2P – 1                      3P + 1
--------  +  6 – P  =   -------- ​Tentukan P
   3                              4

To solve the equation for \( P \), we first rewrite the given expression and combine the fractions.

\[ \frac{2P - 1}{3} + 6 - P = \frac{3P + 1}{4} \]

To eliminate the fractions, we find a common denominator for the fractions. The common denominator of 3 and 4 is 12. So, we multiply every term by 12:

\[ 12 \cdot \left( \frac{2P - 1}{3} \right) + 12 \cdot (6 - P) = 12 \cdot \left( \frac{3P + 1}{4} \right) \]

This simplifies to:

\[ 4(2P - 1) + 12(6 - P) = 3(3P + 1) \]

Next, distribute and simplify each term:

\[ 4 \cdot 2P - 4 \cdot 1 + 12 \cdot 6 - 12 \cdot P = 3 \cdot 3P + 3 \cdot 1 \]

\[ 8P - 4 + 72 - 12P = 9P + 3 \]

Combine like terms:

\[ 8P - 12P + 68 = 9P + 3 \]

\[ -4P + 68 = 9P + 3 \]

Isolate the \( P \) terms on one side of the equation by adding \( 4P \) to both sides:

\[ 68 = 13P + 3 \]

Next, subtract 3 from both sides to further isolate the term containing \( P \):

\[ 65 = 13P \]

Finally, solve for \( P \) by dividing both sides by 13:

\[ P = \frac{65}{13} \]

\[ P = 5 \]

Thus, the value of \( P \) is \( 5 \).

Would you like details or have any questions about the solution?

Here are 5 related questions for further exploration:

1. How would you solve for \( P \) if the equation was \(\frac{2P - 1}{3} + 6 - 2P = \frac{3P + 1}{4}\)?
2. What is the process to combine fractions with different denominators?
3. How can you verify that \( P = 5 \) is the correct solution for the given equation?
4. How do you handle equations with more than two fractions?
5. What are some common mistakes to avoid when solving equations with fractions?

**Tip:** Always check your final solution by substituting it back into the original equation to ensure that both sides are equal.

Explore how to solve the algebraic equation involving fractions: (2P - 1) / 3 + 6 - P = (3P + 1) / 4. This problem demonstrates eliminating fractions using a common denominator, simplifying terms, and solving for P. Ideal for students in Grades 9-10.

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