Drag and drop the points to graph and solve the system of equations.

Question

y = 1/2x - 1 (blue line)

y = -1/2x + 3 (red line)

Answer 1

To graph the system of equations \(y = \frac{1}{2}x - 1\) (blue line) and \(y = -\frac{1}{2}x + 3\) (red line), let's first identify a few points for each line.

For the blue line: \(y = \frac{1}{2}x - 1\)

1. When \(x = 0\): \[ y = \frac{1}{2}(0) - 1 = -1 \quad \Rightarrow (0, -1) \]

2. When \(x = 2\): \[ y = \frac{1}{2}(2) - 1 = 0 \quad \Rightarrow (2, 0) \]

3. When \(x = 4\): \[ y = \frac{1}{2}(4) - 1 = 1 \quad \Rightarrow (4, 1) \]

For the red line: \(y = -\frac{1}{2}x + 3\)

1. When \(x = 0\): \[ y = -\frac{1}{2}(0) + 3 = 3 \quad \Rightarrow (0, 3) \]

2. When \(x = 2\): \[ y = -\frac{1}{2}(2) + 3 = 2 \quad \Rightarrow (2, 2) \]

3. When \(x = 6\): \[ y = -\frac{1}{2}(6) + 3 = 0 \quad \Rightarrow (6, 0) \]

Graphing the Equations:

1. Plot the points for the blue line:

   • (0, -1)
   • (2, 0)
   • (4, 1)

2. Plot the points for the red line:

   • (0, 3)
   • (2, 2)
   • (6, 0)

3. Draw the lines through these points.

Finding the Intersection Point:

To solve the system of equations, we can set the equations equal to each other:

\[ \frac{1}{2}x - 1 = -\frac{1}{2}x + 3 \]

1. Add \(\frac{1}{2}x\) to both sides: \[ \frac{1}{2}x + \frac{1}{2}x - 1 = 3 \] \[ x - 1 = 3 \]

2. Add 1 to both sides: \[ x = 4 \]

3. Substitute \(x = 4\) back into one of the original equations to find \(y\): Using \(y = \frac{1}{2}x - 1\): \[ y = \frac{1}{2}(4) - 1 = 2 - 1 = 1 \]

Solution:

The intersection point, which is the solution to the system of equations, is: \[ \boxed{(4, 1)} \]

This is the point where the blue and red lines intersect.