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solve and show clear working out and steps

Let's solve each of the linear equations step by step.

### a. \(\frac{z - 1}{3} = 5\)

1. Multiply both sides by 3 to get rid of the denominator:
   \[
   z - 1 = 15
   \]
2. Add 1 to both sides to isolate \(z\):
   \[
   z = 16
   \]

### b. \(\frac{z + 1}{4} = 8\)

1. Multiply both sides by 4 to get rid of the denominator:
   \[
   z + 1 = 32
   \]
2. Subtract 1 from both sides to isolate \(z\):
   \[
   z = 31
   \]

### c. \(\frac{z - 4}{2} = -4\)

1. Multiply both sides by 2 to get rid of the denominator:
   \[
   z - 4 = -8
   \]
2. Add 4 to both sides to isolate \(z\):
   \[
   z = -4
   \]

### d. \(\frac{6 - z}{7} = 0\)

1. Multiply both sides by 7 to get rid of the denominator:
   \[
   6 - z = 0
   \]
2. Subtract 6 from both sides:
   \[
   -z = -6
   \]
3. Multiply both sides by -1 to solve for \(z\):
   \[
   z = 6
   \]

### e. \(\frac{3 - z}{2} = 6\)

1. Multiply both sides by 2 to get rid of the denominator:
   \[
   3 - z = 12
   \]
2. Subtract 3 from both sides:
   \[
   -z = 9
   \]
3. Multiply both sides by -1 to solve for \(z\):
   \[
   z = -9
   \]

### f. \(\frac{-z - 50}{22} = -2\)

1. Multiply both sides by 22 to get rid of the denominator:
   \[
   -z - 50 = -44
   \]
2. Add 50 to both sides:
   \[
   -z = 6
   \]
3. Multiply both sides by -1 to solve for \(z\):
   \[
   z = -6
   \]

Here are the solutions:

1. \(z = 16\)
2. \(z = 31\)
3. \(z = -4\)
4. \(z = 6\)
5. \(z = -9\)
6. \(z = -6\)

Do you want any further details or have any questions?

Here are 5 questions you can ask next:
1. Can you explain how to solve quadratic equations?
2. What are the steps to solve a system of linear equations using the substitution method?
3. How do you solve inequalities involving absolute values?
4. Can you explain the method of completing the square?
5. How do you find the derivative of a function using the limit definition?

**Tip:** Always check your solutions by substituting them back into the original equations to verify they work correctly.

Learn how to solve linear equations step-by-step with detailed solutions. This problem covers key algebraic concepts and equation solving methods, suitable for students in Grades 7-9.

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