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The equation you've provided is:

\[
\frac{3y}{2} + \frac{2}{y+1} = 4
\]

Let's solve this equation step by step.

### Step 1: Eliminate the fractions by multiplying every term by the least common denominator (LCD).
The LCD in this case is \(2(y + 1)\).

\[
2(y + 1) \cdot \left(\frac{3y}{2}\right) + 2(y + 1) \cdot \left(\frac{2}{y+1}\right) = 4 \cdot 2(y + 1)
\]

### Step 2: Simplify the equation.
\[
(y + 1) \cdot 3y + 4 = 8(y + 1)
\]

### Step 3: Distribute and expand.
\[
3y^2 + 3y + 4 = 8y + 8
\]

### Step 4: Move all terms to one side of the equation to set it equal to zero.
\[
3y^2 + 3y - 8y + 4 - 8 = 0
\]

Simplify:
\[
3y^2 - 5y - 4 = 0
\]

### Step 5: Solve the quadratic equation.
To solve \(3y^2 - 5y - 4 = 0\), use the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 3\), \(b = -5\), and \(c = -4\).

### Step 6: Plug in the values and solve.
\[
y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-4)}}{2(3)}
\]
\[
y = \frac{5 \pm \sqrt{25 + 48}}{6}
\]
\[
y = \frac{5 \pm \sqrt{73}}{6}
\]

### Final Answer:
The solutions for \(y\) are:
\[
y = \frac{5 + \sqrt{73}}{6} \quad \text{or} \quad y = \frac{5 - \sqrt{73}}{6}
\]

Would you like more details or have any questions?

Here are some related questions for further practice:
1. How do you derive the quadratic formula from a general quadratic equation?
2. What are the conditions for the discriminant in a quadratic equation to yield real or complex solutions?
3. How can you solve quadratic equations by completing the square instead of using the quadratic formula?
4. What are the different types of quadratic equations based on their coefficients?
5. How can you graph a quadratic equation and determine its vertex and axis of symmetry?

**Tip:** Always check your solutions by substituting them back into the original equation to ensure they satisfy it.

Learn how to solve the equation (3y/2) + (2/(y+1)) = 4 with a detailed step-by-step solution. This problem covers key algebraic concepts, including equations involving fractions and quadratic equations. Suitable for students in Grades 10-12.

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