A customer is buying bath towels and hand towels and can spend no more than $100. Each bath towel cost $8, and each hand towel costs $5. The inequality `8x+5y\le100` represents all possible combinations of `x`, the number of bath towels, and `y`, the number of hand towels the customer can buy.

Question

Which graph best represents the solution set for this inequality?

Answer 1

To graph the inequality \( 8x + 5y \leq 100 \), we can start by determining the boundary line represented by the equation \( 8x + 5y = 100 \).

Step 1: Find Points for the Boundary Line

• Finding x-intercept (set \( y = 0 \)): \[ 8x + 5(0) = 100 \implies 8x = 100 \implies x = 12.5 \] So the x-intercept is \( (12.5, 0) \).

• Finding y-intercept (set \( x = 0 \)): \[ 8(0) + 5y = 100 \implies 5y = 100 \implies y = 20 \] So the y-intercept is \( (0, 20) \).

Step 2: Plot the Line

Now we can plot the line connecting the points \( (12.5, 0) \) and \( (0, 20) \).

Step 3: Determine the Region

Since the inequality is \( 8x + 5y \leq 100 \), we are looking for the region that is below the line (including the line itself). To confirm this:

1. Choose a test point not on the line, such as \( (0,0) \): \[ 8(0) + 5(0) \leq 100 \implies 0 \leq 100 \text{ (true)} \]

Thus, the region that satisfies the inequality \( 8x + 5y \leq 100 \) includes the origin and is below the line.

Step 4: Graphing Guidelines

• The boundary line should be solid (indicating that points on the line are included in the solution set).
• The area below the line (where \( 8x + 5y \) is less than or equal to 100) should be shaded.

Summary

To summarize:

• The graph features a solid line segment from \( (12.5, 0) \) to \( (0, 20) \) and the area below (or on the line) should be shaded to represent all combinations of bath towels and hand towels the customer can buy while spending no more than $100.

You can compare with the provided graph options to see which one matches this description.

Answer 2

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

`y\ =\ \frac{2}{3}x-2` (blue line)

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

Answer 3

To graph the system of equations and find the solution, we'll plot each line described by the equations:

1. Equation of the blue line: \[ y = \frac{2}{3}x - 2 \]

   • Find x-intercept (set \( y = 0 \)): \[ 0 = \frac{2}{3}x - 2 \implies \frac{2}{3}x = 2 \implies x = 3 \] So, the x-intercept is \( (3, 0) \).
   • Find y-intercept (set \( x = 0 \)): \[ y = \frac{2}{3}(0) - 2 = -2 \] So, the y-intercept is \( (0, -2) \).
   • Plot Points:
     • (0, -2)
     • (3, 0)

2. Equation of the red line: \[ y = -\frac{1}{2}x + 5 \]

   • Find x-intercept (set \( y = 0 \)): \[ 0 = -\frac{1}{2}x + 5 \implies \frac{1}{2}x = 5 \implies x = 10 \] So, the x-intercept is \( (10, 0) \).
   • Find y-intercept (set \( x = 0 \)): \[ y = -\frac{1}{2}(0) + 5 = 5 \] So, the y-intercept is \( (0, 5) \).
   • Plot Points:
     • (0, 5)
     • (10, 0)

Step 3: Sketch the Lines

1. Blue Line:

   • Connect the points \( (0, -2) \) and \( (3, 0) \) to draw the blue line.
   • Continue the line in both directions, maintaining the slope \( \frac{2}{3} \).

2. Red Line:

   • Connect the points \( (0, 5) \) and \( (10, 0) \) to draw the red line.
   • Continue the line in both directions, maintaining the slope \( -\frac{1}{2} \).

Step 4: Identify the Intersection Point

To find the intersection of the two lines:

Set the equations equal to each other: \[ \frac{2}{3}x - 2 = -\frac{1}{2}x + 5 \]

Multiply everything by 6 (to eliminate fractions): \[ 4x - 12 = -3x + 30 \]

Combine like terms: \[ 4x + 3x = 30 + 12 \] \[ 7x = 42 \implies x = 6 \]

Now substitute \( x = 6 \) back into one of the equations. We'll use the first one: \[ y = \frac{2}{3}(6) - 2 = 4 - 2 = 2 \]

Conclusion

The intersection point of the two lines (which is the solution to the system of equations) is \( (6, 2) \).

Graph Summary:

• Plot the blue line from \( (0, -2) \) to \( (3, 0) \) and extend it.
• Plot the red line from \( (0, 5) \) to \( (10, 0) \) and extend it.
• Shade the intersection point \( (6, 2) \).