We are given the equation:

\[
2c - 5 = -33
\]

### Step 1: Add 5 to both sides to isolate the term with \( c \).
\[
2c - 5 + 5 = -33 + 5
\]
This simplifies to:
\[
2c = -28
\]

### Step 2: Divide both sides by 2 to solve for \( c \).
\[
\frac{2c}{2} = \frac{-28}{2}
\]
This simplifies to:
\[
c = -14
\]

### Final Answer:
\[
c = -14
\]

Would you like further details on any part of the solution?

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Here are 5 related questions for practice:

1. Solve: \( 3x - 7 = -20 \)
2. Solve: \( 5y + 3 = 18 \)
3. Solve: \( 4z - 9 = -5 \)
4. Solve: \( 6p + 4 = -14 \)
5. Solve: \( 7q - 12 = 2 \)

**Tip:** Always perform the inverse operation when moving terms across the equal sign!

Learn how to solve the linear equation 2c - 5 = -33 with a step-by-step solution. This problem demonstrates key algebraic concepts, including isolating variables and applying equality properties. Suitable for students in Grades 6-8.

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