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For all values of x and y, let x y be defined by x y = xy - x + 1. If

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For all values of x and y, let x y be defined by x y = xy - x + 1. If  [#permalink]

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New post 01 Oct 2018, 15:39
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For all values of \(x\) and \(y\), let \(x\blacktriangle y\) be defined by \(x\blacktriangle y = xy - x + 1\). If \(\left( {a - 2} \right)\blacktriangle \,a\,\, = \,\,a\,\blacktriangle \left( {a + 1} \right)\), which of the following numbers is closest to the value of \(a\) ?

(A) 0.66
(B) 0.83
(C) 0.95
(D) 1.14
(E) 1.43

Source: http://www.GMATH.net
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Re: For all values of x and y, let x y be defined by x y = xy - x + 1. If  [#permalink]

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New post 01 Oct 2018, 18:35
(a−2)▲a = a▲(a+1)

(a−2)a-(a−2)+1 = a(a+1)-a+1

\(a^2\)-2a-a+2+1 = \(a^2\)+a-a+1

-3a = -2

Or

a = 2/3 = 0.66

A is the answer.
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Re: For all values of x and y, let x y be defined by x y = xy - x + 1. If  [#permalink]

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New post 02 Oct 2018, 11:40
fskilnik wrote:
For all values of \(x\) and \(y\), let \(x\blacktriangle y\) be defined by \(x\blacktriangle y = xy - x + 1\). If \(\left( {a - 2} \right)\blacktriangle \,a\,\, = \,\,a\,\blacktriangle \left( {a + 1} \right)\), which of the following numbers is closest to the value of \(a\) ?

(A) 0.66
(B) 0.83
(C) 0.95
(D) 1.14
(E) 1.43

Source: http://www.GMATH.net

Thank you for your contribution, Afc0892!
The solution I present below is the same, just to follow my method´s style (for the interested readers).

\(?\,\,\,:\,\,a\,\,\left( {{\rm{approx}}.} \right)\,\)

\({\text{LHS}} = \left( {a - 2} \right)\blacktriangle \,a = \left( {a - 2} \right)a - \left( {a - 2} \right) + 1 = {a^2} - 3a + 3\)

\({\text{RHS}} = a\,\blacktriangle \left( {a + 1} \right) = a\left( {a + 1} \right) - a + 1 = {a^2} + 1\)

\({\rm{LHS}} = {\rm{RHS}}\,\,\,\,\, \Rightarrow \,\,\,\,\, - 3a + 3 = 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,?\,\,\,:\,\,\,a = {2 \over 3} \cong \,\,0.66\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: For all values of x and y, let x y be defined by x y = xy - x + 1. If   [#permalink] 02 Oct 2018, 11:40
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