The equation shown in the image is:

$\frac{1}{x^2 - x - 6} + \frac{3}{x + 2} = \frac{-4}{x - 3}$

Step-by-step solution:

1. Factor the quadratic expression $x^2 - x - 6$: $x^2 - x - 6 = (x - 3)(x + 2)$ So the equation becomes: $\frac{1}{(x - 3)(x + 2)} + \frac{3}{x + 2} = \frac{-4}{x - 3}$

2. Find a common denominator for the terms on the left-hand side:

   • The first term already has the denominator $(x - 3)(x + 2)$.
   • The second term can be rewritten to have the same denominator. Multiply the numerator and denominator by $(x - 3)$: $\frac{3}{x + 2} = \frac{3(x - 3)}{(x - 3)(x + 2)} = \frac{3x - 9}{(x - 3)(x + 2)}$

   Now the equation becomes: $\frac{1 + (3x - 9)}{(x - 3)(x + 2)} = \frac{-4}{x - 3}$ Simplify the numerator on the left-hand side: $\frac{3x - 8}{(x - 3)(x + 2)} = \frac{-4}{x - 3}$

3. Eliminate the denominator by multiplying both sides by $(x - 3)$, assuming $x \neq 3$: $\frac{3x - 8}{x + 2} = -4$

4. Solve the resulting equation: Multiply both sides by $x + 2$ to eliminate the denominator: $3x - 8 = -4(x + 2)$ Expand the right-hand side: $3x - 8 = -4x - 8$

5. Simplify and solve for $x$: Add $4x$ to both sides: $7x - 8 = -8$ Add 8 to both sides: $7x = 0$ Divide by 7: $x = 0$

Final solution:

The solution is $x = 0$, with the restriction that $x \neq 3$ and $x \neq -2$ because these values would make the original denominators zero.

Would you like to see any more details or further clarifications?

Here are 5 related questions:

1. What happens if we substitute $x = 3$ or $x = -2$ in the original equation?
2. Can you explain why we need a common denominator to solve the equation?
3. How do you factor quadratic expressions like $x^2 - x - 6$?
4. Why do we need to check for restrictions on $x$ when solving rational equations?
5. How do you multiply and simplify rational expressions in algebra?

Tip: Always check for restrictions on variables in rational equations to avoid dividing by zero, which leads to undefined expressions.