To solve the system of equations given by:
- \( p = 6 - 2q \) (Equation 1)
- \( p = 4 + q \) (Equation 2)
We can set the two equations for \( p \) equal to each other, since they both equal \( p \):
\[ 6 - 2q = 4 + q \]
Next, let's solve for \( q \):
- Add \( 2q \) to both sides:
\[ 6 = 4 + q + 2q \]
- Combine like terms:
\[ 6 = 4 + 3q \]
- Subtract 4 from both sides:
\[ 2 = 3q \]
- Divide by 3:
\[ q = \frac{2}{3} \]
Now that we have \( q \), we can substitute it back into either of the original equations to find \( p \). We'll use Equation 2:
\[ p = 4 + q \]
Substituting \( q \):
\[ p = 4 + \frac{2}{3} \]
To combine the terms, convert 4 into a fraction:
\[ p = \frac{12}{3} + \frac{2}{3} = \frac{14}{3} \]
Thus, the solution to the system of equations is:
\[ p = \frac{14}{3}, \quad q = \frac{2}{3} \]
In summary, the solution is:
\[ (p, q) = \left( \frac{14}{3}, \frac{2}{3} \right) \]