BeatIsb17 wrote:Given that x ≠0, y ≠0, and (x+y) ≠0, is y/x > y/(x+y)
Statement #1: x > -1
Statement #2: y > -1
I started by simplifying the expression in question
y/x > y/(x+Y)
y(y+x) >xy
The step in red assumes that x and x+y are positive.
If x<0 but x+y>0, then the inequality symbol must flip from > to <.
If x>0 but x+y<0, then the inequality symbol must flip from > to <.
Since the signs of x and y are unknown, I don't recommend this approach.
Alternate approach:
y/x > y/(x+y)
y/x - y/(x+y) > 0
(xy + y²)/[x(x+y)] - xy/[x(x+y)] > 0
y²/[x(x+y)] > 0.
Since y is nonzero, y²>0.
Thus, the inequality in blue will hold true only if x(x+y) > 0.
Question stem, rephrased:
Is x(x+y) > 0?
Statements combined:
If x=1 and y=1, then x(x+y) > 0.
If x=-1/2 and y=2, then x(x+y) < 0.
INSUFFICIENT.
The correct answer is
E.
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