The equation in the image is:

$\frac{6}{5x - 3} = \frac{5}{x + 1}$

To solve for $x$, follow these steps:

1. Cross-multiply the terms to eliminate the fractions:

$6(x + 1) = 5(5x - 3)$

2. Expand both sides:

$6x + 6 = 25x - 15$

3. Move all terms involving $x$ to one side and constants to the other side:

$6x - 25x = -15 - 6$

$-19x = -21$

4. Solve for $x$ by dividing both sides by $-19$:

$x = \frac{-21}{-19} = \frac{21}{19}$

Thus, the solution is:

$x = \frac{21}{19}$

Would you like more details, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. How would you solve if one side had a quadratic term in the denominator?
2. What happens if there is no real solution to an equation like this?
3. How can you verify the solution by plugging the value of $x$ back into the original equation?
4. How do cross-multiplication and clearing denominators help in solving fractional equations?
5. Can you solve this equation using substitution instead of cross-multiplication?

Tip: Always check for extraneous solutions when solving equations involving fractions, as some values may make the denominator zero.