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

Author Message
Manager
Joined: 05 Feb 2007
Posts: 139

Kudos [?]: 8 [0], given: 7

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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Difficulty:

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Question Stats:

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

Kudos [?]: 8 [0], given: 7

Senior Manager
Joined: 19 Apr 2008
Posts: 313

Kudos [?]: 99 [0], given: 0

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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

Kudos [?]: 99 [0], given: 0

CEO
Joined: 29 Mar 2007
Posts: 2553

Kudos [?]: 527 [0], given: 0

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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.

Kudos [?]: 527 [0], given: 0

Manager
Joined: 04 Sep 2007
Posts: 210

Kudos [?]: 20 [0], given: 0

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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.

Kudos [?]: 20 [0], given: 0

Re: PS: Algebra substitution   [#permalink] 15 May 2008, 16:01
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