Question

In the following exercise solve$\frac{v-15}{v^2-9v+18}=\frac{4}{v-3}+\frac{2}{v-6}$

Short answer

The solution is$v=3$

Step-by-step solution

Step 1. Given Information

$\frac{v-15}{v^2-9v+18}=\frac{4}{v-3}+\frac{2}{v-6}$

Step 2. Solve.

The least common denominator of all denominators in the equation is $\left(v-3\right)\left(v-6\right)$

Clear the fraction by multiplying both sides of the equation by the LCD.

$$\begin{gathered} \left(v-3\right)\left(v-6\right)\left(\frac{v-15}{v^2-9v+18}\right)=\left(v-3\right)\left(v-6\right)\left(\frac{4}{v-3}+\frac{2}{v-6}\right) \\ \left(v-3\right)\left(v-6\right)\left(\frac{v-15}{v^2-9v+18}\right)=\left(v-3\right)\left(v-6\right)\left(\frac{4}{v-3}\right)+\left(v-3\right)\left(v-6\right)\left(\frac{2}{v-6}\right) \\ \left(v-15\right)=\left(v-6\right)4+\left(v-3\right)2 \\ v-15=6v-30 \\ 5v=15 \\ v=3 \end{gathered}$$