Let's solve the equation \(2(v - 1) + 4v = 3(v - 1) + 7\).

1. **Distribute** the constants through the parentheses:
   \[
   2 \cdot (v - 1) + 4v = 3 \cdot (v - 1) + 7
   \]
   This simplifies to:
   \[
   2v - 2 + 4v = 3v - 3 + 7
   \]

2. **Combine like terms** on both sides:
   \[
   (2v + 4v) - 2 = 3v + ( -3 + 7)
   \]
   Which simplifies to:
   \[
   6v - 2 = 3v + 4
   \]

3. **Move all terms involving** \(v\) to one side and constants to the other:
   \[
   6v - 3v = 4 + 2
   \]
   Simplifying this gives:
   \[
   3v = 6
   \]

4. **Solve for** \(v\) by dividing both sides by 3:
   \[
   v = \frac{6}{3} = 2
   \]

So, the solution is \(v = 2\).

Would you like more details, or do you have any questions?

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Here are some related questions you might find helpful:

1. How do you verify the solution \(v = 2\) in the original equation?
2. What if the equation had different coefficients for the variables?
3. Can this method be used for systems of linear equations?
4. How do you handle equations where variables are on both sides?
5. What is the next step if the equation has no solutions?
6. How does this change if you add a quadratic term to the equation?
7. What is the significance of the solution in real-world contexts?
8. How do you solve similar equations that involve decimals or fractions?

**Tip:** After solving an equation, substituting your solution back into the original equation is a great way to confirm its accuracy.

Learn how to solve the linear equation 2(v-1) + 4v = 3(v-1) + 7 with a step-by-step solution. This problem involves basic algebraic concepts such as the distributive property and combining like terms. Suitable for students in Grades 6-8.

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