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# Is y < 2x ? (1) y/4 < x/2 (2) (y - 2x)/3 <0

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Re: Is y < 2x ? (1) y/4 < x/2 (2) (y - 2x)/3 <0 [#permalink]
Bunuel wrote:
Is y < 2x ?

(1) y/4 < x/2. Multiply both sides by 4: y < 2x. Sufficient.

(2) (y - 2x)/3 < 0. Multiply both sides by 3: y - 2x < 0. Re-arrange: y < 2x. Sufficient.

Hope it's clear.

Wait when do we know we can multiply or allowed to do so in inequalities or simplify?
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Re: Is y < 2x ? (1) y/4 < x/2 (2) (y - 2x)/3 <0 [#permalink]
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sagnik242 wrote:
Bunuel wrote:
Is y < 2x ?

(1) y/4 < x/2. Multiply both sides by 4: y < 2x. Sufficient.

(2) (y - 2x)/3 < 0. Multiply both sides by 3: y - 2x < 0. Re-arrange: y < 2x. Sufficient.

Hope it's clear.

Wait when do we know we can multiply or allowed to do so in inequalities or simplify?

Hi,
you can always multiply or divide an equality/inequality by a constant number, may it be fraction,+ive or ,-ive..
But you cannot do the same by a variable like X, y etc..
the reason is we do not know the value of variable and it could be 0, but by dividing by a variable, we negate that possiblity..

4/x^2 > 4/x...
here you can eliminate 4 from both sides but not 1/x...
Hope it helps
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Re: Is y < 2x ? (1) y/4 < x/2 (2) (y - 2x)/3 <0 [#permalink]
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Re: Is y < 2x ? (1) y/4 < x/2 (2) (y - 2x)/3 <0 [#permalink]
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