Question

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

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.