Ok, a couple things here.
First, I agree that C is not the right answer and I agree with D.
Second is for srknori. You can not subtract two inequalities when their signs are toward the same direction.
For example, if you have x>5, and also y>3, you cannot subtract the two inequalities and arrive x-y>2. It's clear that it may not be the case. Say x is 6 and y is also 6, then x-y=0, certainly not greater than 2.
You can only do subtraction when the signs for two inequalities are toward different direction. For example, if we know x>5 and y<3, then we know for sure that x-y>2.
Third, to answer Raghavender's question:
Raghavender wrote:
thus, (x-5)(x+3) >=0
<=> x >= 5 or x <= -3
does x+3 >=0, not mean tht x>= -3( when u subtract 3 on both sides...)
please clear this doubt for me.....
When you have something like this, there is a relatively straight forward way to find the solution set. Look at this example:
(x-a)(y-b)<0
Remember for the product to be negative one has to be positive and the other has to be negative. So simply do two cases: x-a>0, y-b<0; and x-a<0, y-b>0. The result would be all possible solutions.
In our case, we have (x-5)(x+3)>=0. For the product to be non-negative, either both have to be non-negative, or both have to be non-positive. (Note here I use non-negative instead of positive because we need to include the case of zero.) So you list two options: x-5>=0, x+3>=0 and x-5<=0, x+3<=0. Now the solution set for the first set is x>=5, and the solution set for the second set is x<=-3.
The sticky math basic principle thread has some discussions about inequalites. I would recommend people who wants to refresh on this topic go spend some time to read it. It may help.
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