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Re: Is lxl <1? (1) lx+1l = 2lx-1l (2) lx-3l does not equal 0 [#permalink]
banerjeea_98 wrote:
"C"

I. Solve for x you get x = 1/3 or 3....insuff

II.X can be 8 or X = 1/2...but X can't be 3 insuff


Combine X = 1/3......which means < 1...suff


how do you come to x = 1/3 or 3 ? i get only 3

lx+1l = 2lx-1l => x+1=2(x-1) => x=3 or => -(x+1)=-2(x-1) => x=3

it would be A) ?
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Re: Is lxl <1? (1) lx+1l = 2lx-1l (2) lx-3l does not equal 0 [#permalink]
christoph wrote:
banerjeea_98 wrote:
"C"

I. Solve for x you get x = 1/3 or 3....insuff

II.X can be 8 or X = 1/2...but X can't be 3 insuff


Combine X = 1/3......which means < 1...suff


how do you come to x = 1/3 or 3 ? i get only 3

lx+1l = 2lx-1l => x+1=2(x-1) => x=3 or => -(x+1)=-2(x-1) => x=3

it would be A) ?



put x = 1/3 in the eq and check u will see that is also a soln.....when u take out the mod, u can't just make it +ve on both sides and -ve on both sides, u r missing 2 cases when one is +ve and other is -ve, that's where x = 1/3
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Re: Is lxl <1? (1) lx+1l = 2lx-1l (2) lx-3l does not equal 0 [#permalink]
Banerjea, how did you get

X= 8 or X= 1/ 2 for stmt 2 could you expalin your approach?????
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Re: Is lxl <1? (1) lx+1l = 2lx-1l (2) lx-3l does not equal 0 [#permalink]
swath20 wrote:
Banerjea, how did you get

X= 8 or X= 1/ 2 for stmt 2 could you expalin your approach?????


he just picked the two numbers to show that "lxl <1" could be wrong => 8 or right => 1/2. both 1/2 and 8 satisfy 2) so it is unsufficient
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Re: Is lxl <1? (1) lx+1l = 2lx-1l (2) lx-3l does not equal 0 [#permalink]
Or you could break it into three ranges, with the two points from x+1=0 and x-1=0. So the three range would be x>-1, -1<x<1, x>1.
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Re: Is lxl <1? (1) lx+1l = 2lx-1l (2) lx-3l does not equal 0 [#permalink]
in 1, u can get -3 or 3...so insuff
in , no constraints given....insuff
but if we combine both, -3 goes out and left value is 3.....suff

Answer C



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