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Glad about that.
Well I was a bit busy so I couldn't post...
@ bob bundy
Well how can x = 0 makes the expression goes to +/- infinity as dividing anything by 0 is undefined,and x is the denominator.
Thanks but I want to try this and the other problems by myself now,but if I got stuck anywhere I will post it here.
x + 2 changes sign at x = -2 so that will divide the cases we need to consider.
There's also something to consider at x = 0, as the expression goes to infinity there. It looks as though this will go to - infinity on one side of zero and + infinity on the other.
To simplify the expression you will have to multiply by x, and that change of sign will influence the inequality sign.
So you need to consider 4 cases:
(i) x < -2
(ii) x = -2
(iii) -2 < x < 0
(iv) 0 < x
Do you want to try this and post what you get or would you like more help with this ?
Yes,I understand this rule.
Well now its a bit more clear,I have to consider/assume the ranges (-3,0) and (0,+3) but I don't understand the logic just because +/-3 makes the denominator 0,in many inequation it happens but we don't have to choose those cases,what my problem was I havent looked at different cases,but still this problem is not clear properly the logic behind choosing the cases ,when I reduced the equation tofrom here why I can't able to solve the inequation ?, may be I need more time .
bob bundy what you did I understood totally and appreciate it.
Well thanks for helping me out
If I call... statement one and ... statement two, then we can work the logic like this.
I got statement one by algebra assuming statement two. So statement one is only true to the extent that it obeys statement two. As statement two is a subset of statement one that means the inequality is satisfied just for the subset ie
So that provides part of the missing answer.
Testing x = 0 by substitution shows it may be added to the set giving
You can finish by considering (0,+3) and showing it is legitimate to add this to the range
Sorry still can't understand...
How can x be +/- 4 ,lets put it in the equation and see
Taking x as -4
So I think x can't be +/- 4
Yes I can understand that,but how you came upto "-3 < x < 0" we just cant guess any range I want to know what are the steps,in your post #2 the case 1 is ok it givesbut case 2 doesn't give any result expect than x is 0 and case 3 which you actually explained in post #5 gives but as already given . But thats not the answer and still where is .
Thanks for the reply
Now in that range for x, (-x-3) is negative
eg. (--2.9 - 3) = -0.1
So when you multiply by that denominator you must reverse the inequality:
This final statement is true in that range so the range is part of the solution set.
I'll look at the other questions once you are happy with this one.
Here are the other questions
Question no 4 I think I may solve it .... but the rest I tried but no result till now.
And thanks bob bundy for the reply
Well still I am having problem to understand it...
How you can take the case (-∞,-3) as x can't be -4 or 4 ?
So we can write
This far I did it... and I can also see that x can't be -3 or 3 as it will make the denominator 0.
But I can't find a proper steps to come to a proper solution set...I am still trying....
(ii) x = 0
(iii) x <0
LATER EDIT. Then I made the graph and realised where the missing answers had gone.
As function tends to +/- infinity at x = +/- 3 we should look at 5 cases
(-∞,-3) (-3,0) 0 (0,3) (3, ∞)
That will produce the answer you want.
Hope that helps.
Well I have four inequations that I can't able to understand properly,what are the steps you guys will take to solve this...
I did it and my solution was
But the correct soultion is
I can understand my solution is wrong ,but don't know what steps are used to get the correct solution set.Can anyone show me the proper steps?