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The operation @ is defined by the equation a@b = (a 1)(b

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sandy
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The operation @ is defined by the equation a@b = (a 1)(b [#permalink] New post   30 Jul 2018, 10:27
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The operation \(@\) is defined by the equation \(a@b = (a – 1)(b – 2)\).

\( x@5 = 3@x\)

Quantity A
Quantity B
x
1


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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B
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Re: The operation @ is defined by the equation a@b = (a 1)(b [#permalink] New post   01 Aug 2018, 09:12
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x@5 = (x-1)(3)
= 3x - 3

3@x = 2(x-2)
= 2x - 4

From question,
3x - 3 = 2x - 4
or, x= -1
B>A

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Re: The operation @ is defined by the equation a@b = (a 1)(b [#permalink] New post   12 Aug 2018, 05:36
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Explanation

This problem defines a function for the made-up symbol @. Use the definition of the new symbol to rewrite the equation x@5 = 3@x without the @ operator:


For x@5, a = x and b = 5: x@5 = (x – 1)(5 – 2) = 3x – 3.
For 3@x, a = 3 and b = x: 3@x = (3 – 1)(x – 2) = 2x – 4.

Equating these two expressions gives you:
\(3x - 3 = 2x - 4\)
\(x = -1\)

Quantity B is greater.
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Re: The operation @ is defined by the equation a@b = (a 1)(b [#permalink] New post   04 Sep 2021, 08:49
Functions and Custom Character Problem

The operation @ is defined by the equation a@b = (a – 1)(b – 2) and we need to find the value of x, given that

x@5 = 3@x
=> (x-1)*(5-2) = (3-1)*(x-2)
=> (x-1)*3 = 2*(x-2)
=> 3x - 3 = 2x - 4
=> 3x-2x = 3 - 4 = -1
=> x = -1

Clearly, Quantity B > Quantity A (x=-1)

So, Answer will be B
Hope it helps!

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Re: The operation @ is defined by the equation a@b = (a 1)(b [#permalink]
04 Sep 2021, 08:49
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