Question

instruction
graph the solution set to the system of
inequalities.
$-3x + 4y > 12$
$3x - 4y \geq 8$

Explanation:

Step1: Rewrite inequalities in slope - intercept form

For the first inequality \(-3x + 4y>12\), solve for \(y\):
Add \(3x\) to both sides: \(4y>3x + 12\).
Divide by \(4\): \(y>\frac{3}{4}x + 3\). The boundary line is \(y=\frac{3}{4}x + 3\) (dashed line since the inequality is strict, \(>\)).
For the second inequality \(3x-4y\geq8\), solve for \(y\):
Subtract \(3x\) from both sides: \(-4y\geq - 3x + 8\).
Divide by \(-4\) (and reverse the inequality sign): \(y\leq\frac{3}{4}x - 2\). The boundary line is \(y = \frac{3}{4}x-2\) (solid line since the inequality is \(\geq\)).

Step2: Analyze the relationship between the two lines

The slopes of both lines \(y=\frac{3}{4}x + 3\) and \(y=\frac{3}{4}x-2\) are equal (\(m = \frac{3}{4}\)), so the lines are parallel.
Now, let's check if there is any overlap in the solution regions.
Take a test point, say \((0,0)\) for the first inequality: \(0>\frac{3}{4}(0)+3\) i.e., \(0 > 3\), which is false. So the solution region for \(y>\frac{3}{4}x + 3\) is above the dashed line.
For the second inequality, test \((0,0)\): \(0\leq\frac{3}{4}(0)-2\) i.e., \(0\leq - 2\), which is false. So the solution region for \(y\leq\frac{3}{4}x - 2\) is below the solid line.
Since the two parallel lines have the region above the top line and below the bottom line, and the distance between \(y=\frac{3}{4}x + 3\) and \(y=\frac{3}{4}x-2\) is positive (the vertical distance between \(y = \frac{3}{4}x+3\) and \(y=\frac{3}{4}x - 2\) is \(|3-(-2)| = 5\) units), there is no intersection between the two solution regions. So the system of inequalities has no solution.

Answer:

The system of inequalities \(

$$\begin{cases}-3x + 4y>12\\3x-4y\geq8\end{cases}$$

\) has no solution (the solution set is empty). To graph it, draw the dashed line \(y=\frac{3}{4}x + 3\) and shade above it, draw the solid line \(y=\frac{3}{4}x-2\) and shade below it; there will be no overlapping shaded region.