We are given the equation:

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

### Step 1: Simplify the denominators
First, observe that \( 2x + 2 \) can be factored:

\[
2x + 2 = 2(x + 1)
\]

So, we rewrite the equation as:

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

### Step 2: Combine the fractions
Both terms on the left side have a common factor \( x + 1 \) in the denominator. Thus, we can combine the fractions:

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

Now, combine the numerators:

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

Simplify the numerator:

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

So the equation becomes:

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

### Step 3: Eliminate the denominator
To eliminate the denominator, multiply both sides of the equation by \( 2(x + 1) \):

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

### Step 4: Solve for \( x \)
Now, expand both sides:

\[
3x - 3 = 2x + 2
\]

Move all terms involving \( x \) to one side and constants to the other:

\[
3x - 2x = 2 + 3
\]

This simplifies to:

\[
x = 5
\]

### Step 5: Check the solution
Substitute \( x = 5 \) back into the original equation to verify:

\[
\frac{5-1}{2(5)+2} + \frac{5-1}{5+1} = \frac{4}{12} + \frac{4}{6} = \frac{1}{3} + \frac{2}{3} = 1
\]

Thus, \( x = 5 \) is the correct solution.

### Final Answer:
\[
x = 5
\]

Would you like further clarification or a deeper explanation of any step?

Here are 5 related questions to practice:
1. Solve \( \frac{x+2}{x-3} + \frac{2x-1}{x-3} = 5 \).
2. Solve \( \frac{2x}{x+1} + \frac{3}{x+1} = 4 \).
3. Solve \( \frac{x-2}{2x+4} + \frac{3x+1}{x+2} = 0 \).
4. Solve \( \frac{x-3}{x+4} + \frac{2}{x+4} = 1 \).
5. Solve \( \frac{3x+1}{x+2} + \frac{5x-4}{x+2} = 7 \).

**Tip:** Always look for common denominators to simplify equations involving fractions.

Learn how to solve the rational equation (x-1)/(2x+2) + (x-1)/(x+1) = 1 through factoring, combining fractions, and simplifying. This step-by-step guide covers essential algebraic concepts, suitable for students in Grades 9-11.

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