https://file.aiquickdraw.com/mathbot/file/1728343112636yx5b29vz.jpg

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

\[
a \cdot (5 - x) = bx - 8
\]

### Step 1: Expand the left side
Distribute the \(a\) on the left-hand side:

\[
a(5 - x) = 5a - ax
\]

So the equation becomes:

\[
5a - ax = bx - 8
\]

### Step 2: Collect like terms
Move all terms involving \(x\) to one side and constants to the other side. Add \(ax\) to both sides and add 8 to both sides:

\[
5a + 8 = ax + bx
\]

### Step 3: Factor out \(x\) on the right-hand side
Factor \(x\) from the terms on the right:

\[
5a + 8 = x(a + b)
\]

### Step 4: Solve for \(x\)
Now, divide both sides by \(a + b\) to isolate \(x\):

\[
x = \frac{5a + 8}{a + b}
\]

Thus, the solution for \(x\) is:

\[
x = \frac{5a + 8}{a + b}
\]

Would you like further details on any of these steps? Let me know if you have any questions!

Here are five related questions:
1. What happens if \(a + b = 0\)? How would that affect the solution?
2. Can this equation be solved using another method, such as substitution or graphing?
3. How would the solution change if a different operation was applied to \(x\), such as squaring both sides?
4. What if the equation involved a quadratic term in \(x\)? How would that impact the solving process?
5. Could we solve for \(x\) if there were more than two variables, such as including an additional parameter \(c\)?

**Tip:** Always check the denominator in the final step to ensure you're not dividing by zero.

Learn how to solve the linear equation a(5 - x) = bx - 8 by using algebraic methods such as distribution and factoring. This problem helps students master linear equations in Grades 9-11.

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