Let's solve the equation:

\[
10(2y + 2) - y = 2(8y - 8)
\]

### Step 1: Distribute the constants on both sides.
For the left-hand side:
\[
10(2y + 2) = 10 \cdot 2y + 10 \cdot 2 = 20y + 20
\]
Thus, the equation becomes:
\[
20y + 20 - y = 2(8y - 8)
\]

For the right-hand side:
\[
2(8y - 8) = 2 \cdot 8y - 2 \cdot 8 = 16y - 16
\]

Now the equation is:
\[
20y + 20 - y = 16y - 16
\]

### Step 2: Simplify both sides.
On the left-hand side:
\[
20y - y + 20 = 19y + 20
\]

The equation becomes:
\[
19y + 20 = 16y - 16
\]

### Step 3: Move all the \(y\) terms to one side and constants to the other side.
Subtract \(16y\) from both sides:
\[
19y - 16y + 20 = -16
\]
\[
3y + 20 = -16
\]

### Step 4: Isolate \(y\).
Subtract 20 from both sides:
\[
3y = -16 - 20
\]
\[
3y = -36
\]

### Step 5: Solve for \(y\).
Divide both sides by 3:
\[
y = \frac{-36}{3} = -12
\]

### Final answer:
\[
y = -12
\]

Would you like more details on any of these steps?

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Here are 5 related questions:
1. What is the next step if the equation had variables on both sides with no constants?
2. How would you solve \(2(3x + 4) = 5(2x - 1)\)?
3. What happens if you isolate the variable on the right side first?
4. Can you solve an equation with fractions like \( \frac{2x + 3}{4} = \frac{5x - 2}{6} \)?
5. How do you handle equations with powers, like \( x^2 + 3x = 4 \)?

**Tip:** Always simplify equations step-by-step to avoid mistakes when distributing and combining like terms.

This page provides a step-by-step guide on solving the linear equation 10(2y + 2) - y = 2(8y - 8). Learn how to apply the distributive property and combine like terms to isolate and solve for y. This example is suitable for students in Grades 7-8.

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