15. ILLUSTRATIVE EXERCISE

Determine all pairs $(x, y)$ of real numbers that satisfy the equation

$$
\frac{4}{x+y}=\frac{1}{x}+\frac{1}{y} .
$$

SOLUTION
15. First, we exclude all pairs $(x, y)$ that make the denominator zero, i.e., we impose the conditions $x \neq 0, y \neq 0$ and $x+y \neq 0$.

Under these conditions, the expression $x y(x+y)$ is different from 0, so multiplying both sides of the given relation by it, we obtain that it is equivalent to $4 x y=(x+y)^{2}$, which simplifies to $(x-y)^{2}=0$, or $x=y$.

Therefore, the pairs we are looking for are all and only those that satisfy $x=y$, except for the pair $(0,0)$.

## SECOND SOLUTION

The given relation is homogeneous, meaning it remains unchanged if $(x, y)$ is replaced by $(\lambda x, \lambda y)$, with $\lambda \neq 0$.

This means that if the relation holds at a point $\left(x_{0}, y_{0}\right)$ different from the origin, it holds for all points on the line passing through the origin and $\left(x_{0}, y_{0}\right)$ (except the origin).

Consequently, after noting that the lines $x=0, y=0$ and $y=-x$ can already be excluded as they make some denominator zero, it suffices to ask for which lines of the form $y=m x$ the given relation is satisfied.

Substituting $m x$ for $y$ in the given relation, we get

$$
\frac{4}{x+m x}=\frac{1}{x}+\frac{1}{m x}
$$

which, since $x \neq 0$, becomes

$$
\frac{4}{1+m}=1+\frac{1}{m}
$$

which has the unique solution $m=1$.

This means that the set of all and only the solutions of the given relation consists of the points on the line $y=x$ except for $(0,0)$.

## THIRD SOLUTION

After recognizing the homogeneity of the relation and proceeding as in the second solution, to show that the only good line is $y=x$, one can also verify that intersecting the given relation with another, such as $x^{2}+y^{2}=1$, which intersects all lines through the origin, yields only solutions with equal abscissa and ordinate.