Let's solve the equation:

\[
8x - 8 + 3ax = 5ax - 2a
\]

### Step 1: Move all terms involving \(x\) to one side and constants to the other side.

First, subtract \(5ax\) from both sides:

\[
8x + 3ax - 5ax - 8 = -2a
\]

Now combine like terms:

\[
8x - 2ax - 8 = -2a
\]

### Step 2: Move constants to the other side.

Add \(8\) to both sides:

\[
8x - 2ax = -2a + 8
\]

### Step 3: Factor out \(x\) from the left-hand side.

\[
x(8 - 2a) = -2a + 8
\]

### Step 4: Solve for \(x\).

Now, divide both sides by \((8 - 2a)\) (assuming \(8 - 2a \neq 0\)):

\[
x = \frac{-2a + 8}{8 - 2a}
\]

Simplifying the fraction, we get:

\[
x = \frac{8 - 2a}{8 - 2a} = 1
\]

Thus, the solution is:

\[
x = 1
\]

---

Would you like further details or explanations? 

Here are 5 related questions for further practice:
1. Solve \(5x + 4ax = 7ax - 6\).
2. If \(2x - ax + 3 = bx\), solve for \(x\) in terms of \(a\) and \(b\).
3. What is the solution to \(3x + 5 = 7x - 9a\)?
4. Factor and solve \(2x - ax + bx = 3a + b\).
5. Solve \(9x + ax - 5 = bx - 10a\) for \(x\).

**Tip:** Always check for factoring opportunities when dealing with terms involving variables like \(x\) on both sides of an equation.

A step-by-step solution to the equation 8x - 8 + 3ax = 5ax - 2a. This problem demonstrates key algebraic concepts like factoring, variable isolation, and solving linear equations with variables on both sides. Suitable for students in Grades 9-11.

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