not that the other problem x+1/x + x-3/2x   = 2x+7/3x

Let's solve the equation:

\[
\frac{x + 1}{x} + \frac{x - 3}{2x} = \frac{2x + 7}{3x}
\]

### Step 1: Combine the left-hand side (LHS) terms
On the left-hand side, we have two fractions. We want to combine them under a common denominator.

The first term is:

\[
\frac{x + 1}{x}
\]

The second term is:

\[
\frac{x - 3}{2x}
\]

The least common denominator (LCD) between \(x\) and \(2x\) is \(2x\). So we rewrite the first fraction:

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

Now, we can combine the fractions on the left-hand side:

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

### Step 2: Set the equation
Now that we've simplified the left-hand side, the equation becomes:

\[
\frac{3x - 1}{2x} = \frac{2x + 7}{3x}
\]

### Step 3: Cross multiply
To eliminate the denominators, we can cross-multiply:

\[
(3x - 1) \cdot 3x = (2x + 7) \cdot 2x
\]

Expanding both sides:

\[
9x(3x - 1) = 2x(2x + 7)
\]

On the left-hand side:

\[
9x(3x - 1) = 27x^2 - 9x
\]

On the right-hand side:

\[
2x(2x + 7) = 4x^2 + 14x
\]

Now the equation is:

\[
27x^2 - 9x = 4x^2 + 14x
\]

### Step 4: Move all terms to one side
Let's move all terms to one side:

\[
27x^2 - 9x - 4x^2 - 14x = 0
\]

Simplify:

\[
23x^2 - 23x = 0
\]

### Step 5: Factor the equation
We can factor out \(23x\):

\[
23x(x - 1) = 0
\]

### Step 6: Solve for \(x\)
Now, set each factor equal to zero:

\[
23x = 0 \quad \text{or} \quad x - 1 = 0
\]

So:

\[
x = 0 \quad \text{or} \quad x = 1
\]

### Step 7: Check for extraneous solutions
We need to check if \(x = 0\) is a valid solution. Substituting \(x = 0\) into the original equation would cause division by zero, which is undefined. Therefore, \(x = 0\) is not a valid solution.

Thus, the only solution is:

\[
\boxed{x = 1}
\]

---

Would you like further details on any part of this process? 

Here are some follow-up questions:
1. How do you find a common denominator for rational expressions?
2. What is cross-multiplication, and why does it work in rational equations?
3. Can you explain how to expand binomials when multiplying terms?
4. How can factoring help in solving quadratic equations?
5. What are extraneous solutions, and how do they arise in rational equations?

**Tip:** Always check for division by zero when solving equations that involve fractions.

Learn how to solve the rational equation x + 1/x + x - 3/2x = 2x + 7/3x with a detailed step-by-step solution. This problem involves key algebraic concepts such as combining rational expressions, cross-multiplication, and factoring. Suitable for students in Grades 9-12.

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