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The image shows a math problem from Chapter 1, Lesson 1.1.2 of the CPM Educational Program, labeled as Problem 1-17. The task involves solving for the variable in different equations. Here are the problems as they appear on the screen:

1. Solve for the given variable. Show the steps leading to your solution. Check your solution.

   a. \(-11z = 77\)

   b. \(5c + 1 = 7c - 8\)

   c. \(\frac{z}{8} = 2\)

   d. \(-12 = 3k + 9\)

Let's solve these equations step by step:

### a. \(-11z = 77\)
To solve for \(z\), divide both sides by \(-11\):
\[
z = \frac{77}{-11} = -7
\]
So, \(z = -7\).

### b. \(5c + 1 = 7c - 8\)
Start by moving all terms involving \(c\) to one side and constants to the other:
\[
5c - 7c = -8 - 1
\]
This simplifies to:
\[
-2c = -9
\]
Now, divide both sides by \(-2\):
\[
c = \frac{-9}{-2} = \frac{9}{2}
\]
So, \(c = \frac{9}{2}\).

### c. \(\frac{z}{8} = 2\)
Multiply both sides by 8 to solve for \(z\):
\[
z = 2 \times 8 = 16
\]
So, \(z = 16\).

### d. \(-12 = 3k + 9\)
First, subtract 9 from both sides:
\[
-12 - 9 = 3k
\]
This simplifies to:
\[
-21 = 3k
\]
Now, divide both sides by 3:
\[
k = \frac{-21}{3} = -7
\]
So, \(k = -7\).

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Would you like further details on any of these steps? 

Here are five related practice questions:
1. Solve \(4x + 3 = 9x - 7\).
2. Solve \(\frac{y}{5} = 3\).
3. Solve \(-15a = 60\).
4. Solve \(2m + 6 = 8 - 4m\).
5. Solve \(-3n - 2 = 4n + 5\).

**Tip:** Always start by simplifying both sides of the equation before moving terms across the equal sign to isolate the variable.

Solve for the given variable. Show the steps leading to your solution. Check your solution for the following equations:
a. -11z = 77
b. 5c + 1 = 7c - 8
c. z/8 = 2
d. -12 = 3k + 9

Learn how to solve linear equations such as -11z = 77 and 5c + 1 = 7c - 8 with a step-by-step guide. This problem focuses on basic algebraic manipulation, inverse operations, and equality properties. Suitable for Grades 7-9 students.

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