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# Let *x* = x^2-1 and $y$ = 6y^2 - (*y*) for all integers x

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Manager
Joined: 05 Feb 2007
Posts: 140
Let *x* = x^2-1 and $y$ = 6y^2 - (*y*) for all integers x [#permalink]

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15 May 2008, 12:49
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions

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Let *x* = x^2-1 and $y$ = 6y^2 - (*y*) for all integers x and y. If m > 0 and *m* = 0, what is the value of $m$?

A. -6
B. -1
C. 0
D. 1
E. 6
Senior Manager
Joined: 19 Apr 2008
Posts: 317
Re: PS: Algebra substitution [#permalink]

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15 May 2008, 13:16
giantSwan wrote:
Let *x* = x^2-1 and $y$ = 6y^2 - (*y*) for all integers x and y. If m > 0 and *m* = 0, what is the value of $m$?

A. -6
B. -1
C. 0
D. 1
E. 6

E .

*m* = 0 , m^2-1=0 ; m+/- 1 , since m>0 m =1

$m$ = 6m^2 - 0 = 6
CEO
Joined: 29 Mar 2007
Posts: 2559
Re: PS: Algebra substitution [#permalink]

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15 May 2008, 14:30
giantSwan wrote:
Let *x* = x^2-1 and $y$ = 6y^2 - (*y*) for all integers x and y. If m > 0 and *m* = 0, what is the value of $m$?

A. -6
B. -1
C. 0
D. 1
E. 6

I also get E.

This question isn't difficult, but the question itself is confusing.
Manager
Joined: 04 Sep 2007
Posts: 213
Re: PS: Algebra substitution [#permalink]

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15 May 2008, 16:01
rpmodi wrote:
giantSwan wrote:
Let *x* = x^2-1 and $y$ = 6y^2 - (*y*) for all integers x and y. If m > 0 and *m* = 0, what is the value of $m$?

A. -6
B. -1
C. 0
D. 1
E. 6

E .

*m* = 0 , m^2-1=0 ; m+/- 1 , since m>0 m =1

$m$ = 6m^2 - 0 = 6

E for me as well. The approach is the same as above.
Re: PS: Algebra substitution   [#permalink] 15 May 2008, 16:01
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# Let *x* = x^2-1 and $y$ = 6y^2 - (*y*) for all integers x

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