Question

which ordered pairs are in the solution set of the system of linear inequalities?
y ≥ -\frac{1}{2}x
y < \frac{1}{2}x + 1
\bigcirc (5, -2), (3, 1), (-4, 2)
\bigcirc (5, -2), (3, -1), (4, -3)
\bigcirc (5, -2), (3, 1), (4, 2)
\bigcirc (5, -2), (-3, 1), (4, 2)

Explanation:

Step1: Test (5,-2) in inequalities

First inequality: $-2 \geq -\frac{1}{2}(5) \implies -2 \geq -2.5$ (True)
Second inequality: $-2 < \frac{1}{2}(5)+1 \implies -2 < 3.5$ (True)

Step2: Test (3,1) in inequalities

First inequality: $1 \geq -\frac{1}{2}(3) \implies 1 \geq -1.5$ (True)
Second inequality: $1 < \frac{1}{2}(3)+1 \implies 1 < 2.5$ (True)

Step3: Test (-4,2) in inequalities

First inequality: $2 \geq -\frac{1}{2}(-4) \implies 2 \geq 2$ (True)
Second inequality: $2 < \frac{1}{2}(-4)+1 \implies 2 < -1$ (False)

Step4: Test (4,-3) in inequalities

First inequality: $-3 \geq -\frac{1}{2}(4) \implies -3 \geq -2$ (False)

Step5: Test (4,2) in inequalities

First inequality: $2 \geq -\frac{1}{2}(4) \implies 2 \geq -2$ (True)
Second inequality: $2 < \frac{1}{2}(4)+1 \implies 2 < 3$ (True)

Step6: Test (-3,1) in inequalities

First inequality: $1 \geq -\frac{1}{2}(-3) \implies 1 \geq 1.5$ (False)

Answer:

(5, -2), (3, 1), (4, 2)