Question

Perform the multiplication or division and simplify

$\frac{x^2+2x-3}{x^2-2x-3}\cdot \frac{3-x}{3+x}$

Short answer

The solution is$\frac{1-x}{1+x}$

Step-by-step solution

Step 1. Apply the concept of multiplying rational expressions.

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

$\frac{A}{B}\cdot \frac{C}{D}=\frac{AC}{BD}$

This says that to multiply two fractions, we multiply their numerators and multiply their denominators.

Step 2. Apply the concept of simplifying rational expressions.

To simplify rational expressions, we factor both numerator and denominator and use the following property of fractions:

$\frac{AC}{BC}=\frac{A}{B}$

This allows us to cancel common factors from the numerator and denominator.

Step 3. Multiply the expression.

Given

$\frac{x^2+2x-3}{x^2-2x-3}\cdot \frac{3-x}{3+x}=\frac{\left(x^2+2x-3\right)\left(3-x\right)}{\left(x^2-2x-3\right)\left(3+x\right)}$

Step 4. Simplify the expression.

From step 3,$\frac{\left(x^2+2x-3\right)\left(3-x\right)}{\left(x^2-2x-3\right)\left(3+x\right)}$

This is a rational expression.

To simplify the rational expression, we factor both numerator and denominator and use this$\frac{AC}{BC}=\frac{A}{B}$

Therefore, to simplify a rational expression, we cancel factors that are common to the numerator and denominator.

Consider

$\begin{array}{l}\frac{\left(x^2+2x-3\right)\left(3-x\right)}{\left(x^2-2x-3\right)\left(3+x\right)}=\frac{\left(x^2+3x-x-3\right)\left(3-x\right)}{\left(x^2-3x+x-3\right)\left(3+x\right)}\mathrm{..............}\left[a^2-b^2=\left(a-b\right)\left(a+b\right)\right]\\ \Rightarrow \frac{\left(x^2+2x-3\right)\left(3-x\right)}{\left(x^2-2x-3\right)\left(3+x\right)}=\frac{\left[x\left(x+3\right)-\left(x+3\right)\right]\left(3-x\right)}{\left[x\left(x-3\right)+\left(x-3\right)\right]\left(3+x\right)}\\ \Rightarrow \frac{\left(x^2+2x-3\right)\left(3-x\right)}{\left(x^2-2x-3\right)\left(3+x\right)}=\frac{\left(x-1\right)\left(x+3\right)\left(3-x\right)}{\left(x-3\right)\left(x+1\right)\left(3+x\right)}\\ \Rightarrow \frac{\left(x^2+2x-3\right)\left(3-x\right)}{\left(x^2-2x-3\right)\left(3+x\right)}=\frac{-\left(x-1\right)\left(3+x\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)\left(3+x\right)}\end{array}$

The common factor to the numerator and denominator is$\left(3+x\right)\left(x-3\right)$

So we will cancel the common factor from the numerator and denominator.

The simplified expression is$\frac{-\left(x-1\right)\left(3+x\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)\left(3+x\right)}=\frac{1-x}{1+x}$