Question

Find$R\left(x\right)=f\left(x\right)-g\left(x\right)$where$f\left(x\right)=\frac{x+1}{x+3}$and$g\left(x\right)=\frac{x+17}{x^2-x-12}$.

Short answer

The resultant expression is$\frac{x-7}{x-4}$.

Step-by-step solution

Step 1. Given Information

The functions are$f\left(x\right)=\frac{x+1}{x+3},g\left(x\right)=\frac{x+17}{x^2-x-12}$

Step 2. Determine the expression

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

$\frac{x+1}{x+3}-\frac{x+17}{x^2-x-12}$

Step 3. Express with common denominator    

• Factor the denominators of the given expression.

$\frac{x+1}{x+3}-\frac{x+17}{(x+3)(x-4)}$

• The LCD of the terms of obtained expression is $(x+3)(x-4)$.
• Multiply the numerator and denominator of first term by $(x-4)$.

$\begin{array}{rcl}\frac{(x+1)(x-4)}{(x+3)(x-4)}-\frac{x+17}{(x+3)(x-4)}& =& \frac{x^2-4x+x-4}{(x+3)(x-4)}-\frac{x+17}{(x+3)(x-4)}\\ & =& \frac{x^2-3x-4}{(x+3)(x-4)}-\frac{x+17}{(x+3)(x-4)}\end{array}$

Step 4. Simplify

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

$\begin{array}{rcl}& & \frac{x^2-3x-4-x-17}{(x+3)(x-4)}\end{array}=\begin{array}{rcl}& & \frac{x^2-4x-21}{(x+3)(x-4)}\end{array}$

• Factor the numerator and cancel the common factors from numerator and denominator.

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

• So, the resultant expression is$\frac{x-7}{x-4}$.