Question

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

Explanation:

Step1: Test point (-1, 3)

Substitute \(x = -1\), \(y = 3\) into \(y < -\frac{1}{2}x + 2\):
Right - hand side: \(-\frac{1}{2}(-1)+2=\frac{1}{2}+2=\frac{5}{2}=2.5\)
Check \(3 < 2.5\)? No.

Step2: Test point (3, -2)

Substitute \(x = 3\), \(y = -2\) into \(y < -\frac{1}{2}x + 2\):
Right - hand side: \(-\frac{1}{2}(3)+2=-\frac{3}{2}+2=\frac{1}{2}=0.5\)
Check \(-2 < 0.5\)? Yes.

Step3: Test point (2, 1)

Substitute \(x = 2\), \(y = 1\) into \(y < -\frac{1}{2}x + 2\):
Right - hand side: \(-\frac{1}{2}(2)+2=-1 + 2=1\)
Check \(1 < 1\)? No (since it's \(y < -\frac{1}{2}x + 2\), not \(y\leqslant-\frac{1}{2}x + 2\)).

Step4: Test point (2, 3)

Substitute \(x = 2\), \(y = 3\) into \(y < -\frac{1}{2}x + 2\):
Right - hand side: \(-\frac{1}{2}(2)+2=-1 + 2=1\)
Check \(3 < 1\)? No.

Answer:

\((3, -2)\)