Question

Simplify the complex rational expression by writing it as division:

$\frac{1-{\displaystyle \frac{3}{c+4}}}{{\displaystyle \frac{1}{c+4}}+{\displaystyle \frac{c}{3}}}$

Short answer

$\frac{3}{c+3}$

Step-by-step solution

Step 1. Given information.

The following complex rational expression is given:

$\frac{1-{\displaystyle \frac{3}{c+4}}}{{\displaystyle \frac{1}{c+4}+\frac{c}{3}}}$

Step 2. Simplification of the complex rational expression.

Find common denominators and simplify the numerator and the denominator:

$$\begin{gathered} \frac{{\displaystyle \frac{1\left(c+4\right)}{\left(c+4\right)}}-{\displaystyle \frac{3}{c+4}}}{{\displaystyle \frac{1.3}{3\left(c+4\right)}+\frac{c\left(c+4\right)}{3\left(c+4\right)}}} \\ \frac{{\displaystyle \frac{c+4-3}{\left(c+4\right)}}}{{\displaystyle \frac{3+c\left(c+4\right)}{3\left(c+4\right)}}} \\ \frac{{\displaystyle \frac{\left(c+1\right)}{\left(c+4\right)}}}{{\displaystyle \frac{3+c^2+4c}{3\left(c+4\right)}}} \\ \frac{{\displaystyle \frac{\left(c+1\right)}{\left(c+4\right)}}}{{\displaystyle \frac{c^2+4c+3}{3\left(c+4\right)}}} \end{gathered}$$

Rewrite as a division problem:

$\frac{\left(c+1\right)}{\left(c+4\right)}÷\frac{c^2+4c+3}{3\left(c+4\right)}$

Rewrite as a product of the first with the reciprocal of the second:

$\frac{\left(c+1\right)}{\left(c+4\right)}.\frac{3\left(c+4\right)}{c^2+4c+3}$

Factor the numerator and the denominator terms (where applicable), cancel the common factors and simplify:

$$\begin{gathered} \frac{\left(c+1\right)}{\left(c+4\right)}.\frac{3\left(c+4\right)}{\left(c+3\right)\left(c+1\right)} \\ \frac{3}{c+3} \end{gathered}$$