Let's solve the equation:

\[
\frac{2}{x} + \frac{3}{x + 2} = 1
\]

### Step 1: Find a common denominator
The common denominator for \(\frac{2}{x}\) and \(\frac{3}{x + 2}\) is \(x(x + 2)\). Rewriting the equation, we get:

\[
\frac{2(x + 2) + 3x}{x(x + 2)} = 1
\]

### Step 2: Expand and simplify
Expanding the numerator:

\[
\frac{2x + 4 + 3x}{x(x + 2)} = \frac{5x + 4}{x(x + 2)} = 1
\]

### Step 3: Clear the fraction by multiplying both sides by \(x(x + 2)\):
\[
5x + 4 = x(x + 2)
\]

### Step 4: Expand and rearrange the equation
\[
5x + 4 = x^2 + 2x
\]
\[
x^2 + 2x - 5x - 4 = 0
\]
\[
x^2 - 3x - 4 = 0
\]

### Step 5: Solve the quadratic equation
The quadratic equation is:
\[
x^2 - 3x - 4 = 0
\]

We can factor this equation as:
\[
(x - 4)(x + 1) = 0
\]

Thus, the solutions are:
\[
x = 4 \quad \text{or} \quad x = -1
\]

### Final Answer
The solutions to the equation are:
\[
x = 4 \quad \text{or} \quad x = -1
\]

Would you like further details or have any questions?

### Related Questions
1. How would you solve \( \frac{3}{x} + \frac{4}{x + 1} = 2 \)?
2. Can you factor other quadratic equations like \(x^2 - 7x + 12 = 0\)?
3. What happens if there’s no real solution to a quadratic equation?
4. How do you complete the square to solve \( x^2 - 3x - 4 = 0 \)?
5. How would the solution change if the original equation were \( \frac{2}{x} + \frac{3}{x+2} = 2 \)?

**Tip:** Always check your solutions in the original equation to verify correctness.

Learn how to solve the rational equation 2/x + 3/(x + 2) = 1 using the common denominator method and the quadratic formula. This step-by-step guide simplifies the equation and provides solutions for x. Ideal for students in Grades 8-10.

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