The inequality provided in the image is:

$\frac{x - 1}{2x + 5} > \frac{-3x}{5 - 2x} - \frac{-2x}{2x - 5}$

To solve this inequality, let's proceed step by step:

Step 1: Simplify the right-hand side

We first need to combine the fractions on the right-hand side.

The right-hand side is:

$\frac{-3x}{5 - 2x} - \frac{-2x}{2x - 5}$

Notice that $5 - 2x$ and $2x - 5$ are negatives of each other:

$5 - 2x = -(2x - 5)$

Thus, we can rewrite the right-hand side as:

$\frac{-3x}{5 - 2x} - \frac{-2x}{-(5 - 2x)} = \frac{-3x}{5 - 2x} + \frac{2x}{5 - 2x}$

Combine the fractions:

$\frac{-3x + 2x}{5 - 2x} = \frac{-x}{5 - 2x}$

Step 2: Rewrite the inequality

Now, the inequality becomes:

$\frac{x - 1}{2x + 5} > \frac{-x}{5 - 2x}$

Step 3: Cross-multiply (taking into account sign changes)

To solve the inequality, we can cross-multiply. However, be careful: the sign of the inequality might change depending on the sign of the expressions $2x + 5$ and $5 - 2x$.

• $2x + 5$ is positive when $x > -\frac{5}{2}$.
• $5 - 2x$ is positive when $x < \frac{5}{2}$.

So, we need to consider the intervals $x < -\frac{5}{2}$, $-\frac{5}{2} < x < \frac{5}{2}$, and $x > \frac{5}{2}$.

Case 1: $x < -\frac{5}{2}$

In this case, both $2x + 5$ and $5 - 2x$ are negative, so we multiply by their absolute values and flip the inequality:

$-(x - 1)(5 - 2x) < -x(2x + 5)$

Simplifying this, we get:

$(x - 1)(5 - 2x) > x(2x + 5)$

Expanding both sides:

$5x - 2x^2 - 5 + 2x > 2x^2 + 5x$

Combine like terms:

$-4x^2 + 7x - 5 > 0$

Case 2: $-\frac{5}{2} < x < \frac{5}{2}$

In this interval, $2x + 5 > 0$ and $5 - 2x > 0$, so we do not flip the inequality:

$(x - 1)(5 - 2x) > -x(2x + 5)$

Expanding:

$5x - 2x^2 - 5 + 2x > -2x^2 - 5x$

Combine like terms:

$-4x^2 + 7x - 5 > 0$

This is the same quadratic as above.

Case 3: $x > \frac{5}{2}$

Here, both $2x + 5 > 0$ and $5 - 2x < 0$, so we flip the inequality:

$(x - 1)(5 - 2x) < -x(2x + 5)$

Expanding:

$5x - 2x^2 - 5 + 2x < -2x^2 - 5x$

This simplifies to:

$-4x^2 + 7x - 5 < 0$

Conclusion

We now need to solve the quadratic inequality $-4x^2 + 7x - 5 > 0$ and $-4x^2 + 7x - 5 < 0$ in the respective intervals.

Would you like to proceed with the solution for the quadratic inequality, or is there any specific step you'd like to delve into further?