Solve the following equation:

6r = 2r + 232

To solve the equation \(6r = 2r + 232\), follow these steps:

1. **Isolate the variable term on one side of the equation:**

   Subtract \(2r\) from both sides of the equation to get all the \(r\) terms on one side:

   \[
   6r - 2r = 2r + 232 - 2r
   \]

   Simplify the equation:

   \[
   4r = 232
   \]

2. **Solve for \(r\):**

   Divide both sides of the equation by 4 to isolate \(r\):

   \[
   \frac{4r}{4} = \frac{232}{4}
   \]

   Simplify the equation:

   \[
   r = 58
   \]

So, the solution to the equation \(6r = 2r + 232\) is \(r = 58\).

Would you like more details or have any questions? 

Here are some related questions for further exploration:

1. How do you solve the equation \(8x = 3x + 125\)?
2. What is the solution to \(5y - 2y = 300\)?
3. Can you solve the equation \(9k = 4k + 150\)?
4. How do you find the value of \(m\) in the equation \(7m = 3m + 100\)?
5. What steps are involved in solving \(10n = 6n + 400\)?
6. Can you solve the equation \(12p = 7p + 150\)?
7. What is the value of \(q\) in the equation \(14q = 8q + 280\)?
8. How do you isolate the variable in the equation \(15a = 5a + 200\)?

**Tip:** When solving linear equations, always aim to isolate the variable on one side of the equation by performing the same operation on both sides.

Learn how to solve the equation 6r = 2r + 232 step-by-step with detailed explanations. This problem involves isolating variables and basic algebraic manipulation. Suitable for students in Grades 6-8.

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