1957 AMC 12 Problem 45

Question

If two real numbers

x

and

y

satisfy the equation

\dfrac xy=x-y,

then:

x\ge4

or

x\le0,

where

x\ge a

means that

x

can take any value greater than

a

or equal to

a

y

can equal

1

both

x

and

y

must be irrational

x

and

y

cannot both be integers

both

x

and

y

must be rational

Clear the denominator and regard the result as a quadratic equation in

y

Real

y

requires the discriminant

x^2-4x

to be nonnegative

Since

y\ne0,

multiplying by

y

gives

x=xy-y^2,

or

y^2-xy+x=0.

For a real value of

y,

its discriminant must satisfy

x^2-4x=x(x-4)\ge0.

Thus

x\le0

or

x\ge4.

Therefore, the correct answer is A.

Problem 45 in other years