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?
2x3> 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:
x2x2x3x3.
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(x3)>3x+1 which gives 2x6>3x+1 continuing, x>7, or x<7 2. First negative, second positive 2(x3)>3x+1 which gives 2x+6>3x+1 continuing, 5x>5 or x<1 3. First positive, second negative 2(x3)>3x1 which gives 2x6>3x1 continuing, 5x>5 or x>1 4. Both negative: 2(x3)>3x1 which gives 2x+6>3x1 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(x3) and y=abs(3x+1) Note that is just the "left" part of y=2abs(x3) 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(x3) 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. 

