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Re: If x -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
jusjmkol740 wrote:
If \(x ≠ -y\), then is \(\frac{x-y}{x + y} > 1\)?


(1) \(x > 0\)

(2) \(y < 0\)
IanStewart chetan2u could you please see if this is right?
Is \(\frac{x-y}{x + y} > 1\) ?
Multiply both sides by \((x+y)^2\), although we don't know the sign of \(x \) and \(y\), we can be sure that \((x+y)^2\) is positive. So, it can be multiplied on both sides.
Is \(\frac{(x+y)^2*x-y}{x + y} > (x+y)^2*1\) ?
Is \((x+y)*(x-y) > (x+y)^2\) ?
Is \((x^2 - y^2) > (x^2 + y^2 + 2xy)\) ?
Is \((- y^2) > (y^2 + 2xy)\)
Is \((- 2y^2) > (2xy)\) ?
Is \((y^2) < (xy)\) ?
Multiply both sides by \(1/y{^2}\),
Is \( 1< (x/y)\) ?
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Re: If x -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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PyjamaScientist wrote:
chetan2u could you please see if this is right?
Is \(\frac{x-y}{x + y} > 1\) ?
Multiply both sides by \((x+y)^2\), although we don't know the sign of \(x \) and \(y\), we can be sure that \((x+y)^2\) is positive. So, it can be multiplied on both sides.
Is \(\frac{(x+y)^2*x-y}{x + y} > (x+y)^2*1\) ?
Is \((x+y)*(x-y) > (x+y)^2\) ?
Is \((x^2 - y^2) > (x^2 + y^2 + 2xy)\) ?
Is \((- y^2) > (y^2 + 2xy)\)
Is \((- 2y^2) > (2xy)\) ?
Is \((y^2) < (xy)\) ?
Multiply both sides by \(1/y{^2}\),
Is \( 1< (x/y)\) ?


It's right up until the point I've highlighted. When you wrote the highlighted line, a negative sign disappeared (I think you were dividing by -2 on both sides, but then the right side should become negative). The other steps all look good -- you can certainly multiply or divide both sides of an inequality by a nonzero square, since squares are never negative.
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If x -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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PyjamaScientist wrote:
jusjmkol740 wrote:
If \(x ≠ -y\), then is \(\frac{x-y}{x + y} > 1\)?


(1) \(x > 0\)

(2) \(y < 0\)
IanStewart chetan2u could you please see if this is right?
Is \(\frac{x-y}{x + y} > 1\) ?
Multiply both sides by \((x+y)^2\), although we don't know the sign of \(x \) and \(y\), we can be sure that \((x+y)^2\) is positive. So, it can be multiplied on both sides.
Is \(\frac{(x+y)^2*x-y}{x + y} > (x+y)^2*1\) ?
Is \((x+y)*(x-y) > (x+y)^2\) ?
Is \((x^2 - y^2) > (x^2 + y^2 + 2xy)\) ?
Is \((- y^2) > (y^2 + 2xy)\)
Is \((- 2y^2) > (2xy)\) ?
Is \((y^2) < (xy)\) ?
Multiply both sides by \(1/y{^2}\),
Is \( 1< (x/y)\) ?


IanStewart has of course told you exactly where you have gone wrong.

Is \((- 2y^2) > (2xy)\) ?
Is \(-y^2>xy\) ? Or
Is \((y^2) <- (xy)\) ?
Multiply both sides by \(1/y{^2}\),
Is \( 1< -(x/y)\) ?

The two statements combined tell us that both are of opposite sign so x/y<0, hence -x/y>0.
Two possibilities
a) 1>x/y>0
b) x/y>1>0
But will it be between 0 and 1 or >1 cannot be told from given information
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Re: If x -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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Re: If x -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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