Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y \geq -\frac{4}{5}x + 8$
$y \geq 2x - 6$
coordinate plane with x from -10 to 10 and y from -10 to 10, grid lines
answer

Explanation:

Step1: Find intersection of two lines

To find the intersection of \( y = -\frac{4}{5}x + 8 \) and \( y = 2x - 6 \), set them equal:
\( -\frac{4}{5}x + 8 = 2x - 6 \)
Multiply both sides by 5 to eliminate fraction: \( -4x + 40 = 10x - 30 \)
Add \( 4x \) to both sides: \( 40 = 14x - 30 \)
Add 30 to both sides: \( 70 = 14x \)
Divide by 14: \( x = 5 \)
Substitute \( x = 5 \) into \( y = 2x - 6 \): \( y = 2(5) - 6 = 4 \)
So intersection point is \( (5, 4) \).

Step2: Graph each inequality

• For \( y \geq -\frac{4}{5}x + 8 \): The line has slope \( -\frac{4}{5} \) and y-intercept 8. Since \( \geq \), shade above the line, line is solid.
• For \( y \geq 2x - 6 \): The line has slope 2 and y-intercept -6. Since \( \geq \), shade above the line, line is solid.

Step3: Determine solution region

The solution region is where both shadings overlap, above both lines, including the intersection and the lines.

Step4: Find a point in solution set

We can test \( (0, 8) \):

• For \( y \geq -\frac{4}{5}x + 8 \): \( 8 \geq -\frac{4}{5}(0) + 8 \) → \( 8 \geq 8 \), true.
• For \( y \geq 2x - 6 \): \( 8 \geq 2(0) - 6 \) → \( 8 \geq -6 \), true.

So \( (0, 8) \) is in the solution set. (Other points like \( (5, 4) \), \( (10, 8) \) also work.)

Answer:

A point in the solution set is \( (0, 8) \) (or other valid points like \( (5, 4) \), \( (10, 8) \), etc.). The graphical solution is the region above both lines \( y = -\frac{4}{5}x + 8 \) and \( y = 2x - 6 \), including the lines, with intersection at \( (5, 4) \).