Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y \leq -\frac{6}{5}x - 4$
$y \geq \frac{2}{5}x + 4$

Explanation:

Step1: Analyze the first inequality \( y \leq -\frac{6}{5}x - 4 \)

The line \( y = -\frac{6}{5}x - 4 \) has a slope of \( -\frac{6}{5} \) and a y - intercept of \( - 4 \). Since the inequality is \( y\leq-\frac{6}{5}x - 4 \), we draw a solid line (because the inequality is non - strict) and shade the region below the line.

Step2: Analyze the second inequality \( y\geq\frac{2}{5}x + 4 \)

The line \( y=\frac{2}{5}x + 4 \) has a slope of \( \frac{2}{5} \) and a y - intercept of \( 4 \). Since the inequality is \( y\geq\frac{2}{5}x + 4 \), we draw a solid line (because the inequality is non - strict) and shade the region above the line.

Step3: Find the intersection of the two regions

To find a point in the solution set, we can find the intersection point of the two lines \( y = -\frac{6}{5}x - 4 \) and \( y=\frac{2}{5}x + 4 \). Set \( -\frac{6}{5}x - 4=\frac{2}{5}x + 4 \).
Add \( \frac{6}{5}x \) to both sides: \( - 4=\frac{2}{5}x+\frac{6}{5}x + 4 \)
Simplify the right - hand side: \( - 4=\frac{8}{5}x + 4 \)
Subtract 4 from both sides: \( - 8=\frac{8}{5}x \)
Multiply both sides by \( \frac{5}{8} \): \( x=-5 \)
Substitute \( x = - 5 \) into \( y=\frac{2}{5}x + 4 \): \( y=\frac{2}{5}\times(-5)+4=-2 + 4 = 2 \)
So the intersection point of the two lines is \( (-5,2) \). We can check if this point satisfies both inequalities:
For \( y\leq-\frac{6}{5}x - 4 \): Substitute \( x=-5,y = 2 \), \( 2\leq-\frac{6}{5}\times(-5)-4=6 - 4 = 2 \), which is true.
For \( y\geq\frac{2}{5}x + 4 \): Substitute \( x=-5,y = 2 \), \( 2\geq\frac{2}{5}\times(-5)+4=-2 + 4 = 2 \), which is true. So \( (-5,2) \) is in the solution set. We can also choose other points. For example, let's test \( x = 0 \) in both inequalities:
For \( y\leq-\frac{6}{5}(0)-4=-4 \), and for \( y\geq\frac{2}{5}(0)+4 = 4 \). There is no \( y \) that satisfies both when \( x = 0 \). Let's test \( x=-5 \), \( y = 2 \) as above. We can also pick a point like \( (-5,2) \) or other points in the overlapping region. For example, if we take \( x=-10 \), for the first inequality \( y\leq-\frac{6}{5}\times(-10)-4 = 12 - 4 = 8 \), for the second inequality \( y\geq\frac{2}{5}\times(-10)+4=-4 + 4 = 0 \). Let's take \( y = 4 \) when \( x=-10 \). Check \( y = 4 \) in first inequality: \( 4\leq-\frac{6}{5}\times(-10)-4=12 - 4 = 8 \) (true), in second inequality: \( 4\geq\frac{2}{5}\times(-10)+4=-4 + 4 = 0 \) (true). But the intersection point \( (-5,2) \) is a convenient point in the solution set.

Answer:

A point in the solution set is \((-5,2)\) (other valid points can also be chosen, for example, \((-10,4)\) etc. as long as they satisfy both inequalities)