Question

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




y =
1
2​x−1 (blue line)

y=−
1
2​x+3 (red line)
What are the points and the graph goes to 8

Answer 1

what is the intersection and the points for the blue line

Answer 2

what is the solution

Answer 3

To graph the two equations and find the point where they intersect, let's first write down the equations clearly:

1. \( y = \frac{1}{2}x - 1 \) (blue line)
2. \( y = -\frac{1}{2}x + 3 \) (red line)

### Step 1: Find the Intersection Point

To find the intersection point, we set the equations equal to each other:

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

### Step 2: Solve for \( x \)

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:

\[
x = 4
\]

### Step 3: Find \( y \)

Now we can find the corresponding \( y \) value by substituting \( x = 4 \) into either equation. Let's use the first equation:

\[
y = \frac{1}{2}(4) - 1 = 2 - 1 = 1
\]

So the intersection point is \( (4, 1) \).

### Step 4: Graph the Equations

To graph these equations:

#### For \( y = \frac{1}{2}x - 1 \):

- The y-intercept is \( -1 \) (0, -1).
- The slope is \( \frac{1}{2} \), meaning for every increase of 1 in \( x \), \( y \) increases by \( \frac{1}{2} \).

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

- The y-intercept is \( 3 \) (0, 3).
- The slope is \( -\frac{1}{2} \), meaning for every increase of 1 in \( x \), \( y \) decreases by \( \frac{1}{2} \).

### Step 5: Plot Points

**For the blue line:**

- When \( x = 0 \), \( y = -1 \) (point: (0, -1))
- When \( x = 4 \), \( y = 1 \) (point: (4, 1))
- When \( x = 8 \), \( y = \frac{1}{2}(8) - 1 = 4 - 1 = 3 \) (point: (8, 3))

**For the red line:**

- When \( x = 0 \), \( y = 3 \) (point: (0, 3))
- When \( x = 4 \), \( y = -1 \) (point: (4, 1))
- When \( x = 8 \), \( y = -\frac{1}{2}(8) + 3 = -4 + 3 = -1 \) (point: (8, -1))

### Conclusion

The graph of these two lines would show that they intersect at the point \( (4, 1) \). Points on the lines would include \( (0, -1) \), \( (4, 1) \), \( (8, 3) \) for the blue line, and \( (0, 3) \), \( (4, 1) \), \( (8, -1) \) for the red line.

Answer 4

The blue line represents the equation:

\[
y = \frac{1}{2}x - 1
\]

### Intersection Point

We previously found that the intersection point of the blue line and the red line \((y = -\frac{1}{2}x + 3)\) is:

\[
(4, 1)
\]

This is the point where the two lines intersect.

### Points on the Blue Line

To find specific points on the blue line, we can choose various values of \(x\) and calculate the corresponding \(y\) values using the equation of the blue line.

#### Examples of Points on the Blue Line

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

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

3. **When \(x = 4\):**
\[
y = \frac{1}{2}(4) - 1 = 2 - 1 = 1
\]
Point: **(4, 1)** (This is also the intersection point with the red line.)

4. **When \(x = 6\):**
\[
y = \frac{1}{2}(6) - 1 = 3 - 1 = 2
\]
Point: **(6, 2)**

5. **When \(x = 8\):**
\[
y = \frac{1}{2}(8) - 1 = 4 - 1 = 3
\]
Point: **(8, 3)**

### Summary of Points on the Blue Line

- (0, -1)
- (2, 0)
- (4, 1) (intersection point)
- (6, 2)
- (8, 3)

These are some of the points on the blue line \(y = \frac{1}{2}x - 1\).

Answer 5

The solution to the system of equations given by the two lines is the intersection point of the two lines.

For the equations:

1. \( y = \frac{1}{2}x - 1 \) (blue line)
2. \( y = -\frac{1}{2}x + 3 \) (red line)

### Intersection Point

We previously calculated the intersection point by setting the two equations equal to each other:

1. Set the equations equal:
\[
\frac{1}{2}x - 1 = -\frac{1}{2}x + 3
\]

2. Solve for \(x\):
\[
\frac{1}{2}x + \frac{1}{2}x = 3 + 1
\]
\[
x = 4
\]

3. Substitute \(x = 4\) back into either equation to find \(y\):
\[
y = \frac{1}{2}(4) - 1 = 2 - 1 = 1
\]

Thus, the intersection (solution) of the system of equations is:

\[
\text{Solution: } (4, 1)
\]

### Summary
The solution to the system of equations is the point **(4, 1)**. This point is where the blue line and the red line intersect.