Let's solve the equation:

\[
6(3x - 5) = 2(2x + 4)
\]

### Step 1: Expand both sides of the equation.

On the left-hand side:
\[
6(3x - 5) = 6 \times 3x - 6 \times 5 = 18x - 30
\]

On the right-hand side:
\[
2(2x + 4) = 2 \times 2x + 2 \times 4 = 4x + 8
\]

So the equation becomes:
\[
18x - 30 = 4x + 8
\]

### Step 2: Move all the terms with \(x\) to one side and constant terms to the other.

Subtract \(4x\) from both sides:
\[
18x - 4x - 30 = 8
\]
\[
14x - 30 = 8
\]

### Step 3: Isolate \(x\).

Add 30 to both sides:
\[
14x = 8 + 30
\]
\[
14x = 38
\]

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

Divide both sides by 14:
\[
x = \frac{38}{14} = \frac{19}{7}
\]

Thus, the solution is:
\[
x = \frac{19}{7}
\]

Would you like further details or have any questions about the steps?

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Here are 5 related questions to expand on this:

1. How do you solve equations with variables on both sides?
2. What happens when equations have no solutions or infinite solutions?
3. Can this process be applied to quadratic equations as well?
4. How would you check if the solution \(x = \frac{19}{7}\) is correct?
5. How do you solve equations involving fractions from the start?

**Tip:** Always check your solution by substituting it back into the original equation to verify its correctness.

Learn how to solve the linear equation 6(3x - 5) = 2(2x + 4) with step-by-step instructions. This algebra problem covers the distributive property, moving variables, and isolating x. Perfect for students in Grades 7-9 looking to master equation solving.

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