Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y < -\frac{6}{5}x - 2$
$y > \frac{1}{5}x + 5$

Explanation:

Step1: Analyze the first inequality

The inequality is \( y < -\frac{6}{5}x - 2 \). This is a linear inequality in slope - intercept form (\( y=mx + b \)) where the slope \( m =-\frac{6}{5}\) and the y - intercept \( b=-2 \). Since the inequality is \( y < \), we will draw a dashed line (because the inequality is strict, \( y\) is not equal to \( -\frac{6}{5}x - 2 \)) and shade the region below the line.

Step2: Analyze the second inequality

The inequality is \( y > \frac{1}{5}x + 5 \). In slope - intercept form, the slope \( m=\frac{1}{5}\) and the y - intercept \( b = 5 \). Since the inequality is \( y>\), we will draw a dashed line (because the inequality is strict, \( y\) is not equal to \( \frac{1}{5}x+5\)) and shade the region above the line.

Step3: Find the intersection of the shaded regions

The solution set of the system of inequalities is the region that is shaded by both inequalities. To find a point in the solution set, we can pick a point and check if it satisfies both inequalities. Let's try \( x=-5 \).
For the first inequality: \( y<-\frac{6}{5}(-5)-2=6 - 2=4 \)
For the second inequality: \( y>\frac{1}{5}(-5)+5=- 1 + 5 = 4 \)
Let's choose \( y = 5 \) when \( x=-5 \).
Check the first inequality: \( 5<-\frac{6}{5}(-5)-2=6 - 2 = 4 \)? No, that's not correct. Let's try \( x=-10 \).
For the first inequality: \( y<-\frac{6}{5}(-10)-2=12 - 2 = 10 \)
For the second inequality: \( y>\frac{1}{5}(-10)+5=-2 + 5=3 \)
Let's choose \( y = 5 \) when \( x=-10 \).
Check the first inequality: \( 5<10 \), which is true.
Check the second inequality: \( 5>3 \), which is true. So the point \((-10,5)\) is in the solution set. (We can also find the intersection of the two lines \( -\frac{6}{5}x - 2=\frac{1}{5}x + 5 \)
\(-\frac{6}{5}x-\frac{1}{5}x=5 + 2\)
\(-\frac{7}{5}x=7\)
\(x=-5\)
Substitute \( x = - 5 \) into \( y=\frac{1}{5}x + 5\), we get \( y=\frac{1}{5}(-5)+5=-1 + 5 = 4 \). The two lines intersect at \((-5,4)\), and the solution region is above \( y=\frac{1}{5}x + 5\) and below \( y=-\frac{6}{5}x - 2\), so any point in that region will work. Another example: let's take \( x=-6 \)
For \( y>\frac{1}{5}(-6)+5=-\frac{6}{5}+5=\frac{-6 + 25}{5}=\frac{19}{5}=3.8 \)
For \( y<-\frac{6}{5}(-6)-2=\frac{36}{5}-2=\frac{36 - 10}{5}=\frac{26}{5}=5.2 \)
Let's choose \( y = 5 \). Check: \( 5>3.8 \) (satisfies the second inequality) and \( 5<5.2 \) (satisfies the first inequality). So the point \((-6,5)\) is in the solution set.

Answer:

A point in the solution set is \((-10,5)\) (or other valid points like \((-6,5)\), \((-5,4.5)\) etc. )