Question

If two real numbers x and y satisfy the equation x/y = x- y, then:

• A. x$\ge$4 and x$\le$0 where x$\ge$ a means that x can take any value greater than a or equal to a
• B. y= can equal 1
• C. both x and y must be irrational
• D. x and y cannot both be integers
• E. both x and y must be rational

Answer 1

The correct option is A x$\ge$4 and x$\le$0 where x$\ge$ a means that x can take any value greater than a or equal to a

We are given that two real numbers $x$ and $y$ satisfy the equation $\frac{x}{y}=x-y$.

We will discuss each option separately.

Option A)

Multiplying the above equation throughout by $y$ and rearranging we get,

$y^2-xy+x=0$. We will treat this as a quadratic equation in the variable $y$. The discriminant of this equation is $x^2-4x$. We know that the above quadratic equation has a real solution in $y$ if and only if the discriminant is greater than or equal to 0. That is, if and only if $x^2-4x\ge 0$. That is, if and only if $x(x-4)\ge 0$. So we observe that the discriminant of the equation is greater than or equal to 0 if and only if $x\ge 4$ or $x\le 0$. So option A is correct.

Option B)

If we put $y=1$ in the equation $\frac{x}{y}=x-y$, then we get $x=x-1$, which has no solution. So option B is wrong.

Option C) and Option D)

We claim that $y=2$ and $x=4$ is a solution for th egiven equation. For this, we write-

$LHS=\frac{x}{y}$

$=\frac{4}{2}=2$, and

$RHS=x-y$

$=4-2=2$.

As 2 and 4 are integers, Option C and Option D are wrong.

Option E)

We write the equation in the form $y^2-xy+x=0$.

After putting $x=5$, we get $y^2-5y+5=0$.

By the quadratic formula, the roots of this equation are :

$y=\frac{5\pm \sqrt{5^2-20}}{2}=\frac{5\pm \sqrt{5}}{2}$.

This means that $x=5$ and $y=\frac{5+\sqrt{5}}{2}$ is a solution of the given equation in which $y$ is irrational. So, option E is wrong.