If 2p +3q = 18 and `4p^(2) + 4pq - 3q^(2) - 36 = 0` them what is (2p + q) equal to

Options

A. (A)6
B. (B)7
C. (C)10
D. (D)20

Answer

To solve the problem step by step, we will use the given equations:

1. **Given Equations:**
- \( 2p + 3q = 18 \) (Equation 1)
- \( 4p^2 + 4pq - 3q^2 - 36 = 0 \) (Equation 2)

2. **Rearranging Equation 1:**
We can express \( 2p \) in terms of \( q \):
\[ 2p = 18 - 3q \]
Let's label this as Equation 3.

3. **Substituting into Equation 2:**
Now, we will substitute \( 2p \) from Equation 3 into Equation 2. First, we rewrite \( 4p^2 \) and \( 4pq \):
\[ 4p^2 = (2p)^2 = (18 - 3q)^2 \]
\[ 4pq = 2(2p)q = 2(18 - 3q)q = 36q - 6q^2 \]
Now substituting these into Equation 2:
\[ (18 - 3q)^2 + (36q - 6q^2) - 3q^2 - 36 = 0 \]

4. **Expanding the Equation:**
Expanding \( (18 - 3q)^2 \):
\[ 324 - 108q + 9q^2 \]
Now substituting this back into the equation:
\[ 324 - 108q + 9q^2 + 36q - 6q^2 - 3q^2 - 36 = 0 \]

5. **Combining Like Terms:**
Combine the terms:
\[ 324 - 36 - 108q + 36q + 9q^2 - 6q^2 - 3q^2 = 0 \]
Simplifying gives:
\[ 288 - 72q = 0 \]

6. **Solving for \( q \):**
Rearranging gives:
\[ 72q = 288 \implies q = \frac{288}{72} = 4 \]

7. **Finding \( p \):**
Now substitute \( q = 4 \) back into Equation 3 to find \( p \):
\[ 2p = 18 - 3(4) = 18 - 12 = 6 \implies p = \frac{6}{2} = 3 \]

8. **Finding \( 2p + q \):**
Now we can find \( 2p + q \):
\[ 2p + q = 2(3) + 4 = 6 + 4 = 10 \]

Thus, the final answer is:
\[ \boxed{10} \]