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How many solutions are there for |x-1|-y=-5 and |x-1|+

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How many solutions are there for |x-1|-y=-5 and |x-1|+ [#permalink] New post 06 Dec 2003, 13:51
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1.How many solutions are there for |x-1|-y=-5 and |x-1|+ |y-5|=1

a.0
b.1
c.2
d 3
e. 4

Last edited by Praetorian on 07 Dec 2003, 23:57, edited 1 time in total.
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 [#permalink] New post 06 Dec 2003, 15:12
Praet--

That's a fantastic question. I tried to simplify and I just got a bee's nest. I ha ve no idea. I wonder if I missed something obvious.


-5(|x-1| + |y-5|)= |x-1|-y
-5|x-1| -5|y-5|= |x-1|-y
-6|x-1| -5|y-5|=-y
-6|x-1| =5|y-5|-y
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Re: PS : Modulus [#permalink] New post 07 Dec 2003, 05:05
praetorian123 wrote:
1.How many solutions are there for |x-1|-y=-5 and |x-1|+ |y-5|=1

a.0
b.1
c.2
d 3
e. 4


This is a good problem:

IMO, the answer is c.
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Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

CEO
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Joined: 15 Aug 2003
Posts: 3470
Followers: 60

Kudos [?]: 665 [0], given: 781

Re: PS : Modulus [#permalink] New post 07 Dec 2003, 05:28
praetorian123 wrote:
1.How many solutions are there for |x-1|-y=-5 and |x-1|+ |y-5|=1

a.0
b.1
c.2
d 3
e. 4


heres what i get .
|x-1|-y=-5 ........(1)

2. |x-1|+ |y-5|=1 .....(2)

from (1) , |x-1|= y-5
Substitute in equation (2).
y-5 + |y-5| = 1
|y-5| = 6 - y

therefore,
y-5 = - (6-y) , which gives no solution
y -5 = 6- y , which gives ... y = 11/2

Substitute back in (1)

|x-1|-y=-5

|x-1|= y-5

x-1 = - y+5

i. x = -y + 6

= -11/2 + 6

= 1/5

ii. x-1 = y-5

x = y -4

= 11/2 - 4

x = 3/2

in other words, we have two solutions.

1. x = 1/5 , y= 11/2
2. x= 3/2 , y=11/2


did i do this right?

thanks
praetorian
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Re: PS : Modulus [#permalink] New post 09 Dec 2003, 04:13
praetorian123 wrote:
is my solution ok?


That is essentially how I solved it, but I didn't take ALL of the steps. Remember, we only needed to know how many solutions there were, not what the solutions actually are!
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Re: PS : Modulus   [#permalink] 09 Dec 2003, 04:13
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