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

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

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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