Question

question
solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y < -\frac{3}{2}x - 7$
$y < \frac{1}{2}x + 1$

Explanation:

Step1: Analyze first inequality

The first inequality is \( y < -\frac{3}{2}x - 7 \). The boundary line is \( y = -\frac{3}{2}x - 7 \), which has a slope of \(-\frac{3}{2}\) and a y - intercept of \(-7\). Since the inequality is \( y < \), we draw a dashed line and shade below the line.

Step2: Analyze second inequality

The second inequality is \( y < \frac{1}{2}x + 1 \). The boundary line is \( y=\frac{1}{2}x + 1 \), with a slope of \(\frac{1}{2}\) and a y - intercept of \(1\). Since the inequality is \( y < \), we draw a dashed line and shade below the line.

Step3: Find the intersection of the shaded regions

To find a point in the solution set, we can pick a point that satisfies both inequalities. Let's try \( x = 0 \):

• For the first inequality: \( y<-\frac{3}{2}(0)-7=-7 \)
• For the second inequality: \( y < \frac{1}{2}(0)+1 = 1 \)

Let's try \( x = 6 \):

• For the first inequality: \( y<-\frac{3}{2}(6)-7=-9 - 7=-16 \)
• For the second inequality: \( y<\frac{1}{2}(6)+1=3 + 1 = 4 \)

Let's try \( x = 8 \):

• For the first inequality: \( y<-\frac{3}{2}(8)-7=-12-7=-19 \)
• For the second inequality: \( y<\frac{1}{2}(8)+1 = 4 + 1=5 \)

A simple way is to find the intersection of the two half - planes. Let's solve the system of equations of the boundary lines \( y = -\frac{3}{2}x - 7 \) and \( y=\frac{1}{2}x + 1 \) to find the point of intersection of the boundary lines (though the solution set is the region below both lines).

Set \( -\frac{3}{2}x-7=\frac{1}{2}x + 1 \)
\(-\frac{3}{2}x-\frac{1}{2}x=1 + 7\)
\(-2x=8\)
\(x=-4\)

Substitute \( x = - 4 \) into \( y=\frac{1}{2}x + 1 \), we get \( y=\frac{1}{2}(-4)+1=-2 + 1=-1 \)

Now, we need a point that is below both lines. Let's take \( x = 0 \), but as we saw \( y < - 7 \) for the first inequality. Let's take \( x = 6 \), \( y=-17 \) (since \(-17<-\frac{3}{2}(6)-7=-16\) and \(-17 < \frac{1}{2}(6)+1 = 4\)). So the point \((6,-17)\) is in the solution set. Another point could be \((4,-13)\):

• For \( y<-\frac{3}{2}(4)-7=-6 - 7=-13 \), so \( y=-14\) satisfies \( y < - 13 \)
• For \( y<\frac{1}{2}(4)+1=2 + 1 = 3 \), and \(-14<3\)

Answer:

A point in the solution set is \((6,-17)\) (other valid points like \((4, - 14)\), \((8,-20)\) etc. are also correct)