I performed a series of algebraic manipulations starting from the formula found at https://en.wikipedia.org/wiki/Celsius, but I can't reach the correct Fahrenheit-to-Celsius formula. Where did I go wrong?
$$\begin{align*} &x^\circ C = \left( \frac{9}{5} x + 32\right)^\circ F \\ &x^\circ C = \frac{9}{5} x^\circ F + 32^\circ F \\ &x^\circ C - 32^\circ F = \frac{9}{5} x^\circ F \end{align*}$$
It would be best to think about the conversion of Celsius to Fahrenheit as follows:
If $x^\circ\text C$ is the equivalent to $y^\circ\text F$, then by the Celsius to Fahrenheit conversion formula:
$$y = \dfrac{9}{5}x + 32$$
Therefore, rearranging for $x$ yields the conversion formula of $y^\circ\text F$ to $x^\circ\text C$.
$$\dfrac{9}{5}x = y - 32 \Longrightarrow \boxed{\therefore x = \dfrac{5}{9}(y - 32)}$$
The wikipedia article that you referenced is accurate, but somewhat confusing.
I will use the variable $~c~$ to refer to temperature measured in Celsius degrees.
I will use the variable $~f~$ to refer to temperature measured in Fahrenheit degrees.
The statement $~x^\circ C \equiv \left[ ~\dfrac{9x}{5} + 32 ~\right]^\circ F~$ should actually be interpreted as
$$f = \frac{9c}{5} + 32 \implies f - 32 = \frac{9c}{5} \implies $$
$$(f - 32) \times \frac{5}{9} = \frac{9c}{5} \times \frac{5}{9} = c.$$
This corresponds to the wikipedia's statement that $~x^\circ F \equiv \left[ ~(x - 32) \times \dfrac{5}{9} ~\right]^\circ C.$
I performed a series of algebraic manipulations starting from the formula found at https://en.wikipedia.org/wiki/Celsius : $$x^\circ C \color\red= \left( \frac{9}{5} x + 32\right)^\circ F\tag1$$
You're misquoting Wikipedia: it actually asserts that $$x^\circ C ≅ \left( \frac{9}{5} x + 32\right)^\circ F,$$ where the symbol ≅ means corresponds to. (As a matter of fact, temperature conversion between Celsius and Fahrenheit is merely a bijection, i.e., one-to-one correspondence.) In other words: $\boxed{x\text{°C} ≅ y\text{°F}\quad\iff\quad y=1.8x+32\,}.\tag*{}$
Statement $(1)$ is incorrect, since it implies that $0\text{°C}\color\red=32\text{°F},\tag*{}$ which, by multiplying both sides, in turn implies that $0\text{°C}\color\red=212\text{°F}\quad!\tag*{}$
Instead of statement $(1)$'s wrong assertion of equality between corresponding Celsius and Fahrenheit temperatures (their non-equality is evidenced by the fact that the zero points of the Celsius and Fahrenheit scales don't coincide) $1\text{°C}\color\red=33.8\text{°F},\tag*{}$ we can equate temperature intervals instead: $\boxed{1\text{°C}=1.8\text{°F}}.\tag*{}$
but I can't reach the correct Fahrenheit-to-Celsius formula.
Yes, it's impossible to manipulate $(1)$ into
$$x^\circ F \color\red= \frac59\left(x - 32\right)^\circ C,\tag2$$ since no algebraic relationship exists between the units °C and °F (in contrast to, for example, metre = 39.37 inch).
Generally, if $U$ and $V$ are measurement units, then $$xU=f(x)V\quad\implies\\x\propto f(x)\quad\text{and}\quad U\propto V.$$
Addendum
Assume, for the sake of contradiction, that corresponding temperatures on the Celsius and Fahrenheit scales are actually equal to each other. Then they exhibit the same scaling behaviour and undergo the same relative change. In particular, $-10 \text{°C}$ rising to $0 \text{°C}$ undergoes the same percentage change as $14 \text{°F}$ rising to $32 \text{°F};$ so, $100\% = 129\%.$ Hence, by contradiction, we conclude that corresponding Celsius and Fahrenheit temperatures are not in fact generally equal to each other.