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 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: 7,274 Thanks: 2435 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: 7,274 Thanks: 2435 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. Thanks from topsquark Last edited by v8archie; November 17th, 2016 at 09:49 AM.
November 17th, 2016, 09:51 AM   #4
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Quote:
 Originally Posted by 10ishq I do not know what mistake I'm doing, because my answer is wrong, . . .
Your answer was incorrect, but can you post your working, so that we can identify any mistake(s) in it?

March 31st, 2017, 02:39 AM   #5
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Quote:
 Originally Posted by 10ishq 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.
__________________________________________________ _______________________________________

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 x-values to the left of -5 (excluding -5) and
x+2 any x-value to the right of -1 (excluding).

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|+x-x-2)/(x+2)>0 (|x+3|-2)/(x+2)>0 (x+2)(|x+3|-2)>0 therefore, x+2>0 or |x+3|-2>0 x+2>0 ; x>-2 |x+3|-2>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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