Question

Which point would not be a solution to the system of linear inequalities shown below?
y, is greater than, minus, start fraction, 3, divided by, 2, end fraction, x, plus, 7, y, is greater than, start fraction, 2, divided by, 3, end fraction, x, plus, 6
y>−
2
3
​
x+7y>
3
2
​
x+6
Answer
Multiple Choice Answers
left parenthesis, 0, comma, 9, right parenthesis(0,9)
left parenthesis, 0, comma, 8, right parenthesis(0,8)
left parenthesis, 0, comma, 10, right parenthesis(0,10)
left parenthesis, minus, 6, comma, minus, 9, right parenthesis(−6,−9)

Answer 1

The system is
y > −(3/2)x + 7 and y > (2/3)x + 6.

For (0,9),(0,8),(0,10): x=0 so the inequalities become y>7 and y>6, which 9, 8, and 10 all satisfy.

For (−6,−9):
−(3/2)(−6)+7 = 9+7 = 16, so we would need y>16 (but y=−9), and
(2/3)(−6)+6 = −4+6 = 2, so we would need y>2 (but y=−9).
Thus (−6,−9) does not satisfy the system.

Answer: (−6, −9).