testing header
Math Goodies is a free math help portal for students, teachers, and parents.
Free Math
Newsletter
 
 
Interactive Math Goodies Software

Buy Math Goodies Software
testing left nav
Math Forums @ Math Goodies
Math Forums @ Math Goodies
Home | Profile | Register | Active Topics | Members | Search | FAQ
 All Forums
 Homework Help Forums
 Pre-Calculus and Calculus
 Solving Absolute Value Inequalities

Note: You must be registered in order to post a reply.
To register, click here. Registration is FREE!

Screensize:
UserName:
Password:
Format Mode:
Format: BoldItalicizedUnderlineStrikethrough Align LeftCenteredAlign Right Horizontal Rule Insert HyperlinkInsert EmailInsert Image Insert CodeInsert QuoteInsert List Insert Special Characters Insert Smilie
   
Message:
* HTML is OFF
* Forum Code is ON

Math Symbols
Squared [squared] Cubed [cubed] Square Root [sqrt] Cube Root [cbrt] Pi [pi]
Alpha [alpha] Beta [beta] Gamma [gamma] Theta [theta] Angle [angle]
Degrees [degrees] Times [times] Divide [divide] Less Than or Equal To [less-than] Greater Than or Equal To [greater-than]
Plus Minus [plus-minus] Integral [integral] Sum [sum] Sub 1 [sub-1] Sub 2 [sub-2]
Element Of [element-of] Union [union] Intersect [intersect] Subset [subset] Empty Set [empty-set]

  Check here to subscribe to this topic.
 
   

T O P I C    R E V I E W
moon1130 Posted - 11/28/2012 : 17:31:06
Hi,

A friend gave me this absolute value inequality to solve. I have not been able to get the correct answer. Could someone show me where I sent wrong?

2|x-3|> |3x+1|. The correct answer is -7 < x < 1.

What I tried was the following:
2(x - 3)> (3x +1) or 2(x - 3) < -(3x + 1). The answer I get is
-7 < x or x < 1.
15   L A T E S T    R E P L I E S    (Newest First)
moon1130 Posted - 12/06/2012 : 00:10:56
the_hill1962, I must revise my answer that I gave earlier as to which method I prefer, the graphing or the algebraic. I have been working with some problems where I used a graphing calculator to find the solution. It was easier, faster, and clearer than using the algebraic method.

For example, consider the following inequality problem:

x-2x-2x-3|x-3|.

As to which method is easier, the algebraic or the graphical, it all depends on the particular problem. Graphing solutions on a graphing calculator are very close approximations of the exact value. I like both methods.
moon1130 Posted - 12/04/2012 : 00:57:43
You are welcome, the_hill1962. And thanks for your responses.
moon1130 Posted - 12/04/2012 : 00:54:34
You are correct Ultraglide. After I figured out how to check my results, I too found that -8 gave the false statement 22 > 23.

Thank you for your help and for showing the algebraic solution to those of us not familiar with it.
Ultraglide Posted - 12/03/2012 : 16:21:26
Moon, I think if you check -8 again you will get 22>23 which is not possible.
the_hill1962 Posted - 12/03/2012 : 15:33:21
Yes, graphing would not give exact solutions.
Luckily, this absolute value inequality had integers.
Thanks for the replies.
moon1130 Posted - 12/02/2012 : 01:26:03
Ultraglide, I figured out how to use test scores to determine what ranges worked or did not work.

Moon1130
moon1130 Posted - 12/01/2012 : 15:53:26
the_hill1962, I like the algebraic method better because it is exact. I prefer to use the graphing method as a visual check.

ultraglide, I am not sure how you are testing -8, 0, 2, and -6. For example, when I substitute -8 for x in the original inequality, I get -22 > -23, which is a true statement. However, -8 is not in the solution range. Could you show me?

What I did, I took Case 1, for example, and following your outline, I assumed that both members of the inequality were positive. Therefore, x>3 for the left member and x > -1/3 for the right member. If both members are positive, we get the solution x < -7, which fails the condition that x > -1/3 and x>3. So we reject Case 1, x < -7 as being in the solution range.
Ultraglide Posted - 11/30/2012 : 18:01:15
It all depends. If you have access to graphing software, such as Graphmatica, or a graphing calculator, that would certainly be easier. You may be able to determine the point(s) of intersection and the ranges this way. If you don't and if the function is not really complex, you could graph manually. In this case, you would sketch each side separately. The only problem with either of these two is the accuracy of the graphs. What if the points that determine the ranges are not integers? The algebraic method is exact. Notice in this case, the solutions occur in pairs, i.e. x>1 and x<1 which are mutually exclusive, so when you find one, the other is eliminated.
the_hill1962 Posted - 11/30/2012 : 17:08:08
moon1130:
Just wondering if you tried graphing and saw what I was referring to...
You only need to solve
-2x+6 > abs(3x+1)
Instead of having to go through the 4 cases.
If I would have thought of how to do it the way Ultraglide states, that is the way I would have done if first.

Ultraglide and/or moon1130:
Would either of you give me an opinion on which method you think is easier? Just wondering.
moon1130 Posted - 11/30/2012 : 16:45:09
Thanks Ultraglide. That's a beautiful solution.

moon1130
the_hill1962 Posted - 11/30/2012 : 12:38:29
Thanks, Ultraglide.
Of course, "4 cases"
Ultraglide Posted - 11/30/2012 : 00:21:19
If you want to solve the problem by using algebra, you have to consider 4 cases since each quantity inside the absolue value signs could be positive or negative. You will end up with some solutions that you could eliminate.

1. Both positive: 2(x-3)>3x+1
which gives 2x-6>3x+1
continuing, -x>7, or x<-7
2. First negative, second positive -2(x-3)>3x+1
which gives -2x+6>3x+1
continuing, -5x>-5 or x<1
3. First positive, second negative 2(x-3)>-3x-1
which gives 2x-6>-3x-1
continuing, 5x>5 or x>1
4. Both negative: -2(x-3)>-3x-1
which gives -2x+6>-3x-1
continuing, x>-7

Now you need to test your results. Try -8,0,2 and -6. You will find that the solutions are as previously mentioned, i.e. -7<x<1.
Note: You should always check the solutions for these types of problems.
moon1130 Posted - 11/29/2012 : 11:55:48
Thank you very much royhaas the_hill1962 for your responses. I will work with the your graphical solution the_hill1962. It's pretty nice.
the_hill1962 Posted - 11/29/2012 : 08:26:33
royhaas, the correct answer is -7<x<1
Your testing of x=3 is outside of the -7<x<1 interval. x=3 is not supposed to be part of the solution.
moon1130,
While I don't know the elegant way of solving this problem purely algebraically, the way that I tackled it is to
first look at the graphs y=2abs(x-3) and y=abs(3x+1)
Note that is just the "left" part of y=2abs(x-3) where it is greater than y=abs(3x+1) so we can simplify the problem. The linear equation for the left part of y=2abs(x-3) is y=-2x+6
So, just solve
-2x+6 > abs(3x+1)
Hopefully you can solve this. If not, here is a hint:
Use the pattern for solving abs(x) < a
The solution to that is -a < x < a
If you need help with this, please let us know.
royhaas Posted - 11/29/2012 : 08:08:04
The correct answer cannot be -7<x<1. Substitute x=3 in the original inequality. I agree with x<1.

Math Forums @ Math Goodies © 2000-2004 Snitz Communications Go To Top Of Page
This page was generated in 0.21 seconds. Snitz Forums 2000
testing footer
About Us | Contact Us | Advertise with Us | Facebook | Blog | Recommend This Page




Copyright © 1998-2014 Mrs. Glosser's Math Goodies. All Rights Reserved.

A Hotchalk/Glam Partner Site - Last Modified 30 Oct 2014