The problem asks us to simplify the expression $\frac{2}{x^2-25} - \frac{1}{x^2-10x+25}$ and solve the equation $6x^2 = 7x + 3$ by factoring.

1. Problem Description

The problem asks us to simplify the expression $\frac{2}{x^2-25} - \frac{1}{x^2-10x+25}$ and solve the equation $6x^2 = 7x + 3$ by factoring.

2. Solution Steps

Part 1: Simplify the expression.

First, factor the denominators:

$x^2 - 25 = (x-5)(x+5)$

$x^2 - 10x + 25 = (x-5)(x-5) = (x-5)^2$

The expression is:

$\frac{2}{(x-5)(x+5)} - \frac{1}{(x-5)^2}$

To combine the fractions, we need a common denominator, which is $(x-5)^2(x+5)$. So, we multiply the first fraction by $\frac{x-5}{x-5}$ and the second fraction by $\frac{x+5}{x+5}$.

$\frac{2(x-5)}{(x-5)^2(x+5)} - \frac{1(x+5)}{(x-5)^2(x+5)} = \frac{2(x-5) - (x+5)}{(x-5)^2(x+5)} = \frac{2x-10 - x - 5}{(x-5)^2(x+5)} = \frac{x-15}{(x-5)^2(x+5)}$

Therefore, the simplified expression is $\frac{x-15}{(x-5)^2(x+5)}$.

Part 2: Solve the equation by factoring.

$6x^2 = 7x + 3$

$6x^2 - 7x - 3 = 0$

We look for two numbers that multiply to $6(-3) = -18$ and add up to $-7$. Those numbers are $-9$ and $2$.

$6x^2 - 9x + 2x - 3 = 0$

$3x(2x - 3) + 1(2x - 3) = 0$

$(3x + 1)(2x - 3) = 0$

So, $3x + 1 = 0$ or $2x - 3 = 0$.

If $3x + 1 = 0$, then $3x = -1$, so $x = -\frac{1}{3}$.

If $2x - 3 = 0$, then $2x = 3$, so $x = \frac{3}{2}$.

3. Final Answer

Simplified expression: $\frac{x-15}{(x-5)^2(x+5)}$

Solutions to the equation: $x = -\frac{1}{3}, \frac{3}{2}$