Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #1 20121122 20:02:33
Linear inequation involving absolute value problem...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? Thanks Last edited by debjit625 (20121122 20:04:31) Debjit Roy ___________________________________________________ The essence of mathematics lies in its freedom  Georg Cantor #2 20121122 20:12:16
Re: Linear inequation involving absolute value problem...hi debjit625 (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. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #3 20121123 01:01:40
Re: Linear inequation involving absolute value problem...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.... Last edited by debjit625 (20121123 19:35:02) Debjit Roy ___________________________________________________ The essence of mathematics lies in its freedom  Georg Cantor #4 20121123 01:13:18
Re: Linear inequation involving absolute value problem...Here are the other questions Question 3 Question 4 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 Debjit Roy ___________________________________________________ The essence of mathematics lies in its freedom  Georg Cantor #5 20121123 03:49:12
Re: Linear inequation involving absolute value problem...Q1. Now in that range for x, (x3) 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. Bob Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #6 20121123 20:13:59
Re: Linear inequation involving absolute value problem...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 gives but 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 Last edited by debjit625 (20121123 20:26:15) Debjit Roy ___________________________________________________ The essence of mathematics lies in its freedom  Georg Cantor #7 20121123 20:34:53
Re: Linear inequation involving absolute value problem...HI debjit625
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 Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #8 20121124 02:31:14
Re: Linear inequation involving absolute value problem...
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 to from 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 Debjit Roy ___________________________________________________ The essence of mathematics lies in its freedom  Georg Cantor #9 20121124 05:44:55
Re: Linear inequation involving absolute value problem...hi debjit625 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 ? Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #10 20121124 18:12:16
Re: Linear inequation involving absolute value problem...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. Debjit Roy ___________________________________________________ The essence of mathematics lies in its freedom  Georg Cantor #11 20121125 01:52:18
Re: Linear inequation involving absolute value problem...@ 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. Debjit Roy ___________________________________________________ The essence of mathematics lies in its freedom  Georg Cantor #12 20121125 03:40:01
Re: Linear inequation involving absolute value problem...hi debjit625 You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #13 20121128 00:17:21
Re: Linear inequation involving absolute value problem...Well I was a bit busy so I couldn't post... Debjit Roy ___________________________________________________ The essence of mathematics lies in its freedom  Georg Cantor 