Question

1. \

$$\begin{cases} 5x - 6y > 42 \\\\ 3x + 2y \\leq 14 \\end{cases}$$

Explanation:

Step1: Analyze the first inequality \(5x - 6y>42\)

Rewrite it in slope - intercept form (\(y = mx + b\)):
\(-6y>-5x + 42\)
Divide both sides by \(- 6\) (and reverse the inequality sign): \(y<\frac{5}{6}x - 7\)
The boundary line is \(y=\frac{5}{6}x - 7\), which is a dashed line (since the inequality is strict, \(>\) or \(<\)) with slope \(\frac{5}{6}\) and \(y\) - intercept \(-7\). We shade the region below this line.

Step2: Analyze the second inequality \(3x + 2y\leq14\)

Rewrite it in slope - intercept form:
\(2y\leq - 3x+14\)
\(y\leq-\frac{3}{2}x + 7\)
The boundary line is \(y =-\frac{3}{2}x + 7\), which is a solid line (since the inequality is \(\leq\)) with slope \(-\frac{3}{2}\) and \(y\) - intercept \(7\). We shade the region below (or on) this line.

Step3: Graph the boundary lines and find the intersection

• For \(y=\frac{5}{6}x - 7\): When \(x = 0\), \(y=-7\); when \(y = 0\), \(x=\frac{42}{5}=8.4\). Plot the points \((0,-7)\) and \((8.4,0)\) and draw a dashed line through them, then shade below the line.
• For \(y =-\frac{3}{2}x + 7\): When \(x = 0\), \(y = 7\); when \(y=0\), \(x=\frac{14}{3}\approx4.67\). Plot the points \((0,7)\) and \((\frac{14}{3},0)\) and draw a solid line through them, then shade below (or on) the line.

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

Answer:

To graph the system \(

$$\begin{cases}5x - 6y>42\\3x + 2y\leq14\end{cases}$$

\), we:

1. Graph \(y<\frac{5}{6}x - 7\) (dashed line, shade below).
2. Graph \(y\leq-\frac{3}{2}x + 7\) (solid line, shade below).

The solution region is the overlapping shaded area of the two inequalities.