Let's solve the equation \( 2c - 5 = c + 4 \).

### Step 1: Move all terms involving \( c \) to one side
Subtract \( c \) from both sides:
\[
2c - 5 - c = c + 4 - c
\]
This simplifies to:
\[
c - 5 = 4
\]

### Step 2: Move constant terms to the other side
Add 5 to both sides:
\[
c - 5 + 5 = 4 + 5
\]
Simplifying this:
\[
c = 9
\]

So, the solution is \( c = 9 \).

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. What happens if you have fractions in the equation?
2. How do you handle equations with variables on both sides?
3. What if there were more terms involving \( c \)?
4. How would the equation change if you had \( c^2 \) instead of \( c \)?
5. How do you solve inequalities involving similar terms?

**Tip:** Always check your solution by substituting it back into the original equation.

Learn how to solve the linear equation 2c - 5 = c + 4 using simple algebraic techniques. This step-by-step guide explains how to manipulate terms and solve for the variable c. Suitable for students in Grades 6-8, covering foundational algebra concepts.

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