Question

Find$R\left(x\right)=f\left(x\right)+g\left(x\right)$where$f\left(x\right)=\frac{x-4}{x+3}$and$g\left(x\right)=\frac{4x+6}{x^2-9}$.

Short answer

The resultant expression is$\frac{x^2-3x+18}{(x-3)(x+3)}$.

Step-by-step solution

Step 1. Given Information

The given functions are$f\left(x\right)=\frac{x-4}{x+3},g\left(x\right)=\frac{4x+6}{x^2-9}$.

Step 2. Determine the expression

Substitute the given functions into $R\left(x\right)=f\left(x\right)+g\left(x\right)$.

$R\left(x\right)=\frac{x-4}{x+3}+\frac{4x+6}{x^2-9}$

Step 3. Express with common denominator    

• Factor the denominators of the obtained expression.

$\frac{x-4}{x+3}+\frac{4x+6}{(x-3)(x+3)}$

• The LCD of the terms of obtained expression is $(x-3)(x+3)$.
• Multiply the numerator and denominator of first term by $(x-3)$.

$\begin{array}{rcl}\frac{(x-4)(x-3)}{(x+3)(x-3)}+\frac{4x+6}{(x-3)(x+3)}& =& \frac{x^2-4x-3x+12}{(x+3)(x-3)}+\frac{4x+6}{(x-3)(x+3)}\\ & =& \frac{x^2-7x+12}{(x+3)(x-3)}+\frac{4x+6}{(x-3)(x+3)}\end{array}$

Step 4. Simplify

• As the denominator is the same for both the terms, perform addition in the numerator and place it over the common denominator.

$\begin{array}{rcl}& & \frac{x^2-7x+12+4x+6}{(x+3)(x-3)}\end{array}=\begin{array}{rcl}& & \frac{x^2-3x+18}{(x+3)(x-3)}\end{array}$

• So, the resultant expression is $\begin{array}{rcl}& & \frac{x^2-3x+18}{(x+3)(x-3)}\end{array}$.