Solve the following equations:

$$
\begin{aligned}
& \text { a) } \frac{1}{x-1}+\frac{3}{x-3}-\frac{9}{x-5}+\frac{5}{x-7}=0 \\
& \text { b) } \frac{3}{x+1}+\frac{5}{x-1}-\frac{3}{x-3}-\frac{5}{x-5}=0 \\
& \text { c) } \frac{19}{x+3}+\frac{21}{x+1}-\frac{27}{x-1}+\frac{35}{x-3}=0 \\
& \text { d) } \frac{3}{x-3}-\frac{1}{x-2}-\frac{1}{x+1}+\frac{3}{x+2}=0
\end{aligned}
$$

(For the first-year students, we consider the complete solution of equations $a, b, c$ as a full-score paper.)
a) The sum of the denominators of the extreme terms, $2x-8$, is equal to the sum of the denominators of the two middle terms. Therefore, we can try to take half of this sum as a new unknown. $x-4=z$, that is, $x=z+4$ substitution, by bringing the extreme and middle term pairs to a common denominator

$$
\begin{gathered}
\left(\frac{1}{z+3}+\frac{5}{z-3}\right)+\left(\frac{3}{z+1}-\frac{9}{z-1}\right)=\frac{6(z+2)}{z^{2}-9}-\frac{6(z+2)}{z^{2}-1}= \\
=6(z+2)\left(\frac{1}{z^{2}-9}-\frac{1}{z^{2}-1}\right)=0
\end{gathered}
$$

Accordingly, $z+2=0$ gives the solution: $z=-2$, and thus $x=2$. The second parenthesis cannot be 0 because $z^{2}-9$ and $z^{2}-1$ are different, and their reciprocals are also different, so their difference is not 0. Therefore, the equation has only one root.

b) Here, too, the relationship observed above holds, but we can also make good use of the fact that the sum of two numerators is 0. Adding these terms in pairs

$$
3\left(\frac{1}{x+1}-\frac{1}{x-3}\right)=5\left(\frac{1}{x-5}-\frac{1}{x-1}\right), \quad \frac{-12}{x^{2}-2 x-3}=\frac{20}{x^{2}-6 x+5}
$$

Assuming that no denominator is 0, and taking the reciprocal of both sides, after rearrangement

$$
x(2 x-7)=0, \quad \text { and from this } \quad x_{1}=0, \quad x_{2}=3.5
$$

neither of which is an excluded value. Indeed, both satisfy the equation.

c) With similar rearrangement as in case a)

$$
\frac{6(9 x+8)}{x^{2}-9}-\frac{6(x+8)}{x^{2}-1}=\frac{48\left(x^{3}+8\right)}{\left(x^{2}-1\right)\left(x^{2}-9\right)}=0
$$

The two-term expression in the numerator can be factored based on the observation that $8=2^{3}$:

$$
x^{3}+2^{3}=(x+2)\left(x^{2}-2 x+2^{2}\right)
$$

Now, $x+2=0$ gives $x=-2$ (with which no denominator is 0), but the second factor cannot be 0 because it is positive for any $x$:

$$
x^{2}-2 x+4=(x-1)^{2}+3 \geq 3
$$

so there are no more roots.

d) Let $x-0.5=z$ as in case a). Thus

$$
\begin{aligned}
& 3\left(\frac{1}{z-2.5}+\frac{1}{z+2.5}\right)-\left(\frac{1}{z-1.5}+\frac{1}{z+1.5}\right)=\frac{6 z}{z^{2}-6.25}-\frac{2 z}{z^{2}-2.25}= \\
& =\frac{4 z(z-0.5)(z+0.5)}{\left(z^{2}-6.25\right)\left(z^{2}-2.25\right)}=0
\end{aligned}
$$

which can only be satisfied by $z_{1}=0, z_{2}=0.5$ and $z_{3}=-0.5$. Each of these gives a solution because with them all four denominators are non-zero numbers, namely

$$
x_{1}=0.5, \quad x_{2}=1, \quad x_{3}=0
$$

Ámon Magdolna (Győr, Zrínyi Ilona Secondary School II. grade)

Note. When setting the practice problem, we recommended to the contestants the solution of practice problem 626, which gave some conditions for the equation

$$
\frac{A}{m x+n}+\frac{B}{p x+q}+\frac{C}{r x+s}+\frac{D}{t x+u}=0
$$

to be solvable (without knowledge of mixed cubic equations). As we saw, all four of our equations belong to this type. In case a)

$$
\frac{A}{m}+\frac{B}{p}+\frac{C}{r}+\frac{frac{D}{t}=1+3-9+5=0
$$

and by combining the terms on the left side - further transforming the above (1) - we do not get a cubic term in the numerator, as we generally saw in the condition before (4) in the cited place. Similarly, for equations b) and d), the condition (4a) there holds:

$$
\begin{gathered}
\frac{A}{n}+\frac{B}{q}+\frac{C}{s}+\frac{D}{u}=0, \quad \text { because } \quad 3-5+1+1=0, \quad \text { or } \\
-1+\frac{1}{2}-1+\frac{3}{2}=0
\end{gathered}
$$

The simplification seen in case c) was mentioned in practice problem 626 without writing down the condition. [^0]

[^0]:    ${ }^{1}$ K.M.L. $22(1961) 67$. p.