Question

1. $\frac{x^{2}-8x}{3x^{2}+8x - 3}cdot\frac{6x - 2}{16 - 2x}$

Explanation:

Step1: Factor the expressions

Factor $x^{2}-8x=x(x - 8)$, $3x^{2}+8x - 3=(3x - 1)(x + 3)$, $6x-2 = 2(3x - 1)$ and $16-2x=2(8 - x)=-2(x - 8)$.

Step2: Rewrite the product of fractions

The original expression $\frac{x^{2}-8x}{3x^{2}+8x - 3}\cdot\frac{6x - 2}{16-2x}$ becomes $\frac{x(x - 8)}{(3x - 1)(x + 3)}\cdot\frac{2(3x - 1)}{-2(x - 8)}$.

Step3: Simplify the fraction

Cancel out the common factors $(x - 8)$ and $(3x - 1)$ and simplify the constants $\frac{2}{-2}$. We get $\frac{x}{x + 3}\cdot(-1)=-\frac{x}{x + 3}$.

Answer:

$-\frac{x}{x + 3}$