Identify the equation for the red line and the blue line. Then, identify the solution to the system of equations.(3 points) Drop target 1 Drop target 2 Drop target 3 y=12x−3(4, 1)(4, −1)y=−x+3(−1, 4)y=−12x+3y=−x−3(−4, −1)

Answer 1

To identify the equations of the lines and the solution to the system of equations, we need to analyze the given equations and points.

1. Analyzing the Given Information:

   • The equations given include:
     • \(y = 12x - 3\) (appears to be one of the lines)
     • \(y = -x + 3\) (appears to be the other line)
   • The points provided include:
     • \((4, 1)\)
     • \((4, -1)\)
     • \((-1, 4)\)
     • \((-4, -1)\)

2. Identifying the Equations of the Lines:

   • Red Line: Based on the equation \(y = 12x - 3\) and evaluating it for point \( (4, 1) \):
     • Substitute \(x = 4\): \(y = 12(4) - 3 = 48 - 3 = 45\) which does not match \((4, 1)\) or \((4, -1)\). Therefore, the first equation does not fit these points.
   • Blue Line: The equation \(y = -x + 3\) can be checked with point \((-1, 4)\):
     • Substitute \(x = -1\): \(y = -(-1) + 3 = 1 + 3 = 4\), so \((-1, 4)\) is on this line.

3. Finding the Intersection Point (Solution):

   • To find the intersection of the two lines, set \(12x - 3 = -x + 3\): \[ 12x + x = 3 + 3 \implies 13x = 6 \implies x = \frac{6}{13} \]
   • Substitute this value into either equation to find \(y\). Using the second equation: \[ y = -\left(\frac{6}{13}\right) + 3 = -\frac{6}{13} + \frac{39}{13} = \frac{33}{13} \]
   • Therefore, the solution to the system of equations is \(\left(\frac{6}{13}, \frac{33}{13}\right)\).

Summary:

• Red Line Equation: \(y = 12x - 3\)
• Blue Line Equation: \(y = -x + 3\)
• Solution to the System: \(\left(\frac{6}{13}, \frac{33}{13}\right)\)