Avoid This Common Mistake Made With Fractions

Clearly,if $\frac{1}{x} + \frac{1}{y} = \frac{1}{x+y}$ we have $(x+y)^2 = xy$ $x^{2} + xy + y^{2} = 0$ .

Note that $x$ or $y$ cannot be $0$ .

$\textbf{Case 1.}$ $x$ and $y$ both have the same signs.

then clearly $x^{2} + xy + y^{2} > 0$ .

$\textbf{Case 2.}$ If $x$ and $y$ both have the opposite signs .Without loss of generality let us consider that $x$ is negative.

Let us take $x = -a$ where $a>0$ .

So the equation becomes $a^{2} -ay + y^{2} =0$ $a^{2} + y^{2} = ay$ --(1)

But since $a,y > 0$ ,by AM-Gm inequality we have $a^{2} + y^{2} \ge 2ay$

So equation (1) can never hold true.

Hence there are no solutions

$\frac{1}{x} + \frac{1}{y} = \frac{1}{x+y} \implies x^2 +xy+y^2=0$ considering $x$ or $y$ or $(x+y)$ are not zero.

Completing squares we get

$x^2 + 2 x \frac{y}{2} + \frac{y^2}{4} + \frac{3y^2}{4} =0 \implies (x+ \frac{y}{2})^2 + (\frac{ \sqrt{3} y}{2})^2 =0 \implies$ each term is $0$

From this, we infer that $(y=0)$ and $(x= - \frac{y}{2} =0)$ which violates our assumptions.

Thus, there are no such $(x,y)$ to satisfy the given equations.

According to the given equation: 1/x + 1/y = 1/(x+y)

if we take lcm of the left hand equation we get (x+y)/xy = 1/(x+y)
on multiplying by (x+y) both sides we get $(x+y)^{2}$ = xy
on opening and simplification we get $x^{2}$ + $y^{2}$ + xy = 0
on dividing by y^{2} we get
$\frac{x^{2}}{y^{2}}$ + $\frac{x}{y}$ + 1=0
the above equation is never zero as its discriminant is negative
Hence the given equation has no solution for $\frac{x}{y}$ . Hence no value for x and y is possible;

The equation is simplified to x^2+ xy+y^2=0.This can be treated as a quadratic of variable x with a=1,b=y and c=y^2.Putting these in x= [-b+/-sqr root(b^2-4ac)]/2a gives [y^2+/-sqr root(-3y^2)]/2.But sqr root of -3y^2 will always be imaginary unless y=0 for which x becomes =0.Hence (0,0)is the only instance this is possible.

taking LCM on LHS, we get

$\frac { x+y }{ xy } =\frac { 1 }{ x+y }$

now comparing denominator and numerator, we get

$x+y=1\quad and\quad x \times y=x+y$

from which we conclude that $x \times y = 1$ i.e., both are reciprocals and also both sum up as $1$

but given that x and y $\epsilon$ $Z$ we cannot have a reciprocal existing in integer hence, answer is $\boxed{0}$

Note: given that $x+y = 1$ we cannot have $x=y=1$ and also, there are no reciprocals whose sum is $1$ [This is according to me as I can't think of reciprocals whose sum is $1$ ]

First, get rid of the denominators: $\frac{1}{x} + \frac{1}{y} = \frac{1}{x+y}$ $x(x+y) + y(x+y) = xy$ $(x+y)(x+y) = xy$ $x^2+xy+y^2 = 0$

In order for the above equation to be true, either x or y must be negative. WLOG, let x be postive and y be negative. Therefore: $x^2+y^2 = -xy$

Lets first test some values: (1, -1), (1, -2), (2, -2), (1,-100) Then: $2 = 1$ $5 = 2$ $8 = 4$ $10,001 = 100$

Testing these values, the left hand side increases more than the right hand side. Therefore, $x^2+y^2$ cannot equal $xy$

If 1/x + 1/y = 1/x+y,
both x and y need to be 0. That's not possible as problem doesn't allow it !

Hence, no pairs exist.