Question

perform the following operation and express in simplest form.\\(dfrac{3x}{2x + 2} cdot dfrac{x^2 - 1}{x^2 + 6x - 7}\\)

Explanation:

Step1: Factor all expressions

Factor numerators and denominators:
$2x+2=2(x+1)$
$x^2-1=(x+1)(x-1)$
$x^2+6x-7=(x+7)(x-1)$
The expression becomes:
$\frac{3x}{2(x+1)} \cdot \frac{(x+1)(x-1)}{(x+7)(x-1)}$

Step2: Cancel common factors

Cancel $(x+1)$ and $(x-1)$ from numerator and denominator:
$\frac{3x}{2\cancel{(x+1)}} \cdot \frac{\cancel{(x+1)}\cancel{(x-1)}}{(x+7)\cancel{(x-1)}} = \frac{3x}{2} \cdot \frac{1}{x+7}$

Step3: Multiply remaining terms

Multiply the simplified fractions:
$\frac{3x}{2(x+7)}$

Answer:

$\frac{3x}{2(x+7)}$ (where $x
eq -1, 1, -7$ to avoid division by zero)