Paper test question cordinate geometry involving inequality

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ifthyder: Master | Next Rank: 500 Posts; Posts: 118; Joined: Fri Apr 18, 2008 4:58 am; Thanked: 1 times

by ifthyder » Wed Sep 10, 2008 1:28 am

answer is E .please clarify the stragedy for solving it.

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Source: Beat The GMAT — Problem Solving | Wed Sep 10, 2008 1:28 am

bharathaitha: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Thu May 15, 2008 12:51 am; Thanked: 1 times

by bharathaitha » Wed Sep 10, 2008 2:56 am

Convert the equation into y=mx+c form. Thus

2x-3y <= -6
2x+6 <= 3y
3y >= 2x+6
y >= 2/3 x + 2

Here the slope (m=2/3) is positive and intercepting the y-axis at y=2. Hence the line will go thru Quadrant III, II, and I.

So the answer is E i.e. Quadrant II

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Shrinidhi: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sat Aug 02, 2008 12:39 pm

by Shrinidhi » Wed Sep 10, 2008 6:10 pm

bharathaitha wrote:Convert the equation into y=mx+c form. Thus

2x-3y <= -6
2x+6 <= 3y
3y >= 2x+6
y >= 2/3 x + 2

Here the slope (m=2/3) is positive and intercepting the y-axis at y=2. Hence the line will go thru Quadrant III, II, and I.

So the answer is E i.e. Quadrant II

I am sure, you mean, answer E, i.e. Quadrant IV right??

Anyways, another way to solve this problem is by substituting values.

Lets take Quadrant IV first. In quadrant IV, x value is always positive and y value is always negative. Lets take (1, -1)

so the equation becomes,
2(1) - 3(-1)
= 2 + 3
= 5,
which is clearly does not satisfy our condition.
Just to be sure, we can plugin other values. We can see that it will always be positive.
Hence, Answer is E

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manulath: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Tue Sep 09, 2008 10:12 pm; Location: Lko; Thanked: 3 times

by manulath » Mon Sep 15, 2008 4:07 am

lets rearrange the equation

2x - 3y =< -6
2x =< 3y - 6
2x =< 3(y-2)

Quadrant IV, x positive, y negetive

the right side 3(y-2) will always be negetive
and x will be positive

Hence, not possible

ans: E

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