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# PS: Old question

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Manager
Joined: 19 Jun 2003
Posts: 151

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11 Jul 2004, 21:02
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|B+6|â€“|Bâ€“5|=0

Could somebody explain the solution?

Thank you.
CIO
Joined: 09 Mar 2003
Posts: 463

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11 Jul 2004, 23:37
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.
|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
Manager
Joined: 08 Jun 2004
Posts: 245
Location: INDIA

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12 Jul 2004, 06:11
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
_________________

the whole worldmakes way for the man who knows wer he's going... good luck

Manager
Joined: 16 May 2004
Posts: 64
Location: columbus

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15 Jul 2004, 16:53
|B+6|â€“|Bâ€“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
Manager
Joined: 21 Jun 2004
Posts: 52

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15 Jul 2004, 20:42
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.
|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
15 Jul 2004, 20:42
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