Question

writing an inequality in two variables given a table of values
which linear inequality could represent the given table of values?

x	y
-2	4
3	-3
3	-4

\\(\circ\\) \\(y < -2x + 3\\)
\\(\circ\\) \\(y \leq -2x + 3\\)
\\(\circ\\) \\(y > -\frac{1}{2}x - 3\\)
\\(\circ\\) \\(y \leq -\frac{1}{2}x - 3\\)

Explanation:

Step1: Test first point in options

Take $(x=-4,y=-1)$:

• Option1: $-1 < -2(-4)+3 \implies -1 < 11$ (True)
• Option2: $-1 \leq -2(-4)+3 \implies -1 \leq 11$ (True)
• Option3: $-1 > -\frac{1}{2}(-4)-3 \implies -1 > -1$ (False, eliminate)
• Option4: $-1 \leq -\frac{1}{2}(-4)-3 \implies -1 \leq -1$ (True)

Step2: Test second point

Take $(x=-2,y=4)$:

• Option1: $4 < -2(-2)+3 \implies 4 < 7$ (True)
• Option2: $4 \leq -2(-2)+3 \implies 4 \leq 7$ (True)
• Option4: $4 \leq -\frac{1}{2}(-2)-3 \implies 4 \leq -2$ (False, eliminate)

Step3: Test third point

Take $(x=3,y=-3)$:

• Option1: $-3 < -2(3)+3 \implies -3 < -3$ (False, eliminate)
• Option2: $-3 \leq -2(3)+3 \implies -3 \leq -3$ (True)

Step4: Verify fourth point

Take $(x=3,y=-4)$:

• Option2: $-4 \leq -2(3)+3 \implies -4 \leq -3$ (True)

Answer:

$\boldsymbol{y \leq -2x + 3}$ (Option 2)