Name: 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. \[ \text { 1. } \quad \begin{array}{l} y=x+2 \\ y=2 x-1 \end{array} \] 2. \[ \begin{array}{l} y=3 x+5 \\ y=4 x+8 \end{array} \] 4. \[ \begin{array}{l} y=3-2 x \\ y=1+2 x \end{array} \] 6. \( y=-\frac{2}{3} x+4 \) \[ y=\frac{1}{3} x-2 \] 7. \[ \begin{array}{l} y=4.5-x \\ y=-2 x+6 \end{array} \] 8. \[ \begin{array}{l} y=4 x \\ y=x+1 \end{array} \]

To find the points of intersection for each pair of equations using the Equal Values Method, set the equations equal to each other and solve for \(x\). 1. **For the first pair:** \[ x + 2 = 2x - 1 \implies 2 + 1 = 2x - x \implies x = 3 \] Substitute \(x = 3\) back into either equation to find \(y\): \[ y = 3 + 2 = 5 \] Thus, the P.O.I. is \((3, 5)\). 2. **For the second pair:** \[ 3x + 5 = 4x + 8 \implies 5 - 8 = 4x - 3x \implies x = -3 \] Substitute \(x = -3\) into either equation: \[ y = 3(-3) + 5 = -9 + 5 = -4 \] So, the P.O.I. is \((-3, -4)\). 4. **For the third pair:** \[ 3 - 2x = 1 + 2x \implies 3 - 1 = 2x + 2x \implies 2 = 4x \implies x = \frac{1}{2} \] Substitute \(x = \frac{1}{2}\) into either equation: \[ y = 3 - 2 \left(\frac{1}{2}\right) = 3 - 1 = 2 \] Hence, the P.O.I. is \(\left(\frac{1}{2}, 2\right)\). 6. **For the fourth pair:** \[ -\frac{2}{3}x + 4 = \frac{1}{3}x - 2 \implies 4 + 2 = \frac{1}{3}x + \frac{2}{3}x \implies 6 = x \] Substitute \(x = 6\): \[ y = -\frac{2}{3}(6) + 4 = -4 + 4 = 0 \] The P.O.I. is \((6, 0)\). 7. **For the fifth pair:** \[ 4.5 - x = -2x + 6 \implies 4.5 - 6 = -x + 2x \implies -1.5 = x \] Substitute \(x = -1.5\): \[ y = 4.5 - (-1.5) = 4.5 + 1.5 = 6 \] Therefore, the P.O.I. is \((-1.5, 6)\). 8. **For the sixth pair:** \[ 4x = x + 1 \implies 4x - x = 1 \implies 3x = 1 \implies x = \frac{1}{3} \] Substitute \(x = \frac{1}{3}\): \[ y = 4 \left(\frac{1}{3}\right) = \frac{4}{3} \] Thus, the P.O.I. is \(\left(\frac{1}{3}, \frac{4}{3}\right)\).