AD\# \( \qquad \) Equal Values Method Find the point of intersection (P.O.I.) for each pair of lines by using the Equal Values Method. 1. \( y=x+2 \) \( y=2 x-1 \) \( (3,5) \) 3. \( y=11-2 x \) \( y=x+2 \) 2. \( \quad y=3 x+5 \) \[ y=4 x+8 \] 4. \[ \begin{array}{l} y=3-2 x \\ y=1+2 x \end{array} \] 6. \( y=-\frac{2}{3} x+4 \) \[ y=\frac{1}{3} x-2 \] \( \frac{\left(\frac{1}{2}, 2\right)}{y=-\frac{2}{3} x+4} \) \( y=3 x-4 \) \( y=.5 x+6 \) \( (0,512) \) \( y=4.5-x \) \( y=-2 x+6 \) \[ \begin{array}{l} y=4.5-x \\ y=-2 x+6 \end{array} \] \[ \begin{array}{l} y=4 x \\ y=x+1 \end{array} \]

To find the points of intersection (P.O.I.) for each pair of lines using the Equal Values Method, we will set the equations equal to each other and solve for the variable. Let's go through each pair step by step. ### 1. Pair: \( y = x + 2 \) and \( y = 2x - 1 \) Set the equations equal to each other: \[ x + 2 = 2x - 1 \] Now, solve for \( x \): \[ 2 + 1 = 2x - x \\ 3 = x \] Now substitute \( x = 3 \) back into one of the original equations to find \( y \): \[ y = 3 + 2 = 5 \] **Point of Intersection:** \( (3, 5) \) ### 2. Pair: \( y = 11 - 2x \) and \( y = x + 2 \) Set the equations equal to each other: \[ 11 - 2x = x + 2 \] Now, solve for \( x \): \[ 11 - 2 = x + 2x \\ 9 = 3x \\ x = 3 \] Now substitute \( x = 3 \) back into one of the original equations to find \( y \): \[ y = 3 + 2 = 5 \] **Point of Intersection:** \( (3, 5) \) ### 3. Pair: \( y = 3x + 5 \) and \( y = 4x + 8 \) Set the equations equal to each other: \[ 3x + 5 = 4x + 8 \] Now, solve for \( x \): \[ 5 - 8 = 4x - 3x \\ -3 = x \] Now substitute \( x = -3 \) back into one of the original equations to find \( y \): \[ y = 3(-3) + 5 = -9 + 5 = -4 \] **Point of Intersection:** \( (-3, -4) \) ### 4. Pair: \( y = 3 - 2x \) and \( y = 1 + 2x \) Set the equations equal to each other: \[ 3 - 2x = 1 + 2x \] Now, solve for \( x \): \[ 3 - 1 = 2x + 2x \\ 2 = 4x \\ x = \frac{1}{2} \] Now substitute \( x = \frac{1}{2} \) back into one of the original equations to find \( y \): \[ y = 3 - 2\left(\frac{1}{2}\right) = 3 - 1 = 2 \] **Point of Intersection:** \( \left(\frac{1}{2}, 2\right) \) ### 5. Pair: \( y = -\frac{2}{3}x + 4 \) and \( y = \frac{1}{3}x - 2 \) Set the equations equal to each other: \[ -\frac{2}{3}x + 4 = \frac{1}{3}x - 2 \] Now, solve for \( x \): \[ 4 + 2 = \frac{1}{3}x + \frac{2}{3}x \\ 6 = 1x \\ x = 6 \] Now substitute \( x = 6 \) back into one of the original equations to find \( y \): \[ y = -\frac{2}{3}(6) + 4 = -4 + 4 = 0 \] **Point of Intersection:** \( (6, 0) \) ### 6. Pair: \( y = 4.5 - x \) and \( y = -2x + 6 \) Set the equations equal to each other: \[ 4.5 - x = -2x + 6 \] Now, solve for \( x \): \[ 4.5 - 6 = -2x + x \\ -1.5 = -x \\ x = 1.5 \] Now substitute \( x = 1.5 \) back into one of the original equations to find \( y \): \[ y = 4.5 - 1.5 = 3 \] **Point of Intersection:** \( (1.5, 3) \) ### 7. Pair: \( y = 4x \) and \( y = x + 1 \) Set the equations equal to each other: \[ 4x = x + 1 \] Now, solve for \( x \): \[ 4x - x = 1 \\ 3x = 1 \\ x = \frac{1}{3} \] Now substitute \( x = \frac{1}{3} \) back into one of the original equations to find \( y \): \[ y = 4\left(\frac{1}{3}\right) = \frac{4}{3} \] **Point of Intersection:** \( \left(\frac{1}{3}, \frac{4}{3}\right) \) ### Summary of Points of Intersection: 1. \( (3, 5) \) 2. \( (3, 5) \) 3. \( (-3, -4) \) 4. \( \left(\frac{1}{2}, 2\right) \) 5. \( (6, 0) \) 6. \( (1.5, 3) \) 7. \( \left(\frac{1}{3}, \frac{4}{3}\right) \)