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What values of x have a corresponding value of y that satisfies both x y > 0 and x y = x + y ?

Explanation	Calculations	Help
$We first rewrite the given relationship to express y in terms of x by isolating y.$	$x \times y = x + y x \times y - y = x y \times (x - 1) = x y = \frac{x}{x - 1}$
$We then substitute this expression for y into the condition that the product of x and y is positive.$	$x \times y > 0 x \times \frac{x}{x - 1} > 0 \frac{x ^{2}}{x - 1} > 0$
$We note that x cannot equal 0 or 1. Because the numerator of the fraction is always positive when x is not zero, the fraction will be positive exactly when the denominator is positive.$	$x - 1 > 0 x > 1$	Theory & Tactics Method Card INEQ2-C Click to view full details
$We conclude that the conditions are satisfied exactly when x is greater than 1.$	$x > 1$

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x > 1

What values of x have a corresponding value of y that satisfies both x y > 0 and x y = x + y ?

Explanation	Calculations	Help
$We first rewrite the given relationship to express y in terms of x by isolating y.$	$x \times y = x + y x \times y - y = x y \times (x - 1) = x y = \frac{x}{x - 1}$
$We then substitute this expression for y into the condition that the product of x and y is positive.$	$x \times y > 0 x \times \frac{x}{x - 1} > 0 \frac{x ^{2}}{x - 1} > 0$
$We note that x cannot equal 0 or 1. Because the numerator of the fraction is always positive when x is not zero, the fraction will be positive exactly when the denominator is positive.$	$x - 1 > 0 x > 1$	Theory & Tactics Method Card INEQ2-C Click to view full details
$We conclude that the conditions are satisfied exactly when x is greater than 1.$	$x > 1$

Scroll horizontally to view all columns

x > 1