We are given the equation:

\[
\frac{x+1}{3} = \frac{x-3}{5}
\]

### Step 1: Eliminate the denominators

To eliminate the fractions, multiply both sides of the equation by the least common denominator (LCD) of 3 and 5, which is 15:

\[
15 \cdot \frac{x+1}{3} = 15 \cdot \frac{x-3}{5}
\]

Simplifying both sides:

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

### Step 2: Distribute

Now, distribute the 5 and the 3:

\[
5x + 5 = 3x - 9
\]

### Step 3: Move variables to one side

Subtract \(3x\) from both sides:

\[
5x - 3x + 5 = -9
\]

This simplifies to:

\[
2x + 5 = -9
\]

### Step 4: Isolate \(x\)

Subtract 5 from both sides:

\[
2x = -14
\]

### Step 5: Solve for \(x\)

Now divide both sides by 2:

\[
x = -7
\]

### Final Answer:

\[
x = -7
\]

Would you like more details on any step? Here are five related questions:

1. How do you find the least common denominator in equations?
2. How does multiplying both sides by the LCD eliminate fractions?
3. Why is it important to move variables to one side of the equation?
4. What are common mistakes when solving equations with fractions?
5. How can you check if your solution is correct by plugging it back into the original equation?

**Tip:** Always double-check your solution by substituting the value of \(x\) back into the original equation.

Learn how to solve the linear equation with fractions \frac{x+1}{3} = \frac{x-3}{5} step by step. This problem involves eliminating fractions using the least common denominator (LCD) and basic algebraic manipulation. Suitable for students in Grades 6-8.

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