To solve the equation \( ax + 3x = bx + 5 \), we start by combining like terms and isolating \( x \). Here's the step-by-step solution:

### Step 1: Combine like terms
Combine \( ax \) and \( 3x \) on the left side of the equation:

\[
ax + 3x = (a + 3)x
\]

Now the equation looks like:

\[
(a + 3)x = bx + 5
\]

### Step 2: Move all terms involving \( x \) to one side
Subtract \( bx \) from both sides:

\[
(a + 3)x - bx = 5
\]

Simplify this by factoring \( x \) out:

\[
((a + 3) - b)x = 5
\]

\[
(a + 3 - b)x = 5
\]

### Step 3: Solve for \( x \)
Divide both sides by \( (a + 3 - b) \):

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

This is the solution for \( x \), provided that \( a + 3 - b \neq 0 \) (otherwise, the equation would be undefined).

### Final Solution:
\[
\boxed{x = \frac{5}{a + 3 - b}}
\]

Would you like further clarification on any part of this solution?

### Related Questions:
1. What happens if \( a + 3 = b \)?
2. How would you solve if the equation was \( ax + 3x = bx - 5 \)?
3. Can you graph the solution \( x = \frac{5}{a + 3 - b} \)?
4. How does changing the values of \( a \) and \( b \) affect the value of \( x \)?
5. What are some real-world applications of solving equations like this?

**Tip:** Always ensure that the denominator is not zero when solving equations involving fractions to avoid undefined solutions.

This detailed solution explains how to solve the linear equation ax + 3x = bx + 5 step by step. Learn key algebraic concepts such as combining like terms and solving for x. Perfect for students in Grades 8-10.

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