Let's solve the rational equation step by step.

The given equation is:

$\frac{3}{y + 3} + \frac{2}{y - 2} = \frac{y^2}{y^2 + y - 6}$

Step 1: Factor the denominator on the right-hand side

We will first factor the quadratic expression in the denominator on the right-hand side:

$y^2 + y - 6 = (y + 3)(y - 2)$

Thus, the equation becomes:

$\frac{3}{y + 3} + \frac{2}{y - 2} = \frac{y^2}{(y + 3)(y - 2)}$

Step 2: Determine the Least Common Denominator (LCD)

The LCD for the fractions on both sides of the equation is $(y + 3)(y - 2)$. This is because it is the product of the factors in the denominators.

Step 3: Rewrite each term with the LCD

We now rewrite each fraction so that all terms have the same denominator, $(y + 3)(y - 2)$:

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

Substitute these back into the equation:

$\frac{3(y - 2)}{(y + 3)(y - 2)} + \frac{2(y + 3)}{(y + 3)(y - 2)} = \frac{y^2}{(y + 3)(y - 2)}$

Step 4: Combine the numerators

Now that all the fractions have the same denominator, we can combine the numerators on the left-hand side:

$\frac{3(y - 2) + 2(y + 3)}{(y + 3)(y - 2)} = \frac{y^2}{(y + 3)(y - 2)}$

Simplify the numerators:

$3(y - 2) = 3y - 6$ $2(y + 3) = 2y + 6$

Now combine:

$(3y - 6) + (2y + 6) = 5y$

So the equation becomes:

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

Step 5: Eliminate the denominators

Since the denominators are the same on both sides, we can eliminate them and solve the resulting equation:

$5y = y^2$

Step 6: Solve the quadratic equation

Rearrange the equation:

$y^2 - 5y = 0$

Factor the quadratic equation:

$y(y - 5) = 0$

Thus, the solutions are:

$y = 0 \quad \text{or} \quad y = 5$

Step 7: Check for extraneous solutions

We need to check if these solutions make any denominators equal to zero. The denominators are $y + 3$ and $y - 2$.

• For $y = 0$, the denominators are: $y + 3 = 0 + 3 = 3 \quad \text{(not zero)}, \quad y - 2 = 0 - 2 = -2 \quad \text{(not zero)}$ So, $y = 0$ is valid.

• For $y = 5$, the denominators are: $y + 3 = 5 + 3 = 8 \quad \text{(not zero)}, \quad y - 2 = 5 - 2 = 3 \quad \text{(not zero)}$ So, $y = 5$ is valid.

Final Answer:

The solutions are $y = 0$ and $y = 5$.

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Would you like any additional explanations or details?

Here are some related questions to further practice:

1. Solve $\frac{1}{x-1} + \frac{2}{x+1} = \frac{x}{x^2-1}$.
2. Solve $\frac{5}{z+4} - \frac{3}{z-2} = \frac{z+1}{z^2+2z-8}$.
3. Factor and solve $y^2 + 6y + 9 = 0$.
4. Solve $\frac{4}{x+2} - \frac{3}{x-3} = 1$.
5. Solve $\frac{7}{y+1} + \frac{2}{y-2} = \frac{5}{y^2-2y-3}$.

Tip: Always check for extraneous solutions in rational equations by verifying that the solutions do not make any denominator zero.