Question

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

$$\begin{cases}-4x + 3y < 12\\\\3x + 5y \\geq 15\\end{cases}$$

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

Explanation:

Part (a) Graphing the System of Inequalities

Step 1: Rewrite the first inequality in slope - intercept form

We have the inequality \(-4x + 3y<12\). First, solve for \(y\):
Add \(4x\) to both sides: \(3y<4x + 12\).
Divide both sides by 3: \(y<\frac{4}{3}x + 4\).
The boundary line is \(y=\frac{4}{3}x + 4\) (dashed line because the inequality is strict, \(y <\) ). To find two points on this line, when \(x = 0\), \(y=4\); when \(x=- 3\), \(y=\frac{4}{3}\times(-3)+4=-4 + 4 = 0\).

Step 2: Rewrite the second inequality in slope - intercept form

We have the inequality \(3x + 5y\geq15\). Solve for \(y\):
Subtract \(3x\) from both sides: \(5y\geq - 3x+15\).
Divide both sides by 5: \(y\geq-\frac{3}{5}x + 3\).
The boundary line is \(y =-\frac{3}{5}x + 3\) (solid line because the inequality is non - strict, \(y\geq\)). To find two points on this line, when \(x = 0\), \(y = 3\); when \(x = 5\), \(y=-\frac{3}{5}\times5+3=-3 + 3=0\).

Step 3: Determine the regions to shade

For the inequality \(y<\frac{4}{3}x + 4\), we can test the point \((0,0)\). Substitute \(x = 0\) and \(y = 0\) into \(y<\frac{4}{3}x + 4\): \(0<0 + 4\), which is true. So we shade the region below the dashed line \(y=\frac{4}{3}x + 4\).

For the inequality \(y\geq-\frac{3}{5}x + 3\), test the point \((0,0)\). Substitute \(x = 0\) and \(y = 0\) into \(y\geq-\frac{3}{5}x + 3\): \(0\geq0 + 3\), which is false. So we shade the region above the solid line \(y=-\frac{3}{5}x + 3\).

The solution to the system is the region that is shaded for both inequalities.

Part (b) Finding a Point in the Solution Set

We can choose a point that satisfies both inequalities. Let's try \(x = 0\).

For the first inequality \(y<\frac{4}{3}(0)+4=4\).

For the second inequality \(y\geq-\frac{3}{5}(0)+3 = 3\).

So we can choose \(y = 3\) (since \(3\geq3\) and \(3<4\)). Let's check the point \((0,3)\):

• For \(-4x + 3y<12\): Substitute \(x = 0\), \(y = 3\): \(-4(0)+3(3)=9<12\), which is true.
• For \(3x + 5y\geq15\): Substitute \(x = 0\), \(y = 3\): \(3(0)+5(3)=15\geq15\), which is true.

Answer:

(Part b):
The coordinates of one point in the solution set are \((0,3)\) (other valid points could also be, for example, \((5,0)\) needs to be checked: For \(-4(5)+3(0)=-20<12\) (true), for \(3(5)+5(0)=15\geq15\) (true), so \((5,0)\) is also valid. But we will use \((0,3)\) here).
So the point in the solution set is \(\boldsymbol{(0,3)}\)