November 17th, 2016, 09:12 AM  #1 
Newbie Joined: Nov 2016 From: India Posts: 1 Thanks: 0  Absolute value inequality
Solve f(x)=(x+3+x)/(x+2)>1 My solution: x+3>2 my answer (∞,5)U(1,∞) except 2 I do not know what mistake I'm doing,because my answer is wrong, please cope with me as it is my first question. 
November 17th, 2016, 09:45 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,939 Thanks: 2266 Math Focus: Mainly analysis and algebra 
Well 2 is not part of the union, so there's probably no need to mention it.

November 17th, 2016, 09:46 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,939 Thanks: 2266 Math Focus: Mainly analysis and algebra 
Also, consider what happens if $x+2 \lt 0$. The answer you have assumes $x+2 \gt 0$, which has an impact on the answer you quoted. Last edited by v8archie; November 17th, 2016 at 09:49 AM. 
November 17th, 2016, 09:51 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1385  
March 31st, 2017, 02:39 AM  #5  
Newbie Joined: Mar 2017 From: Portland, OR Posts: 5 Thanks: 0  Quote:
Alright so first things first. Isolate the absolute on the left side. x+3+x > 1  > multiply by (x+2) on both sides x+2 x+3+x > x+2 > subtract X from left side and also from right side x x x+3 > 2 > Now Greater than means OR. So.. x+3 >2 OR x+3 < 2 subtract 3 from both sides in both inequalities. Isolate x. x+3 > 2 OR x+3 < 2 3 3 3 3 x > 1 OR x < 5 and written in notation form it is (∞, 5)U(1, ∞). Since 2 isn't covered anywhere in our ranges, we don't have to mention 2. At 2, it is undefined. To graph this, y1= x+3+x and y2= 1 Since it's greater than, find the 2 points where both lines intersect.  Your answers are all the xvalues to the left of 5 (excluding 5) and x+2 any xvalue to the right of 1 (excluding). Your answer is correct except you don't need to mention 2. Last edited by c0rnel; March 31st, 2017 at 02:42 AM.  
March 31st, 2017, 03:22 AM  #6 
Newbie Joined: Mar 2017 From: Ireland Posts: 1 Thanks: 0 
(x+3+x)/(x+2)>1 (x+3+x)/(x+2)1>0 (x+3+xx2)/(x+2)>0 (x+32)/(x+2)>0 (x+2)(x+32)>0 therefore, x+2>0 or x+32>0 x+2>0 ; x>2 x+32>0 x+3>2 x+3>2 or x+3<2 x>1 or x<5 Hence, x>2 and x<5 Which give rise to (∞, 5)U(2, ∞) This is what i got 

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absolute, absolute value, calculus, functions, inequality 
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