Chapter 11.4, Problem 20MCQ

To determine

Add or subtract $\frac{2x}{x-5}+\frac{9}{x+4}$

Answer to Problem 20MCQ

  $\frac{2x^2+17x-45}{\left(x-5\right)\left(x+4\right)}$

Explanation of Solution

Given:

  $\frac{2x}{x-5}+\frac{9}{x+4}$ ,

Concept used:

The LCD (Least Common Denominator) is the smallest number that is divisible by all the denominators.

Calculation:

The LCD (Least Common Denominator) is the smallest number that is divisible by all the denominators.

The given expression can be written as

  $\frac{2x}{x-5}+\frac{9}{x+4}$

Here observe that the denominators are not same, so first we will make both denominators equal to LCD. The LCD of $\left(x-5\right)\text{and}\left(x+4\right)$ is $\left(x-5\right)\left(x+4\right)$ .

So in order to subtract the fractions, we make both the denominators equal to LCD by multiplying the first fraction by $\left(x+4\right)$ up and down and second fraction by $\left(x-5\right)$ . And then when denominators are equal, we directly subtract the numerators and put over the common denominator. So, we get

  $\begin{array}{c}\frac{2x}{x-5}+\frac{9}{x+4}=\frac{2x}{x-5}\cdot \frac{\left(x+4\right)}{\left(x+4\right)}+\frac{9}{x+4}\cdot \frac{\left(x-5\right)}{\left(x-5\right)}\\ =\frac{2x\cdot x+2x\cdot 4}{\left(x-5\right)\left(x+4\right)}+\frac{9\cdot x+9\cdot \left(-5\right)}{\left(x-5\right)\left(x+4\right)}\\ =\frac{2x^2+8x+9x-45}{\left(x-5\right)\left(x+4\right)}\\ =\frac{2x^2+17x-45}{\left(x-5\right)\left(x+4\right)}\end{array}$

Thus, the given expression simplifies to $\frac{2x^2+17x-45}{\left(x-5\right)\left(x+4\right)}$ .