We need to simplify the expression: $\frac{3x-1}{x^2+4x} + \frac{7-8x}{x^2+5x+4}$.

1. Problem Description

We need to simplify the expression:

$\frac{3x-1}{x^2+4x} + \frac{7-8x}{x^2+5x+4}$.

2. Solution Steps

First, we factor the denominators:

$x^2 + 4x = x(x+4)$

$x^2 + 5x + 4 = (x+1)(x+4)$

So the expression becomes:

$\frac{3x-1}{x(x+4)} + \frac{7-8x}{(x+1)(x+4)}$

To add the two fractions, we need to find a common denominator. The least common denominator is $x(x+1)(x+4)$. Thus, we multiply the first fraction by $\frac{x+1}{x+1}$ and the second fraction by $\frac{x}{x}$:

$\frac{(3x-1)(x+1)}{x(x+1)(x+4)} + \frac{(7-8x)x}{x(x+1)(x+4)}$

$= \frac{(3x^2+3x-x-1)}{x(x+1)(x+4)} + \frac{7x-8x^2}{x(x+1)(x+4)}$

$= \frac{3x^2+2x-1}{x(x+1)(x+4)} + \frac{7x-8x^2}{x(x+1)(x+4)}$

$= \frac{3x^2+2x-1 + 7x-8x^2}{x(x+1)(x+4)}$

$= \frac{-5x^2+9x-1}{x(x+1)(x+4)}$

$= \frac{-5x^2+9x-1}{x(x^2+5x+4)}$

$= \frac{-5x^2+9x-1}{x^3+5x^2+4x}$

3. Final Answer

$\frac{-5x^2+9x-1}{x(x+1)(x+4)} = \frac{-5x^2+9x-1}{x^3+5x^2+4x}$