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

Answer: A

Concepts:rational equation quadratic inequality

Difficulty rating: 1830

Small Hint:

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

Big Hint:

Real $y$ requires the discriminant $x^2-4x$ to be nonnegative

Solution:

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.

1957 AMC 12 Problem 45

Answer: A

Difficulty rating: 1830

Small Hint:

Big Hint:

Solution:

Problem 45 in Other Years