Simplify the given expression: $\frac{20x^2 - 13xy - 15y^2}{25y^2 - 16x^2}$.

1. Problem Description

Simplify the given expression: $\frac{20x^2 - 13xy - 15y^2}{25y^2 - 16x^2}$.

2. Solution Steps

First, factor the numerator $20x^2 - 13xy - 15y^2$. We are looking for two binomials of the form $(ax + by)(cx + dy)$ such that $ac = 20$, $bd = -15$, and $ad + bc = -13$.

We can try $a = 4$ and $c = 5$. Then we need $4d + 5b = -13$ and $bd = -15$. If we choose $b = 3$ and $d = -5$, then $4(-5) + 5(3) = -20 + 15 = -5$, which is not $-13$.

If we choose $b = -3$ and $d = 5$, then $4(5) + 5(-3) = 20 - 15 = 5$, which is not $-13$.

Let's try $a = 5$ and $c = 4$. Then we need $5d + 4b = -13$ and $bd = -15$. If we choose $b = -3$ and $d = 5$, then $5(5) + 4(-3) = 25 - 12 = 13$. This is the opposite sign, so we can change the signs of $b$ and $d$.

Try $b = 3$ and $d = -5$. Then $5(-5) + 4(3) = -25 + 12 = -13$. This works.

So, $20x^2 - 13xy - 15y^2 = (5x + 3y)(4x - 5y)$.

Next, factor the denominator $25y^2 - 16x^2$. This is a difference of squares, so we have:

$25y^2 - 16x^2 = (5y - 4x)(5y + 4x) = -(4x - 5y)(4x + 5y)$.

Therefore, the expression becomes:

$\frac{(5x + 3y)(4x - 5y)}{-(4x - 5y)(4x + 5y)} = \frac{(5x + 3y)(4x - 5y)}{-(4x - 5y)(4x + 5y)}$.

We can cancel the common factor $(4x - 5y)$ from the numerator and denominator, assuming $4x - 5y \neq 0$:

$\frac{5x + 3y}{-(4x + 5y)} = -\frac{5x + 3y}{4x + 5y}$.

3. Final Answer

$-\frac{5x+3y}{4x+5y}$