System of Equations:

p=6−2q
p=4+q

1 answer

To solve the system of equations given by:

  1. \( p = 6 - 2q \) (Equation 1)
  2. \( p = 4 + q \) (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 \( q \):

  1. Add \( 2q \) to both sides:

\[ 6 = 4 + q + 2q \]

  1. Combine like terms:

\[ 6 = 4 + 3q \]

  1. Subtract 4 from both sides:

\[ 2 = 3q \]

  1. 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:

\[ (p, q) = \left( \frac{14}{3}, \frac{2}{3} \right) \]