Linear inequation involving absolute value problem...

debjit625 · 2012-11-21 21:02:33

Well I have four inequations that I can't able to understand properly,what are the steps you guys will take to solve this...

Question 1

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 (2012-11-21 21:04:31)

Bob · 2012-11-21 21:12:16

hi debjit625

Absolute value problems can be reduced to non absolute by considering values of x on each side of the critcal value.

|x| is like x when x>0 and like -x when x < 0

So I would split the problem into three cases (LATER EDIT see at end)

(i) x> 0

(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

debjit625 · 2012-11-22 02:01:40

Well still I am having problem to understand it...

As function tends to +/- infinity at x = +/- 3 we should look at 5 cases
(-∞,-3) (-3,0) 0 (0,3) (3, ∞)

How you can take the case (-∞,-3) as x can't be -4 or 4 ?

I did it a different way I just assumed two cases when x>= 0 and x<0
Here is what I am doing

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 (2012-11-22 20:35:02)

debjit625 · 2012-11-22 02:13:18

Here are the other questions
Question 2

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

Bob · 2012-11-22 04:49:12

Q1.

Look at the graph I posted. You can see that + / - 4 is calculable and that values between -3 and + 3 satisfy the inequality.

So for example

-3 < x < 0

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.

Bob

debjit625 · 2012-11-22 21:13:59

Sorry still can't understand...

You can see that + / - 4 is calculable

How can x be +/- 4 ,lets put it in the equation and see
Taking x as 4

Taking x as -4

So I think x can't be +/- 4

between -3 and + 3 satisfy the inequality.

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 (2012-11-22 21:26:15)

Bob · 2012-11-22 21:34:53

HI debjit625

Sorry, I just misunderstood what you were saying about + / - 4

What you have just said is correct, no problem.

In post 1 you had two ranges, in words below -5 and above + 5

Those are both correct and my graph shows these.

What was missing from your answer was another range of possible values for x, between -3 and + 3

Look at my graph and you can see that, in this range, 'y' is always negative, so certainly < 1/2

So why did these values 'go missing' ?

The graph has a discontinuity at -3 and again at +3 because of the division by zero.

So you need to check in this range too.

I tried to demonstrate why (-3,0) is another set of values that fit the inequality. Twice I had to use the rule that 'if you multiply or divide by a negative, you must reverse the inequality'. Do you know and understand this rule?

I have recently posted an explanation at

[url]http://www.mathisfunforum.com/viewtopic.php?id=18411[/math]

You might like to have a look at that, especially post 7 in that thread.

I've got to log off for a short while. I'll be back on in about 15 mins.

ADDITIONAL EDIT:

case 3 which you actually explained in post #5 gives
but as already given

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

debjit625 · 2012-11-23 03:31:14

'if you multiply or divide by a negative, you must reverse the inequality'. Do you know and understand this rule?

Yes,I understand this rule.

You can finish by considering (0,+3) and showing it is legitimate to add this to the range

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

Bob · 2012-11-23 06:44:55

hi debjit625

Let's move on to Q2 and maybe it will become clearer.

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

debjit625 · 2012-11-23 19:12:16

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.

Thanks...

debjit625 · 2012-11-24 02:52:18

@ bob bundy
Ok I did all the problems,but I have some question regrding post #9 where you posted about question 2.

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.

Well how can x = 0 makes the expression goes to +/- infinity as dividing anything by 0 is undefined,and x is the denominator.

What solution set did you got for question 2 , 3 and 4?

Thanks

Bob · 2012-11-24 04:40:01

hi debjit625

At x = 0, the expression becomes (0 + 2 - 0)/0

The rule about dividing by 0 is

(i) (non zero)/0 ---> infinity

(ii) 0/0 is indeterminate (although you may be able to find a limit as x tends to zero)

I've made the graph below.

So you should have got all x except when 0<x<1

Bob

debjit625 · 2012-11-27 01:17:21

Well I was a bit busy so I couldn't post...

@bob bundy ,thanks I have also got the same answer.

Bob · 2012-11-27 02:04:06

Glad about that.

Bob

Math Is Fun Forum

#1 2012-11-21 21:02:33

Linear inequation involving absolute value problem...

#2 2012-11-21 21:12:16

Re: Linear inequation involving absolute value problem...

#3 2012-11-22 02:01:40

Re: Linear inequation involving absolute value problem...

#4 2012-11-22 02:13:18

Re: Linear inequation involving absolute value problem...

#5 2012-11-22 04:49:12

Re: Linear inequation involving absolute value problem...

#6 2012-11-22 21:13:59

Re: Linear inequation involving absolute value problem...

#7 2012-11-22 21:34:53

Re: Linear inequation involving absolute value problem...

#8 2012-11-23 03:31:14

Re: Linear inequation involving absolute value problem...

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Re: Linear inequation involving absolute value problem...

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