Solve the following rational equations. Note that your solutions will be graded based on the correctness of your LCD (1 point), correctness of the steps prior to the final answer (1 point), correctness of the solutions/roots (1 point) and correctness of the checking part (1 point). (16 points)

1.    3          2       y^2
   --------  + ------ =  --------------
   y + 3     y - 2    y^2 + y - 6

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 \( y = 5 \).

---

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.

Learn how to solve a rational equation by factoring the quadratic denominator and finding the least common denominator (LCD). Follow a detailed step-by-step approach for solving and checking solutions. Suitable for students in Grades 9-12.

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