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Manager
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PS: Old question [#permalink] New post 11 Jul 2004, 20:02
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
|B+6|–|B–5|=0

Could somebody explain the solution?

Thank you. :help2
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 [#permalink] New post 11 Jul 2004, 22:37
any algebraic equation with absolute value must be rewritten in its positive and negative form without the absolute value.

For example,
|x| = 6 so
x = 6 and x = -6

|x-3| = 5 becomes
x-3 = 5 and x-3 = -5 so
x=8 and x=-2, and both work when you plug them back in.

Same thing here.
Start with |b+6|-|b-5|=0, and make it
|b+6|=|b-5|

So b+6 = b-5 or b+6 = -(b-5)
Solve each one, and see that in the first one, b cancels and it's impossible, and in the second one, b=-.5
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 [#permalink] New post 12 Jul 2004, 05:11
let us try one another way

sqauring BOTH the absolute values...


we have... B^2+12B +36 = B^2-10B +25

22 b = -(11)
b = -.5

hope that helps!

have fun :)
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Manager
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ans [#permalink] New post 15 Jul 2004, 15:53
|B+6|–|B–5|=0

I think mod problems should be tackled taking into consideration the ranges in which the exp are valid

B+6 > 0 if B > -6

B- 5 > 0 if B > 5

so there are 3 regions for which we have to consider the expression
B> 5 and -6 > B > 5 andB < -6

for B > 5 and D < -6 the equation has no soln
for the 2nd range B + 6 -(-(B - 5)) = 0
= B + 6 + B -5 = 0
2B = -1
or B= -.5
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 [#permalink] New post 15 Jul 2004, 19:42
ian7777 wrote:
any algebraic equation with absolute value must be rewritten in its positive and negative form without the absolute value.

For example,
|x| = 6 so
x = 6 and x = -6

|x-3| = 5 becomes
x-3 = 5 and x-3 = -5 so
x=8 and x=-2, and both work when you plug them back in.

Same thing here.
Start with |b+6|-|b-5|=0, and make it
|b+6|=|b-5|

So b+6 = b-5 or b+6 = -(b-5)
Solve each one, and see that in the first one, b cancels and it's impossible, and in the second one, b=-.5


Taking time in to consideration, this will be the quickest method to solve the problem
  [#permalink] 15 Jul 2004, 19:42
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