For $x≠y$ and $2005(x+y) = 1$; Show that $\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$

Question

Problem:

Let $x$ and $y$ two real numbers such that $x≠0$ ; $y≠0$ ; $x≠y$ and $2005(x+y) = 1$

1. Show that $$\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$$

2. Calculate $l$:

$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$

For the first question, I tried to work it out with algebra; I solved for x through the equation given, then multiplied it by y and I got the value of $\frac{1}{xy} = 2005\left(\frac{1}{y-2005y^2}\right) $. Then I tried proving that $\frac{1}{y-2005y^2} =\frac{1}{x} + \frac{1}{y} $ but I failed at this.

Answer 1

1. $$\frac{1}{xy} = 2005(\frac{1}{x}+\frac{1}{y}) \iff \frac{1}{xy}=\frac{2005(x+y)}{xy}$$ which follows immediately from the condition

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2. $$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2=$$$$= \frac{y}{y-x}-({1}-\frac{x}{y})-\frac{x}{y-x}-(-1+\frac{y}{x})+\frac{y}{x} - \frac{x}{y}+2=$$$$=\frac{y-x}{y-x}+2=3$$

Explanation:

• First divide the fraction into two fractions (like $\frac{y-x}{y}=1-\frac{x}{y}$)
• Cancel out the opposite terms

Answer 2

For the second part of your question,

For $y \neq x$, $$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$ $$\implies l = \left(\frac{y}{y-x} - \frac{x}{y-x}\right) - \frac{y-x}{y} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$ $$\implies l= -(y-x)\left(\frac{1}{y} + \frac{1}{x}\right) + \frac{(y+x)(y-x)}{xy} + 3$$ $$\implies l = (y-x)\left(\frac{y+x}{xy} - \frac1y - \frac1x\right) + 3$$ $$\implies l =3$$

Answer 3

For part 1, if $xy\not=0$, then

$$\begin{align} 1=2005(x+y)\implies{1\over xy}&=2005(x+y){1\over xy}\\ &=2005\left({x\over xy}+{y\over xy}\right)\\ &=2005\left({1\over y}+{1\over x}\right)\\ &=2005\left({1\over x}+{1\over y}\right) \end{align}$$

For part 2, regroup the terms with denominators $y-x$, $y$, and $x$ and simplify:

$$\begin{align} {y\over y-x}-{y-x\over y}-{x\over y-x}-{y-x\over x}+{y\over x}-{x\over y}+2 &=\left({y\over y-x}-{x\over y-x} \right)-\left({y-x\over y}+{x\over y} \right)+\left({y\over x}-{y-x\over x} \right)+2\\ &=\left(y-x\over y-x\right)-\left(y\over y\right)+\left(x\over x\right)+2\\ &=1-1+1+2\\ &=3 \end{align}$$

Answer 4

According to the problem, $x, y$ $\in$ $\mathbb{R}$ and $x \neq y$.

$\bullet$ For the first part,

(I) From the given, we have that \begin{align*} &2005(x + y) = 1\\ \implies & 2005 \cdot \frac{(x + y)}{xy} = \frac{1}{xy}\\ \implies & 2005 \cdot \bigg( \frac{1}{x} + \frac{1}{y} \bigg) = \frac{1}{xy} \end{align*} Hence, done!

$\bullet$ For the second part,

(II) According to question, \begin{align*} l =& ~\frac{y}{y-x} - \frac{y - x}{y} - \frac{x}{y - x} - \frac{y - x}{x} + \frac{y}{x} - \frac{x}{y} + 2\\ = & ~ \bigg[ \frac{y}{y-x} - \frac{x}{y - x} \bigg] - \bigg[ \frac{y - x}{y} + \frac{x}{y} \bigg] + \bigg[ \frac{y}{x} - \frac{y - x}{x} \bigg] + 2\\ = & ~ 1 - 1 + 1 + 2\\ = &~ 3 \end{align*} Hence, the value of $l$ is $3$.