Question

Perform the indicated operation and simplify.

$\frac{-3a}{3a-3}+\frac{5a}{a^2+3a-4}$

Short answer

The solution is $\frac{-a}{a+4}$.

Step-by-step solution

Step 1. Given information.

The given expression is$\frac{-3a}{3a-3}+\frac{5a}{a^2+3a-4}$.

Step 2. Simplify the expression.

$$\begin{gathered} \frac{-3a}{3a-3}+\frac{5a}{a^2+3a-4}=\frac{-3a}{3\left(a-1\right)}+\frac{5a}{a^2+4a-a-4} \\ \frac{-3a}{3a-3}+\frac{5a}{a^2+3a-4}=\frac{-a}{a-1}+\frac{5a}{a\left(a+4\right)-1\left(a+4\right)} \\ \frac{-3a}{3a-3}+\frac{5a}{a^2+3a-4}=\frac{-a}{a-1}+\frac{5a}{\left(a-1\right)\left(a+4\right)} \\ \frac{-3a}{3a-3}+\frac{5a}{a^2+3a-4}=\frac{-a\left(a+4\right)+5a}{\left(a-1\right)\left(a+4\right)} \\ \frac{-3a}{3a-3}+\frac{5a}{a^2+3a-4}=\frac{-a^2-4a+5a}{\left(a-1\right)\left(a+4\right)} \\ \frac{-3a}{3a-3}+\frac{5a}{a^2+3a-4}=\frac{-a^2+a}{\left(a-1\right)\left(a+4\right)} \\ \frac{-3a}{3a-3}+\frac{5a}{a^2+3a-4}=\frac{-a\left(a-1\right)}{\left(a-1\right)\left(a+4\right)}(Removethecommonfactor) \\ \frac{-3a}{3a-3}+\frac{5a}{a^2+3a-4}=\frac{-a}{a+4} \end{gathered}$$

Step 3. Simplified answer.

Hence, the simplified expression is$\frac{-a}{a+4}$.