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

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Math Expert
Joined: 02 Sep 2009
Posts: 58991
The operator @ is defined by the following expression: a@b = |(a + 1)|  [#permalink]

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27 Jun 2018, 21:14
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Difficulty:

95% (hard)

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: $$a@b = |\frac{a+1}{a}| - \frac{b+1}{b}$$ 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

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Joined: 02 Aug 2009
Posts: 8150
The operator @ is defined by the following expression: a@b = |(a + 1)|  [#permalink]

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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) $$\frac{(x+1)}{x}=-3.....x+1=-3x.....x=-\frac{1}{4}$$

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

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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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Joined: 08 Oct 2019
Posts: 1
Re: The operator @ is defined by the following expression: a@b = |(a + 1)|  [#permalink]

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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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Joined: 10 Mar 2018
Posts: 60
Location: India
Concentration: Entrepreneurship, Marketing
WE: Design (Retail)
Re: The operator @ is defined by the following expression: a@b = |(a + 1)|  [#permalink]

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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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