Question

Simplify the compound fractional expression.

$\frac{\frac{x-3}{x-4}-\frac{x+2}{x+1}}{x+3}$

Short answer

The solution is$\frac{5}{\left(x-4\right)\left(x+1\right)\left(x+3\right)}$

Step-by-step solution

Step 1. Apply the concept of addition of rational expressions.

To add or subtract rational expressions, we first find a common denominator and then use the following property of fractions:

$\frac{A}{C}+\frac{B}{C}=\frac{A+B}{C}$

The LCD is found by factoring each denominator and taking the product of the distinct factors, using the highest power that appears in any of the factors.

Step 2. Apply the concept of dividing rational expressions.

To divide rational expressions, we use the following property of fractions:

$\frac{A}{B}÷\frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C}=\frac{AD}{BC}$

Step 3. Analyze the expression.

Given

$\frac{\frac{x-3}{x-4}-\frac{x+2}{x+1}}{x+3}$

The LCD of terms in numerator is$\left(x-4\right)\left(x+1\right)$

So, we have:

$\begin{array}{l}\frac{\frac{x-3}{x-4}-\frac{x+2}{x+1}}{x+3}=\frac{\frac{\left(x-3\right)\left(x+1\right)}{\left(x-4\right)\left(x+1\right)}-\frac{\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+1\right)}}{x+3}\\ \Rightarrow \frac{\frac{x-3}{x-4}-\frac{x+2}{x+1}}{x+3}=\frac{\frac{x^2-2x-3}{\left(x-4\right)\left(x+1\right)}-\frac{x^2-2x-8}{\left(x-4\right)\left(x+1\right)}}{x+3}\\ \Rightarrow \frac{\frac{x-3}{x-4}-\frac{x+2}{x+1}}{x+3}=\frac{\frac{x^2-2x-3-x^2+2x+8}{\left(x-4\right)\left(x+1\right)}}{x+3}\\ \Rightarrow \frac{\frac{x-3}{x-4}-\frac{x+2}{x+1}}{x+3}=\frac{\frac{5}{\left(x-4\right)\left(x+1\right)}}{x+3}\end{array}$

Step 4. Simplify the expression.

Consider$\frac{\frac{5}{\left(x-4\right)\left(x+1\right)}}{x+3}$

So, the simplified expression is given by:

$\begin{array}{l}\frac{\frac{5}{\left(x-4\right)\left(x+1\right)}}{x+3}=\frac{5}{\left(x-4\right)\left(x+1\right)}\cdot \frac{1}{x+3}\\ \Rightarrow \frac{\frac{5}{\left(x-4\right)\left(x+1\right)}}{x+3}=\frac{5}{\left(x-4\right)\left(x+1\right)\left(x+3\right)}\end{array}$