To graph the system of equations:

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

### Step 1: Graph each equation.

#### Equation 1: \( p = 6 - 2q \)
This is a linear equation in slope-intercept form \( p = m q + b \), where:
- The slope (\( m \)) is \( -2 \)
- The y-intercept (\( b \)) is \( 6 \)

#### Equation 2: \( p = 1 + q \)
This is also in slope-intercept form, where:
- The slope (\( m \)) is \( 1 \)
- The y-intercept (\( b \)) is \( 1 \)

### Step 2: Find the point of intersection
To find where the two lines intersect, set the two equations equal to each other:

\[
6 - 2q = 1 + q
\]

Now solve for \( q \):

\[
6 - 1 = q + 2q
\]
\[
5 = 3q
\]
\[
q = \frac{5}{3}
\]

Now substitute \( q = \frac{5}{3} \) into one of the original equations to find \( p \). Let's use \( p = 1 + q \):

\[
p = 1 + \frac{5}{3} = \frac{8}{3}
\]

So, the point of intersection is \( \left( \frac{5}{3}, \frac{8}{3} \right) \).

### Step 3: Graph
1. For \( p = 6 - 2q \), start at \( (0, 6) \) and use the slope of \( -2 \) to go down 2 units for each step right.
2. For \( p = 1 + q \), start at \( (0, 1) \) and use the slope of \( 1 \) to go up 1 unit for each step right.
3. Plot both lines and their point of intersection \( \left( \frac{5}{3}, \frac{8}{3} \right) \).

Do you want me to graph it for you?

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Here are 5 related questions:
1. What is the significance of the point of intersection in solving systems of equations?
2. Can systems of equations have more than one solution?
3. How do you solve a system of equations graphically if the slopes are the same?
4. What is the algebraic method to solve systems of linear equations?
5. How do you check if a point is a solution to a system of equations?

**Tip:** For systems with parallel lines, there is no solution since they never intersect!

Learn how to graph and solve the system of linear equations p=6-2q and p=1+q. This problem covers key algebraic concepts such as slope-intercept form and finding the point of intersection of two lines. Suitable for students in Grades 8-10.

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