Question

olve by graphing: \

$$\begin{cases} -3x + y < 3 \\\\ x + y \\geq -1 \\end{cases}$$

Explanation:

Step1: Rewrite inequalities as lines

For \(-3x + y < 3\), rewrite as \(y < 3x + 3\). The boundary line is \(y = 3x + 3\) (dashed, since \(<\)). For \(x + y \geq -1\), rewrite as \(y \geq -x - 1\). The boundary line is \(y = -x - 1\) (solid, since \(\geq\)).

Step2: Graph boundary lines

• \(y = 3x + 3\): Slope \(3\), y - intercept \(3\). Draw dashed line.
• \(y = -x - 1\): Slope \(-1\), y - intercept \(-1\). Draw solid line.

Step3: Shade regions

• For \(y < 3x + 3\), shade below the dashed line.
• For \(y \geq -x - 1\), shade above the solid line.

Step4: Find intersection

The solution is the overlapping shaded region (where both inequalities are satisfied).

Answer:

The solution is the region that is below the dashed line \(y = 3x + 3\) and above the solid line \(y=-x - 1\) (the overlapping shaded area when both inequalities are graphed). To present the final answer in terms of the graph, it's the area where these two shaded regions overlap. If we were to describe the vertices (by solving \(3x + 3=-x - 1\), we get \(4x=-4\), \(x = - 1\), \(y=0\); and intercepts: for \(y = 3x + 3\), x - intercept \(-1\), y - intercept \(3\); for \(y=-x - 1\), x - intercept \(-1\), y - intercept \(-1\)), the solution region is bounded by these lines and the overlapping area. But graphically, it's the region with the dashed line \(y = 3x+3\) (above? No, below) wait, no: \(y < 3x + 3\) is below the dashed line \(y = 3x + 3\), and \(y\geq -x - 1\) is above the solid line \(y=-x - 1\). So the overlapping region is between these two lines, with the solid line included and the dashed line excluded. The key is the graph of the system's solution is the intersection of the two shaded regions.