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

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

15 May 2008, 12:49

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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

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Re: PS: Algebra substitution [#permalink]

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

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Re: PS: Algebra substitution [#permalink]

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.

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Re: PS: Algebra substitution [#permalink]

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

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

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