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Let's solve the given equation for \(x\):

The equation is:

\[
a - (a + b)x = (b - a)x - (c + bx), \quad \text{if} \, b \neq 0.
\]

### Step 1: Expand both sides
Distribute the terms on both sides to simplify:

\[
a - (a + b)x = (b - a)x - (c + bx)
\]

Left-hand side:
\[
a - (a + b)x
\]

Right-hand side:
\[
(b - a)x - (c + bx) = (b - a)x - c - bx
\]

Thus, the equation becomes:

\[
a - (a + b)x = (b - a)x - c - bx
\]

### Step 2: Combine like terms
First, collect all the \(x\)-terms on the left side and constant terms on the right side. The equation is:

\[
a - (a + b)x = (b - a - b)x - c
\]

Simplifying the right-hand side:

\[
a - (a + b)x = -a x - c
\]

### Step 3: Move all terms involving \(x\) to one side
Move all the \(x\)-terms to one side and constants to the other side:

\[
a + c = (a + b)x - ax
\]

### Step 4: Factor the \(x\)-terms
Factor \(x\) from the terms on the right-hand side:

\[
a + c = x[(a + b) - a]
\]

Simplify:

\[
a + c = x[b]
\]

### Step 5: Solve for \(x\)
Solve for \(x\) by dividing both sides by \(b\):

\[
x = \frac{a + c}{b}, \quad \text{where} \, b \neq 0.
\]

### Final Answer:
\[
x = \frac{a + c}{b}
\]

---

Would you like more details or have any questions?

Here are some related questions to explore further:

1. How do you handle equations where the variables appear on both sides?
2. What happens if the condition \(b = 0\) is applied?
3. How would this solution change if there were additional constants involved?
4. Can we verify this solution by substituting back into the original equation?
5. How can we approach solving equations with more complex polynomial terms?

**Tip:** When solving equations, always aim to collect all like terms and isolate the variable step by step for clarity!

Solve the equation a - (a + b)x = (b - a)x - (c + bx) if b ≠ 0

This guide provides a step-by-step solution to solve the equation a - (a + b)x = (b - a)x - (c + bx), involving variables on both sides. Suitable for students in Grades 9-12, covering key algebraic concepts such as simplifying expressions and isolating the variable.

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