frac{x}{x + 3} + \frac{x}{5}$ .

Move the negative in front of the fraction.

$- \frac{3}{x + 3} - \frac{x}{x + 3} + \frac{x}{5} = 0$

Combine the numerators over the common denominator.

$\frac{- 3 - x}{x + 3} + \frac{x}{5} = 0$

Cancel the common factor of $- 3 - x$ and $x + 3$ .

Rewrite $- 3$ as $- 1 (3)$ .

$\frac{- 1 (3) - x}{x + 3} + \frac{x}{5} = 0$

Factor $- 1$ out of $- x$ .

$\frac{- 1 (3) - (x)}{x + 3} + \frac{x}{5} = 0$

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

$\frac{- 1 (3 + x)}{x + 3} + \frac{x}{5} = 0$

Reorder terms.

$\frac{- 1 (x + 3)}{x + 3} + \frac{x}{5} = 0$

Cancel the common factor.

$\frac{- 1 (x + 3)}{x + 3} + \frac{x}{5} = 0$

Divide $- 1$ by $1$ .

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

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 the Excluded Values -3/(x+3)=x/(x+3)-x/5

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