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18 May 2021, 04:09
Kudos
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jashshah
VeritasKarishma for st 1 as per graph below i get following reults:
Attachment:
veritas (2).jpg [ 72.15 KiB | Viewed 1395 times ]
For range x ≥ 1
-(1-x)-x+1 = 0
-1+x-x+1 =0
0=0 (doesn`t it mean infinite number of solutions ?
For range -1≤x<1
1-x-x+1=0
2-2x=0
2x=2
x =1
For range x<-1
1-x- -(x+1) =0
1-x- -x-1 =0
1-x+x-1=0
again 0=0
so i get either 0=0 or x=1 whats wrong with my solution / reasoning.... i see others get x=0 but 0=0 is not the same as x=0 and how about x=1
19 May 2021, 00:15
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dave13
You are forgetting the bracket of the expression.
For range x ≥ 1
-(1-x) - (x+1) = 0
The negative sign is outside the entire expression of (x + 1). See stmnt 1: |1 - x| - |x + 1| = 0
You get -2 = 0 which is not valid.
Now check other equations too.
19 May 2021, 05:58
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VeritasKarishma
VeritasKarishma thanks.
For range x<-1
i get this
(1-x)-(-x-1) = 0
1-x+x+1=0
2=0 is this result correct ? i ask this question because as per other posts two ranges yield the same value of 0
as per one post in this thread when we have |1 - x| - |x + 1| = 0 we need to change from|1 - x| to | x -1| Is it a must to switch places? And Why? And when ?
so in this case |x-1| - |x + 1| = 0
For range x<-1
I get (x-1) - (-x-1) = 0
x-1+x+1 = 0
2x=0
x=0
As For range -1≤x<1
i get (1-x)-(x+1) =0
1-x-x-1=0
-2x=0
x = 0
solution for this range looks correct ...hopefully
