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The operator @ is defined by the following expression: a@b = |(a + 1)|

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The operator @ is defined by the following expression: a@b = |(a + 1)|  [#permalink]

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New post 27 Jun 2018, 21:14
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A
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Question Stats:

46% (03:11) correct 54% (03:06) wrong based on 196 sessions

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The operator @ is defined by the following expression: a@b = |(a + 1)|  [#permalink]

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New post 28 Jun 2018, 05:08
1
Bunuel wrote:
The operator \(@\) is defined by the following expression: \(a@b = |\frac{a+1}{a}| - \frac{b+1}{b}\) where \(ab\neq{0}\). What is the sum of the solutions to the equation \(x@2 = \frac{x@(-1)}{2}\) ?


A. -1

B. -0.75

C. -0.25

D. 0.25

E. 0.75


Before we go to actual calculations, let's look at what we have...
1) We have x on each side, with function of x on left side and \(\frac{(function of x)}{2}\) on right side..
So final will be \(\frac{(function of x)}{2}\)on left side...
2) @-1 will leave 0 as -1+1 is zero

So \(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......
\(|\frac{x+1}{x}|=3\)....
Two cases..
1) \(\frac{(x+1)}{x}=3......x+1=3x....x=\frac{1}{2}\)
2) \(\frac{(x+1)}{x}=-3.....x+1=-3x.....x=-\frac{1}{4}\)
Sum = \(\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\) or 0.25

even if you solve it completely
\(x@2 = |\frac{x+1}{x}| - \frac{2+1}{2}=|\frac{x+1}{x}| - \frac{3}{2}\).....
\(\frac{x@-1}{2} = \frac{1}{2}*|\frac{x+1}{x}| - \frac{-1+1}{1}=\frac{1}{2}*|\frac{x+1}{x}|\)
so \(|\frac{x+1}{x}| - \frac{3}{2}=\frac{1}{2}*|\frac{x+1}{x}|........................\frac{1}{2}|\frac{x+1}{x}| = \frac{3}{2}...................|\frac{x+1}{x}| = 3\)
Two cases..
1) \(\frac{(x+1)}{x}=3......x+1=3x....x=\frac{1}{2}\)
2) \(\frac{(x+1)}{x}=-3.....x+1=-3x.....x=-\frac{1}{4}\)

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Re: The operator @ is defined by the following expression: a@b = |(a + 1)|  [#permalink]

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New post 13 Oct 2019, 14:59
Hi! I didn't understand the solution in the discussion. Is there another way to solve this?
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Re: The operator @ is defined by the following expression: a@b = |(a + 1)|  [#permalink]

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New post 18 Oct 2019, 01:16
So \(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......
\(|\frac{x+1}{x}|=3\)....

Can anyone explain how \(|\frac{x+1}{x}|=3\)?
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Re: The operator @ is defined by the following expression: a@b = |(a + 1)|  [#permalink]

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New post 19 Oct 2019, 09:00
1
harshit121 wrote:
So \(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......
\(|\frac{x+1}{x}|=3\)....

Can anyone explain how \(|\frac{x+1}{x}|=3\)?



\(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......

=> \(\frac{1}{2}*(|\frac{x+1}{x}|) = \frac{3}{2}\)

Multiply L.H.S. and R.H.S. by 2

\(|\frac{x+1}{x}|=3\)
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Re: The operator @ is defined by the following expression: a@b = |(a + 1)|   [#permalink] 19 Oct 2019, 09:00
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