Question

which ordered pairs are in the solution set of the system of linear inequalities?
$y \geq -\frac{1}{3}x + 2$
$y < 2x + 3$
\bigcirc (2, 2), (3, 1), (4, 2)
\bigcirc (2, 2), (3, -1), (4, 1)
\bigcirc (2, 2), (1, -2), (0, 2)

Explanation:

Step1: Check the first inequality \( y \geq -\frac{1}{3}x + 2 \) for each ordered pair in the options.

For option 1:

• For \((2,2)\): \( 2 \geq -\frac{1}{3}(2)+2 = -\frac{2}{3}+2=\frac{4}{3}\), \(2\geq\frac{4}{3}\) is true.
• For \((3,1)\): \( 1 \geq -\frac{1}{3}(3)+2=-1 + 2 = 1\), \(1\geq1\) is true.
• For \((4,2)\): \( 2 \geq -\frac{1}{3}(4)+2=-\frac{4}{3}+2=\frac{2}{3}\), \(2\geq\frac{2}{3}\) is true.

For option 2:

• For \((3,-1)\): \( -1 \geq -\frac{1}{3}(3)+2=-1 + 2 = 1\), \(-1\geq1\) is false. So option 2 is out.

For option 3:

• For \((1,-2)\): \( -2 \geq -\frac{1}{3}(1)+2=-\frac{1}{3}+2=\frac{5}{3}\), \(-2\geq\frac{5}{3}\) is false. So option 3 is out.

Step2: Check the second inequality \( y < 2x + 3 \) for the remaining ordered pairs in option 1.

• For \((2,2)\): \( 2 < 2(2)+3=4 + 3 = 7\), \(2<7\) is true.
• For \((3,1)\): \( 1 < 2(3)+3=6 + 3 = 9\), \(1<9\) is true.
• For \((4,2)\): \( 2 < 2(4)+3=8 + 3 = 11\), \(2<11\) is true.

Answer:

\((2, 2), (3, 1), (4, 2)\) (the first option)