Question

which point is a solution to the linear inequality $y < -\frac{1}{2}x + 2$?
\bigcirc\\ (2, 3)
\bigcirc\\ (2, 1)
\bigcirc\\ (3, -2)
\bigcirc\\ (-1, 3)

Explanation:

Step1: Test (2,3)

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

Step2: Test (2,1)

Substitute \(x = 2\), \(y = 1\):
\(1<-\frac{1}{2}(2)+2\) → \(1<-1 + 2\) → \(1<1\), false (since it’s \(<\), not \(\leq\)).

Step3: Test (3,-2)

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

Step4: Test (-1,3)

Substitute \(x = -1\), \(y = 3\):
\(3<-\frac{1}{2}(-1)+2\) → \(3<\frac{1}{2}+2\) → \(3<\frac{5}{2}\), false.

Answer:

C. (3, -2)