Question

question 1 of 14 step 3 of 3 solve the system of two linear inequalities graphically. step 3 of 3: find the region with points that satisfy both inequalities. {7x + 6y < 42, x ≥ 3} select the region you wish to be shaded: a b c d

Explanation:

Step1: Rewrite the second - inequality

Rewrite \(7x + 6y<42\) as \(y<-\frac{7}{6}x + 7\). The boundary line is \(y =-\frac{7}{6}x + 7\), and since the inequality is strict (\(<\)), the line will be dashed.

Step2: Analyze the first - inequality

The inequality \(x\geq3\) has a boundary line \(x = 3\), which is a vertical solid line.

Step3: Test points

For \(y<-\frac{7}{6}x + 7\), we can test the point \((0,0)\): \(0<-\frac{7}{6}(0)+7\), which is true, so we shade the region below the line \(y =-\frac{7}{6}x + 7\). For \(x\geq3\), we test a point to the right and left of \(x = 3\). Points to the right of \(x = 3\) (e.g., \((4,0)\)) satisfy \(x\geq3\). The region that satisfies both \(x\geq3\) and \(y<-\frac{7}{6}x + 7\) is the region to the right of \(x = 3\) and below \(y=-\frac{7}{6}x + 7\).

Answer:

The region that is to the right of the vertical line \(x = 3\) (including the points on the line \(x = 3\)) and below the dashed line \(y=-\frac{7}{6}x + 7\) (not including the points on the line \(y=-\frac{7}{6}x + 7\)). Without seeing the specific options A, B, C, D, if we assume a standard multiple - choice setup where regions are labeled on a graph, we would choose the region that meets these criteria.