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If x = 2t and y = t/3, what is the value of x^2 – y^2?

(1) t^2 – 3 = 6 (2) t^3 = −27

Kudos for a correct solution.

Given : x = 2t and y = t/3

Question : x^2 – y^2=(x+y)*(x-y) = ?

Statement 1: t^2 – 3 = 6 i.e. t^2 = 9 i.e. t = +3 i.e. y =+1 respectively for respective values of t=+3 i.e. x =+6 respectively for respective values of t=+3 i.e. x^2 – y^2 = (+6)^2 - (+1)^2 = 36-1 = 35 SUFFICIENT

Statement 2: t^3 = −27 i.e. i.e. t = -3 i.e. y =-1 respectively for respective values of t=+3 i.e. x =-6 respectively for respective values of t=+3 i.e. x^2 – y^2 = (-6)^2 - (-1)^2 = 36-1 = 35 SUFFICIENT

Answer: Option D
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Re: If x = 2t and y = t/3, what is the value of x^2 – y^2? [#permalink]

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19 Jul 2017, 23:52

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I went a little too far while solving this question.

I got: (7t/3) (5t/3)

From statement 1: t^2= 9 So t can be positive 3 or negative 3.

When I plugged in the value the answer was 35 in both the cases..

My query is when we get the same answer, do we have to worry about the two values of t because we usually take one? Or we have to just consider unique value of x^2-y^2?

Also when its given that t^3= -27

We have to just take the negative because (negative)^odd = negative

and we have to take both positive and negative values into consideration when its (value)^even....
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Help me make my explanation better by providing a logical feedback.

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