Question

1. \

$$\begin{cases}3x + y < -2 \\\\ x - y \\leq -2\\end{cases}$$

Explanation:

Step1: Rewrite inequalities in slope - intercept form

For the first inequality \(3x + y<-2\), solve for \(y\):
Subtract \(3x\) from both sides, we get \(y < - 3x-2\). The boundary line is \(y=-3x - 2\), which is a dashed line (since the inequality is strict, \(y < - 3x - 2\)) with a slope of \(-3\) and a \(y\) - intercept of \(-2\).

For the second inequality \(x - y\leq - 2\), solve for \(y\):
Subtract \(x\) from both sides: \(-y\leq - x - 2\). Multiply both sides by \(- 1\) (and reverse the inequality sign) to get \(y\geq x + 2\). The boundary line is \(y=x + 2\), which is a solid line (since the inequality is non - strict, \(y\geq x + 2\)) with a slope of \(1\) and a \(y\) - intercept of \(2\).

Step2: Graph the boundary lines

• Graph of \(y=-3x - 2\):

Plot the \(y\) - intercept \((0,-2)\). Using the slope \(-3\) (rise \(-3\), run \(1\)), we can find another point. From \((0,-2)\), moving down \(3\) units and right \(1\) unit gives the point \((1,-5)\). Draw a dashed line through these points.

• Graph of \(y = x+2\):

Plot the \(y\) - intercept \((0,2)\). Using the slope \(1\) (rise \(1\), run \(1\)), from \((0,2)\) moving up \(1\) unit and right \(1\) unit gives the point \((1,3)\). Draw a solid line through these points.

Step3: Shade the regions

• For \(y < - 3x - 2\), we shade the region below the dashed line \(y=-3x - 2\). We can test a point not on the line, say \((0,0)\). Substitute into \(3x + y\): \(3(0)+0 = 0\), and \(0<-2\) is false. So we shade the opposite side of the line from \((0,0)\), which is the region where \(y < - 3x - 2\).
• For \(y\geq x + 2\), we shade the region above the solid line \(y=x + 2\). Test the point \((0,0)\): \(0\geq0 + 2\) is false. So we shade the opposite side of the line from \((0,0)\), which is the region where \(y\geq x + 2\).

Step4: Find the intersection of the two shaded regions

The solution to the system of inequalities is the region that is shaded for both \(y < - 3x - 2\) and \(y\geq x + 2\) simultaneously.

(Note: Since the problem is to graph the system of inequalities, the final answer is the graph with the dashed line \(y=-3x - 2\), solid line \(y = x + 2\), and the overlapping shaded region of \(y < - 3x - 2\) and \(y\geq x + 2\).)

Answer:

The solution is the region that lies above the solid line \(y = x+2\) and below the dashed line \(y=-3x - 2\). The graph has a dashed line \(y=-3x - 2\) (slope \(-3\), \(y\) - intercept \(-2\)) and a solid line \(y=x + 2\) (slope \(1\), \(y\) - intercept \(2\)), with the overlapping shaded region of the two inequalities' solution sets.