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The image shows a process of checking the solution to the equation \( z - 8 = 16 \) by substituting the solution back into the original equation.

### Original equation:
\[
z - 8 = 16
\]

To solve for \( z \):
\[
z = 16 + 8 = 24
\]

Now, we substitute \( z = 24 \) back into the original equation to verify:

\[
24 - 8 = 16
\]
\[
16 = 16
\]

Since both sides of the equation are equal, the check is correct, and the solution \( z = 24 \) is valid.

Would you like further details on the solution process?

Here are 5 related questions to help deepen understanding:
1. How would you solve the equation if it was \( z - 8 = -16 \)?
2. What steps would you take if the equation had a fraction, such as \( z - \frac{8}{2} = 16 \)?
3. How can you verify solutions for more complex equations?
4. How do you solve an equation if the unknown appears on both sides, like \( z - 8 = 2z - 10 \)?
5. What if the equation involved multiplication, such as \( 2z - 8 = 16 \)?

**Tip:** Always verify your solutions by substituting them back into the original equation to ensure both sides are equal.

Check your answer by substituting it into the original equation: z - 8 = 16. After solving for z, verify the solution by substituting z back into the equation.

Learn how to solve the linear equation z - 8 = 16 by isolating the variable and verifying the solution by substitution. This example demonstrates basic algebraic techniques, suitable for students in Grades 6-8.

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