Question
Drag and drop the points to graph and solve the system of equations.
y =
2
3x−2 (blue line)
y=−
1
2x+5 (red line)
what are the points and the graph only goes up to 8
y =
2
3x−2 (blue line)
y=−
1
2x+5 (red line)
what are the points and the graph only goes up to 8
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To graph the system of equations given, we will analyze each equation. The equations are:
1. \( y = \frac{2}{3}x - 2 \) (blue line)
2. \( y = -\frac{1}{2}x + 5 \) (red line)
### Step 1: Determine Points for Each Equation
**For the blue line \( y = \frac{2}{3}x - 2 \):**
To find points, we can select a few values for \( x \) and calculate \( y \):
- When \( x = 0 \):
\[
y = \frac{2}{3}(0) - 2 = -2 \quad \Rightarrow \text{point } (0, -2)
\]
- When \( x = 3 \):
\[
y = \frac{2}{3}(3) - 2 = 0 \quad \Rightarrow \text{point } (3, 0)
\]
- When \( x = 6 \):
\[
y = \frac{2}{3}(6) - 2 = 2 \quad \Rightarrow \text{point } (6, 2)
\]
**For the red line \( y = -\frac{1}{2}x + 5 \):**
Again, select values for \( x \):
- When \( x = 0 \):
\[
y = -\frac{1}{2}(0) + 5 = 5 \quad \Rightarrow \text{point } (0, 5)
\]
- When \( x = 6 \):
\[
y = -\frac{1}{2}(6) + 5 = 2 \quad \Rightarrow \text{point } (6, 2)
\]
- When \( x = 10 \):
\[
y = -\frac{1}{2}(10) + 5 = 0 \quad \Rightarrow \text{point } (10, 0) \quad (\text{but this is outside the graph range: max y = 8})
\]
### Summary of Points:
- **Blue line points:**
- (0, -2)
- (3, 0)
- (6, 2)
- **Red line points:**
- (0, 5)
- (6, 2) (This point is also on the blue line.)
### Step 2: Finding the Intersection Point
To find the intersection point, we can set the two equations equal to each other:
\[
\frac{2}{3}x - 2 = -\frac{1}{2}x + 5
\]
### Solve for \( x \):
Multiply through by 6 (the least common multiple of the denominators):
\[
6\left(\frac{2}{3}x\right) - 6(2) = 6\left(-\frac{1}{2}x\right) + 6(5)
\]
This simplifies to:
\[
4x - 12 = -3x + 30
\]
Add \( 3x \) to both sides:
\[
4x + 3x - 12 = 30
\]
Combine:
\[
7x - 12 = 30
\]
Add 12 to both sides:
\[
7x = 42
\]
Divide by 7:
\[
x = 6
\]
### Step 3: Substitute for \( y \):
Plug \( x = 6 \) back into either equation (using the red line):
\[
y = -\frac{1}{2}(6) + 5 = -3 + 5 = 2
\]
### Solution of the System of Equations:
The solution to the system, or the intersection point, is \( (6, 2) \).
### Sketching the Graph:
- The blue line passes through (0, -2), (3, 0), and (6, 2).
- The red line passes through (0, 5), (6, 2).
- Both lines intersect at (6, 2).
### Final Summary:
**Intersection Point:** \( (6, 2) \)
You can now take these points to graph the lines on your graph paper, ensuring your graph's y-values do not exceed 8.
1. \( y = \frac{2}{3}x - 2 \) (blue line)
2. \( y = -\frac{1}{2}x + 5 \) (red line)
### Step 1: Determine Points for Each Equation
**For the blue line \( y = \frac{2}{3}x - 2 \):**
To find points, we can select a few values for \( x \) and calculate \( y \):
- When \( x = 0 \):
\[
y = \frac{2}{3}(0) - 2 = -2 \quad \Rightarrow \text{point } (0, -2)
\]
- When \( x = 3 \):
\[
y = \frac{2}{3}(3) - 2 = 0 \quad \Rightarrow \text{point } (3, 0)
\]
- When \( x = 6 \):
\[
y = \frac{2}{3}(6) - 2 = 2 \quad \Rightarrow \text{point } (6, 2)
\]
**For the red line \( y = -\frac{1}{2}x + 5 \):**
Again, select values for \( x \):
- When \( x = 0 \):
\[
y = -\frac{1}{2}(0) + 5 = 5 \quad \Rightarrow \text{point } (0, 5)
\]
- When \( x = 6 \):
\[
y = -\frac{1}{2}(6) + 5 = 2 \quad \Rightarrow \text{point } (6, 2)
\]
- When \( x = 10 \):
\[
y = -\frac{1}{2}(10) + 5 = 0 \quad \Rightarrow \text{point } (10, 0) \quad (\text{but this is outside the graph range: max y = 8})
\]
### Summary of Points:
- **Blue line points:**
- (0, -2)
- (3, 0)
- (6, 2)
- **Red line points:**
- (0, 5)
- (6, 2) (This point is also on the blue line.)
### Step 2: Finding the Intersection Point
To find the intersection point, we can set the two equations equal to each other:
\[
\frac{2}{3}x - 2 = -\frac{1}{2}x + 5
\]
### Solve for \( x \):
Multiply through by 6 (the least common multiple of the denominators):
\[
6\left(\frac{2}{3}x\right) - 6(2) = 6\left(-\frac{1}{2}x\right) + 6(5)
\]
This simplifies to:
\[
4x - 12 = -3x + 30
\]
Add \( 3x \) to both sides:
\[
4x + 3x - 12 = 30
\]
Combine:
\[
7x - 12 = 30
\]
Add 12 to both sides:
\[
7x = 42
\]
Divide by 7:
\[
x = 6
\]
### Step 3: Substitute for \( y \):
Plug \( x = 6 \) back into either equation (using the red line):
\[
y = -\frac{1}{2}(6) + 5 = -3 + 5 = 2
\]
### Solution of the System of Equations:
The solution to the system, or the intersection point, is \( (6, 2) \).
### Sketching the Graph:
- The blue line passes through (0, -2), (3, 0), and (6, 2).
- The red line passes through (0, 5), (6, 2).
- Both lines intersect at (6, 2).
### Final Summary:
**Intersection Point:** \( (6, 2) \)
You can now take these points to graph the lines on your graph paper, ensuring your graph's y-values do not exceed 8.
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