Question

(a) graph the solution to the following system of inequalities. \

$$\begin{cases} -7x + 4y < 24 \\\\ y > \\frac{7}{4}x + 2 \\end{cases}$$

(b) give the coordinates of one point in the solution set. point in the solution set:

Explanation:

Step1: Analyze the inequalities

First, we have the system of inequalities:

1. \(-7x + 4y < 24\)
2. \(y > \frac{7}{4}x + 2\)

Let's rewrite the first inequality in slope - intercept form (\(y=mx + b\)):
Starting with \(-7x + 4y < 24\), add \(7x\) to both sides: \(4y<7x + 24\), then divide by 4: \(y<\frac{7}{4}x+6\)

The second inequality is \(y > \frac{7}{4}x + 2\)

So the solution set is the region between the two lines \(y=\frac{7}{4}x + 2\) (dashed line, since \(y>\frac{7}{4}x + 2\)) and \(y=\frac{7}{4}x+6\) (dashed line, since \(y <\frac{7}{4}x + 6\))

Step2: Find a test point

Let's choose a point that is between these two lines. Let's pick \(x = 0\)

For the second inequality \(y>\frac{7}{4}(0)+2=2\)

For the first inequality (rewritten as \(y<\frac{7}{4}(0)+6 = 6\))

So when \(x = 0\), we can choose \(y = 4\) (since \(2<4<6\))

Let's check if \((0,4)\) satisfies both original inequalities:

1. For \(-7x + 4y < 24\): Substitute \(x = 0,y = 4\), we get \(-7(0)+4(4)=16<24\), which is true.
2. For \(y>\frac{7}{4}x + 2\): Substitute \(x = 0,y = 4\), we get \(4>\frac{7}{4}(0)+2=2\), which is true.

Answer:

\((0,4)\) (Other valid points like \((4,9)\) (check: For \(x = 4\), \(y>\frac{7}{4}(4)+2=7 + 2=9\)? No, wait. Wait, for \(x = 4\), \(y>\frac{7}{4}(4)+2=9\) and \(y<\frac{7}{4}(4)+6=7 + 6 = 13\). So \(y = 10\) would work. \((4,10)\) also satisfies: \(-7(4)+4(10)=-28 + 40 = 12<24\) and \(10>\frac{7}{4}(4)+2=9\)) But \((0,4)\) is a simple one.)