If x and y are non-zero real numbers such that \frac{x - y}{x} > 2 and x y^2 < 0, which of… — GMAT

If $x$ and $y$ are non-zero real numbers such that $\frac{x - y}{x} > 2$ and $x y^2 < 0$ , which of the following inequalities must be true?

A
$x + y < 0$
B
$x y > 0$
$x + y > 0$ Cevap
D
$x - y > 0$
E
$\frac{y}{x} > -1$

Cevap

x + y > 0

From

x y^2 < 0

, since

y \neq 0

y^2

is strictly positive, which forces

x < 0

. Next, expanding

\frac{x - y}{x} > 2

by multiplying both sides by negative

x

reverses the inequality direction to

x - y < 2x

. Subtracting

x

from both sides gives

-y < x

, which is equivalent to

x + y > 0

. Thus,

x + y > 0

must always be true.

Adım Adım Çözüm

Determine the sign of

x

using

x y^2 < 0

Since

y \neq 0

y^2 > 0

. For the product

x y^2

to be negative,

x

must be negative (

x < 0

A positive quantity multiplied by a negative quantity yields a negative product.

Simplify the inequality

\frac{x - y}{x} > 2

taking the sign of

x

into account.

Multiplying both sides by

x

(and reversing the inequality because

x < 0

) gives

x - y < 2x

Multiplying or dividing an inequality by a negative value reverses the inequality sign.

Rearrange the terms of

x - y < 2x

to isolate the relationship between

x

and

y

Subtracting

x

from both sides yields

-y < x

, which rearranges to

x + y > 0

(or

y > -x

Adding

y

to both sides converts

-y < x

directly into

x + y > 0

Anahtar Kavram

Deducing signs and algebraic bounds in inequalities involving negative variables

Tahmini Süre:1m 30s

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