To solve the system of equations given by:
- \( p = 6 - 2q \)
- \( p = 4 + q \)
we can find the point of intersection (the values of \( p \) and \( q \) that solve these equations simultaneously) by setting the two equations equal to each other:
\[ 6 - 2q = 4 + q \]
Now, we'll solve for \( q \):
- Rearranging: \[ 6 - 4 = 2q + q \] \[ 2 = 3q \] \[ q = \frac{2}{3} \]
Next, we substitute the value of \( q \) back into either of the original equations to find \( p \). Let's use the second equation:
\[ p = 4 + q = 4 + \frac{2}{3} \] To add \( 4 \) and \( \frac{2}{3} \), we convert \( 4 \) to a fraction: \[ 4 = \frac{12}{3} \] So, \[ p = \frac{12}{3} + \frac{2}{3} = \frac{14}{3} \]
Thus, the solution to the system of equations is: \[ (p, q) = \left(\frac{14}{3}, \frac{2}{3}\right) \]
Details about the Graph
Slopes of the Lines:
-
For \( p = 6 - 2q \): This can be rearranged to \( p = -2q + 6 \). The slope is \( -2 \).
-
For \( p = 4 + q \): This can be rearranged to \( p = q + 4 \). The slope is \( 1 \).
Inverse of the Slopes:
-
The inverse of the slope of the first line (slope = -2) is: \[ \text{Inverse of } -2 = -\frac{1}{2} \]
-
The inverse of the slope of the second line (slope = 1) is: \[ \text{Inverse of } 1 = 1 \]
The Intercepts:
-
For \( p = 6 - 2q \):
- When \( q = 0 \), \( p = 6 \) --> y-intercept: \((0, 6)\)
- When \( p = 0 \), \( 0 = 6 - 2q \Rightarrow 2q = 6 \Rightarrow q = 3\) --> x-intercept: \((3, 0)\)
-
For \( p = 4 + q \):
- When \( q = 0 \), \( p = 4 \) --> y-intercept: \((0, 4)\)
- When \( p = 0 \), \( 0 = 4 + q \Rightarrow q = -4\) --> x-intercept: \((-4, 0)\)
Summary:
- Point of Intersection: \(\left(\frac{14}{3}, \frac{2}{3}\right)\)
- Slopes: Line 1: \(-2\), Line 2: \(1\)
- Inverses of Slopes: Line 1: \(-\frac{1}{2}\), Line 2: \(1\)
- Intercepts:
- Line 1: y-intercept \((0, 6)\), x-intercept \((3, 0)\)
- Line 2: y-intercept \((0, 4)\), x-intercept \((-4, 0)\)