Question

which of the following is a solution to this inequality?
y < \frac{2}{3}x + 2
\bigcirc\\ (0, 3)
\bigcirc\\ (-3, 1)
\bigcirc\\ (3, 5)
\bigcirc\\ (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 \(\frac{8}{3}\approx2.666\)), true.

Answer:

(1, 2)