Question

which of the following is a solution to this inequality?
$y < \frac{2}{3}x + 2$
\\(\circ\\) (0, 3)
\\(\circ\\) (-3, 1)
\\(\circ\\) (3, 5)
\\(\circ\\) (1, 2)

Explanation:

Step1: Test (0, 3)

Substitute \(x = 0\), \(y = 3\) into \(y < \frac{2}{3}x + 2\):
\(3 < \frac{2}{3}(0) + 2\) → \(3 < 2\), false.

Step2: Test (-3, 1)

Substitute \(x = -3\), \(y = 1\) into \(y < \frac{2}{3}x + 2\):
\(1 < \frac{2}{3}(-3) + 2\) → \(1 < -2 + 2\) → \(1 < 0\), false.

Step3: Test (3, 5)

Substitute \(x = 3\), \(y = 5\) into \(y < \frac{2}{3}x + 2\):
\(5 < \frac{2}{3}(3) + 2\) → \(5 < 2 + 2\) → \(5 < 4\), false.

Step4: Test (1, 2)

Substitute \(x = 1\), \(y = 2\) into \(y < \frac{2}{3}x + 2\):
\(2 < \frac{2}{3}(1) + 2\) → \(2 < \frac{2}{3} + 2\) → \(2 < \frac{8}{3}\) (since \(\frac{8}{3} \approx 2.666\)), true.

Answer:

(1, 2)