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moon1130
Average Member
USA
18 Posts |
 Posted - 11/28/2012 : 17:31:06
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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. |
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royhaas
Moderator
USA
3039 Posts |
Posted - 11/29/2012 : 08:08:04
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| The correct answer cannot be -7<x<1. Substitute x=3 in the original inequality. I agree with x<1. |
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the_hill1962
Advanced Member
USA
1439 Posts |
Posted - 11/29/2012 : 08:26:33
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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.
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Edited by - the_hill1962 on 11/29/2012 08:34:15 |
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moon1130
Average Member
USA
18 Posts |
Posted - 11/29/2012 : 11:55:48
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| Thank you very much royhaas the_hill1962 for your responses. I will work with the your graphical solution the_hill1962. It's pretty nice. |
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Ultraglide
Advanced Member
Canada
278 Posts |
Posted - 11/30/2012 : 00:21:19
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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. |
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the_hill1962
Advanced Member
USA
1439 Posts |
Posted - 11/30/2012 : 12:38:29
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Thanks, Ultraglide.
Of course, "4 cases" |
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moon1130
Average Member
USA
18 Posts |
Posted - 11/30/2012 : 16:45:09
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Thanks Ultraglide. That's a beautiful solution.
moon1130 |
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the_hill1962
Advanced Member
USA
1439 Posts |
Posted - 11/30/2012 : 17:08:08
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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. |
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Ultraglide
Advanced Member
Canada
278 Posts |
Posted - 11/30/2012 : 18:01:15
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| 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. |
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moon1130
Average Member
USA
18 Posts |
Posted - 12/01/2012 : 15:53:26
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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. |
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moon1130
Average Member
USA
18 Posts |
Posted - 12/02/2012 : 01:26:03
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Ultraglide, I figured out how to use test scores to determine what ranges worked or did not work.
Moon1130 |
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the_hill1962
Advanced Member
USA
1439 Posts |
Posted - 12/03/2012 : 15:33:21
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Yes, graphing would not give exact solutions.
Luckily, this absolute value inequality had integers.
Thanks for the replies. |
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Ultraglide
Advanced Member
Canada
278 Posts |
Posted - 12/03/2012 : 16:21:26
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| Moon, I think if you check -8 again you will get 22>23 which is not possible. |
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moon1130
Average Member
USA
18 Posts |
Posted - 12/04/2012 : 00:54:34
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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. |
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moon1130
Average Member
USA
18 Posts |
Posted - 12/04/2012 : 00:57:43
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| You are welcome, the_hill1962. And thanks for your responses. |
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moon1130
Average Member
USA
18 Posts |
Posted - 12/06/2012 : 00:10:56
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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. |
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Interactive Math Goodies Software
Solving Absolute Value Inequalities
