Question

graph the solution set of the system of linear inequalities in the coordinate plane.
$y > -\frac{3}{4}x + 4$
$y < \frac{3}{2}x - 5$

Explanation:

Step1: Analyze the first inequality \( y > -\frac{3}{4}x + 4 \)

The inequality is in slope - intercept form \( y=mx + b \), where the slope \( m =-\frac{3}{4}\) and the y - intercept \( b = 4\). Since the inequality is \( y >-\frac{3}{4}x + 4\), the line should be dashed (because the inequality is strict, \(>\) not \(\geq\)) and we shade above the line.

Step2: Analyze the second inequality \( y<\frac{3}{2}x - 5\)

The inequality is in slope - intercept form \( y = mx + b\), where the slope \( m=\frac{3}{2}\) and the y - intercept \( b=- 5\). Since the inequality is \( y<\frac{3}{2}x - 5\), the line should be dashed (because the inequality is strict, \(<\) not \(\leq\)) and we shade below the line.

Step3: Find the intersection of the two shaded regions

To graph the solution set of the system of inequalities, we first graph each inequality separately and then find the region that is common to both. The solution set is the region that is above the line \( y =-\frac{3}{4}x + 4\) (dashed line) and below the line \( y=\frac{3}{2}x - 5\) (dashed line).

To find the intersection point of the two lines \( y =-\frac{3}{4}x + 4\) and \( y=\frac{3}{2}x - 5\), we set them equal to each other:

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Substitute \( x = 4\) into \( y=\frac{3}{2}x-5\), we get \( y=\frac{3}{2}(4)-5=6 - 5 = 1\). So the two lines intersect at the point \((4,1)\).

The solution set is the region bounded by the two dashed lines, above \( y =-\frac{3}{4}x + 4\) and below \( y=\frac{3}{2}x - 5\).

Answer:

The solution set is the region that is above the dashed line \( y =-\frac{3}{4}x + 4\) and below the dashed line \( y=\frac{3}{2}x - 5\), with the intersection point of the two lines at \((4,1)\). To graph it, draw two dashed lines (one with slope \(-\frac{3}{4}\) and y - intercept 4, the other with slope \(\frac{3}{2}\) and y - intercept - 5) and shade the region that satisfies both inequalities.