Question

which ordered pair makes both inequalities true?
y > -3x + 3
y ≥ 2x - 2

options:
(1,0)
(-1,1)
(2,2)
(0,3)

Explanation:

Step1: Test (1,0) in first inequality

Substitute $x=1, y=0$ into $y > -3x + 3$:
$0 > -3(1) + 3 \implies 0 > 0$ (False)

Step2: Test (-1,1) in first inequality

Substitute $x=-1, y=1$ into $y > -3x + 3$:
$1 > -3(-1) + 3 \implies 1 > 6$ (False)

Step3: Test (2,2) in first inequality

Substitute $x=2, y=2$ into $y > -3x + 3$:
$2 > -3(2) + 3 \implies 2 > -3$ (True)

Step4: Test (2,2) in second inequality

Substitute $x=2, y=2$ into $y \geq 2x - 2$:
$2 \geq 2(2) - 2 \implies 2 \geq 2$ (True)

Step5: Verify (0,3) in second inequality

Substitute $x=0, y=3$ into $y \geq 2x - 2$:
$3 \geq 2(0) - 2 \implies 3 \geq -2$ (True), but we already found (2,2) works, and we confirm it satisfies both.

Answer:

C. (2,2)