Question

If $xy=2(x+y),x\le y$ and $x,y\in \mathbb{N}$, then the number of solutions of the equations are:
a) Two
b) Three
c) no solution
d) infinitely many solutions

Answer 1

Hint: Here, first convert the equation $xy=2(x+y)$ in terms of y. Now, we will get a fraction and the denominator should be non-zero. Then, put that value of y in the condition, $x\le y$, we will get the inequality $x\le 4$ and also apply the condition $x,y\in \mathbb{N}$, we will get the number of solutions of the equation.

Complete step-by-step answer:

Here, we are given that $xy=2(x+y),x\le y$ and $xy\in \mathbb{N}$.

Now, we have to find the number of solutions of the equations.

We have the equation:

$\begin{align}

  & xy=2(x+y) \\

 & \Rightarrow xy=2x+2y \\

\end{align}$

Next, take 2y to the left side, then 2y becomes -2y, the equation,

$\Rightarrow xy-2y=2x$

In the above equation y is the common factor for both the terms. So take y outside,

$\Rightarrow y(x-2)=2x$

In the next step, we have to do the cross multiplication,

$\Rightarrow y=\dfrac{2x}{x-2}$

We know that a fraction is defined if the denominator is non-zero.

Therefore, we can say that:

$x-2\ne 0$

Next, take 2 to the right side, 2 becomes -2,

$\Rightarrow x\ne 2$

Now also have the inequality,

$x\le y$

Next, substitute the value of $y=\dfrac{2x}{x-2}$ in the above inequality.

$x\le \dfrac{2x}{x-2}$

In the next step, by doing the cross multiplication we get the inequality,

$\Rightarrow x(x-2)\le 2x$

Now, by taking 2x to the left side, 2x becomes -2x,

$\begin{align}

  & \Rightarrow x(x-2)-2x\le 0 \\

 & \Rightarrow x\times x-x\times 2-2x\le 0 \\

 & \Rightarrow {{x}^{2}}-2x-2x\le 0 \\

 & \Rightarrow {{x}^{2}}-4x\le 0 \\

\end{align}$

Now, by again taking -4x to the right side, we get the equation:

$\Rightarrow {{x}^{2}}\le 4x$

Here, we have to do the cross multiplication,

$\Rightarrow \dfrac{{{x}^{2}}}{x}\le 4$

Hence, by cancellation,

$\Rightarrow x\le 4$

We are given that x and y are natural numbers, which means they cannot be zero or negative

integers and $x\ne 2$.

Therefore we can say that x can take the values 1, 3 and 4.

Hence, we get x = 1, 3, 4

So, the number of solutions of the equations are three.

Therefore, the correct answer for this question is option (b).

Note: Here, in the question it is given that $x,y\in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers, that is

$\mathbb{N}=\{1,2,3,4,5,...........\}$. Here, zero is not included, only the positive integers are included. Since, for $x\le 4$ we can’t take zero and negative values and $x\ne 2$, x can only take the values 1, 3 and 5.