Question

which ordered pair is in the solution set of $y \geq \frac{1}{3}x + 4$? a $(-6, 1)$ b $(-1, 6)$ c $(6, -1)$ d $(1, -6)$

Explanation:

Step1: Substitute \(x=-6\) and \(y = 1\) into the inequality

Substitute \(x=-6\) and \(y = 1\) into \(y\geq\frac{1}{3}x + 4\).
Left - hand side \(y = 1\), right - hand side \(\frac{1}{3}\times(-6)+4=-2 + 4=2\).
Since \(1<2\), \((-6,1)\) is not a solution.

Step2: Substitute \(x=-1\) and \(y = 6\) into the inequality

Substitute \(x=-1\) and \(y = 6\) into \(y\geq\frac{1}{3}x + 4\).
Left - hand side \(y = 6\), right - hand side \(\frac{1}{3}\times(-1)+4=-\frac{1}{3}+4=\frac{-1 + 12}{3}=\frac{11}{3}\approx3.67\).
Since \(6>\frac{11}{3}\), \((-1,6)\) is a solution.

Step3: Substitute \(x = 6\) and \(y=-1\) into the inequality

Substitute \(x = 6\) and \(y=-1\) into \(y\geq\frac{1}{3}x + 4\).
Left - hand side \(y=-1\), right - hand side \(\frac{1}{3}\times6+4=2 + 4=6\).
Since \(-1<6\), \((6,-1)\) is not a solution.

Step4: Substitute \(x = 1\) and \(y=-6\) into the inequality

Substitute \(x = 1\) and \(y=-6\) into \(y\geq\frac{1}{3}x + 4\).
Left - hand side \(y=-6\), right - hand side \(\frac{1}{3}\times1+4=\frac{1+12}{3}=\frac{13}{3}\approx4.33\).
Since \(-6<\frac{13}{3}\), \((1,-6)\) is not a solution.

Answer:

B. \((-1,6)\)