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

Question

y =
2
3​x−2 (blue line)

y=−
1
2​x+5 (red line)

Answer 1

To graph and solve the system of equations given by:

1. \( y = \frac{2}{3}x - 2 \) (blue line)
2. \( y = -\frac{1}{2}x + 5 \) (red line)

Step 1: Graph the equations

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

• Y-intercept: Set \( x = 0 \): \[ y = \frac{2}{3}(0) - 2 = -2 \] So the point is \( (0, -2) \).

• X-intercept: Set \( y = 0 \): \[ 0 = \frac{2}{3}x - 2 \implies \frac{2}{3}x = 2 \implies x = 3 \] So the point is \( (3, 0) \).

Using these two points, draw the blue line.

For the red line: \( y = -\frac{1}{2}x + 5 \)

• Y-intercept: Set \( x = 0 \): \[ y = -\frac{1}{2}(0) + 5 = 5 \] So the point is \( (0, 5) \).

• X-intercept: Set \( y = 0 \): \[ 0 = -\frac{1}{2}x + 5 \implies \frac{1}{2}x = 5 \implies x = 10 \] So the point is \( (10, 0) \).

Using these two points, draw the red line.

Step 2: Find the intersection point (solution to the system)

To find the solution to the system, set the two equations equal to each other:

\[ \frac{2}{3}x - 2 = -\frac{1}{2}x + 5 \]

To eliminate the fractions, multiply through by 6 (the least common multiple of 3 and 2):

\[ 6 \left(\frac{2}{3}x\right) - 6(2) = 6 \left(-\frac{1}{2}x\right) + 6(5) \]

This simplifies to:

\[ 4x - 12 = -3x + 30 \]

Combine like terms:

\[ 4x + 3x = 30 + 12 \] \[ 7x = 42 \] \[ x = 6 \]

Now substitute \( x = 6 \) back into either equation to find \( y \):

Using the blue line equation: \[ y = \frac{2}{3}(6) - 2 = 4 - 2 = 2 \]

So the intersection point is \( (6, 2) \).

Summary

The solution to the system of equations is \( (6, 2) \), which is the point where the blue line intersects the red line.

Answer 2

what are the points

Answer 3

The points related to the system of equations \(y = \frac{2}{3}x - 2\) (blue line) and \(y = -\frac{1}{2}x + 5\) (red line) are:

Key Points for Each Line

1. Blue Line \(y = \frac{2}{3}x - 2\):

   • Y-intercept: \( (0, -2) \)
   • X-intercept: \( (3, 0) \)

2. Red Line \(y = -\frac{1}{2}x + 5\):

   • Y-intercept: \( (0, 5) \)
   • X-intercept: \( (10, 0) \)

Intersection Point

• Point of Intersection: \( (6, 2) \)

Summary of Points

• Blue Line Points: \( (0, -2) \) and \( (3, 0) \)
• Red Line Points: \( (0, 5) \) and \( (10, 0) \)
• Intersection Point: \( (6, 2) \)

If you need more specific points or details, feel free to ask!

Answer 4

the graph only goes to 8

Answer 5

If the graph dimensions are limited and only extend to an axis value of 8, you can still effectively graph the two equations by using the relevant points that fall within the graph's boundaries. Let's highlight the points again, but ensure that they fit within the range of your graph, which goes up to (and includes) 8.

Points for the Blue Line \(y = \frac{2}{3}x - 2\)

1. Y-intercept:

   • \( (0, -2) \) (this point is below the graph)

2. X-intercept:

   • \( (3, 0) \)

3. A point further for the graph:

   • Choose \( x = 6 \): \[ y = \frac{2}{3}(6) - 2 = 4 - 2 = 2 \] So the point \( (6, 2) \) is within the graph.

Points for the Red Line \(y = -\frac{1}{2}x + 5\)

1. Y-intercept:

   • \( (0, 5) \)

2. X-intercept:

   • \( (10, 0) \) (this point is outside your graph range)

3. A point further for the graph:

   • Choose \( x = 4 \): \[ y = -\frac{1}{2}(4) + 5 = -2 + 5 = 3 \] So the point \( (4, 3) \) is within the graph.

Summary of Points Within Graph Limits

• Blue Line Points (valid within graph limits):

  • \( (3, 0) \)
  • \( (6, 2) \)

• Red Line Points (valid within graph limits):

  • \( (0, 5) \)
  • \( (4, 3) \)

• Intersection Point:

  • \( (6, 2) \) (this point is the solution to the system)

Using these points, you should be able to graph both lines effectively on your limited graph. If you need more specific points or adjustments, let me know!