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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. \[ \begin{array}{l} y=4 x \\ y=x+1 \end{array} \]

Ask by Sandoval Huang. in the United States
Feb 19,2025

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Answer

Points of Intersection: 1. (3, 5) 2. (-3, -4) 3. (0.5, 2) 4. (6, 0) 5. (1.5, 3) 6. (0.333..., 1.333...) Note: The fractions are approximated for simplicity.

Solution

To find the point 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 \(x\). Once we find \(x\), we will substitute it back into one of the original equations to find \(y\). ### 1. \[ \begin{array}{l} y = x + 2 \\ y = 2x - 1 \end{array} \] **Step 1:** Set the equations equal to each other: \[ x + 2 = 2x - 1 \] **Step 2:** Solve for \(x\): \[ 2 + 1 = 2x - x \\ 3 = x \] **Step 3:** Substitute \(x\) back into one of the equations to find \(y\): \[ y = 3 + 2 = 5 \] **Point of Intersection:** \( (3, 5) \) --- ### 2. \[ \begin{array}{l} y = 3x + 5 \\ y = 4x + 8 \end{array} \] **Step 1:** Set the equations equal to each other: \[ 3x + 5 = 4x + 8 \] **Step 2:** Solve for \(x\): \[ 5 - 8 = 4x - 3x \\ -3 = x \] **Step 3:** Substitute \(x\) back into one of the equations to find \(y\): \[ y = 3(-3) + 5 = -9 + 5 = -4 \] **Point of Intersection:** \( (-3, -4) \) --- ### 4. \[ \begin{array}{l} y = 3 - 2x \\ y = 1 + 2x \end{array} \] **Step 1:** Set the equations equal to each other: \[ 3 - 2x = 1 + 2x \] **Step 2:** Solve for \(x\): \[ 3 - 1 = 2x + 2x \\ 2 = 4x \\ x = \frac{1}{2} \] **Step 3:** Substitute \(x\) back into one of the equations to find \(y\): \[ y = 3 - 2\left(\frac{1}{2}\right) = 3 - 1 = 2 \] **Point of Intersection:** \( \left(\frac{1}{2}, 2\right) \) --- ### 6. \[ \begin{array}{l} y = -\frac{2}{3}x + 4 \\ y = \frac{1}{3}x - 2 \end{array} \] **Step 1:** Set the equations equal to each other: \[ -\frac{2}{3}x + 4 = \frac{1}{3}x - 2 \] **Step 2:** Solve for \(x\): \[ 4 + 2 = \frac{1}{3}x + \frac{2}{3}x \\ 6 = x \] **Step 3:** Substitute \(x\) back into one of the equations to find \(y\): \[ y = -\frac{2}{3}(6) + 4 = -4 + 4 = 0 \] **Point of Intersection:** \( (6, 0) \) --- ### 7. \[ \begin{array}{l} y = 4.5 - x \\ y = -2x + 6 \end{array} \] **Step 1:** Set the equations equal to each other: \[ 4.5 - x = -2x + 6 \] **Step 2:** Solve for \(x\): \[ 4.5 - 6 = -2x + x \\ -1.5 = -x \\ x = 1.5 \] **Step 3:** Substitute \(x\) back into one of the equations to find \(y\): \[ y = 4.5 - 1.5 = 3 \] **Point of Intersection:** \( (1.5, 3) \) --- ### 8. \[ \begin{array}{l} y = 4x \\ y = x + 1 \end{array} \] **Step 1:** Set the equations equal to each other: \[ 4x = x + 1 \] **Step 2:** Solve for \(x\): \[ 4x - x = 1 \\ 3x = 1 \\ x = \frac{1}{3} \] **Step 3:** Substitute \(x\) back into one of the 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. \( (-3, -4) \) 3. \( \left(\frac{1}{2}, 2\right) \) 4. \( (6, 0) \) 5. \( (1.5, 3) \) 6. \( \left(\frac{1}{3}, \frac{4}{3}\right) \)

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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 \(\left(\frac{1}{3}, \frac{4}{3}\right)\).

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