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Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?

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Posts: 50730
Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?  [#permalink]

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New post 20 Dec 2017, 05:18
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

67% (01:32) correct 33% (02:03) wrong based on 37 sessions

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Re: Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?  [#permalink]

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New post 20 Dec 2017, 05:20
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Re: Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?  [#permalink]

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New post 21 Dec 2017, 10:34
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Bunuel wrote:
Is \(\frac{x}{p}(p^2 + q^2 + r^2) = xp + yq + zr\) ?


(1) \(yq = r^2\) and \(zr = q^2\)

(2) \(x = p\)


We'll go for a Precise approach of simplification since this looks like a very technical question.

Multiplying by p (p cannot be zero as it is in denominator) gives:
\(xp^2 + xq^2 + xr^2 = xp^2 + yqp + zrp\). Cancelling common factors and changing sides gives:
\(xq^2 - yqp + xr^2 - rzp = 0\)
Substituting the equations in (1) gives \(xzr - yqp + xyq - zrp = zr(x - p) + yq(x - p) = (zr + yq)(x-p)\)
Without knowing if \(zr + yq = 0\) or if \(p = x\) we cannot know if this is 0.
Insufficient!

Substituting the equations in (2) gives
\(xq^2 - yqx + xr^2 - rzx = x(q^2 - yq + r^2 - rz)\)
Since we don't know if either side is 0, this is also insufficient.

But, if we substitute \(x = p\) into our result from (1) we get a result of 0!
This means that the original equation is true.
(C) is our answer.
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Re: Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ? &nbs [#permalink] 21 Dec 2017, 10:34
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