Linear Programming Question 48
Question: The true statement for the graph of inequations $ 3x+2y\le 6 $ and $ 6x+4y\ge 20 $ , is
Options:
A) Both graphs are disjoint
B) Both do not contain origin
C) Both contain point (1, 1)
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
The equations, corresponding to inequalities $ 3x+2y\le 6 $ and $ 6x+4y\ge 20 $ , are $ 3x+2y=6 $ and $ 6x+4y=20 $ .
So the lines represented by these equations are parallel. Hence the graphs are disjoint.
BETA
