As percusse and GEdgar point-out in there comments that the reason this seemingly simple equation is not solvable using simple algebra lies in the fact that that the LHS of $2^x = x^2$ is a transcendental function. i.e. it cannot be expressed as a polynomial. Actually the closest it can come to in a "polynomial" form is its Maclaurin series form (see below).
Using pre-calculus techniques you can show, for instance, that you can take log of both sides as in
$$2^x = x^2$$
$$\implies ln(2^x) = ln(x^2) \quad \forall x \ne 0 $$
$$\implies x ln(2) = 2 ln(x) $$
$$\implies ln(x) = {2x \over ln(2)} \quad \textbf {(A)}$$
So the solution to our problem are all values of $x$ that are the roots of equation $\textbf{(A)}$ ... Pre-calculus you can use graphing techniques to determine the answer.
Solving transcendental functions, in general, requires a lot of different calculus techniques, that are probably beyond the scope of this answer.
Infinite Series for ${2^x}$
Using Taylor's Theorem (which is part of calculus) we can show that:
$$e^u = \sum_{n=0}^{ \infty } {u^n \over n!} = 1 + {u^1 \over 1!} + {u^2 \over 2!} + {u^3 \over 3!} + {u^4 \over 4!} + \cdots \quad \textbf{(B)}$$
For considerable historical reasons $\textbf{(B)}$ is called Maclaurin series for $e^u$. You can find Maclaurin series for a large number of functions that have certain properties.
For purpose of this discussion, assume that (B) is provable. We can use it to express the infinite series for $2^x$ by noting that $2 \equiv e^{ln(2)}$, and that $(a^x)^y = (a)^{xy}$.
$$ [2]^x = [e^{ln(2)}]^x = [e^{ln(2) \dot x}]$$
Substituting $u$ with $2^x$ in $\textbf{(B)}$ power-series we get:
$$2^x = ln(2) \sum_{n=0}^{ \infty } {x^n \over n!} = ln(2) \left( 1 + {x^1 \over 1!} + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots \right)$$
As you can see solving (A) without knowing some more properties about behavior of $e^x$ becomes intractable. That is what calculus is all about ;) once you get into it, you wil see that these problems become solvable. Although, the solutions are, by no means, trivial.
$y=\dfrac{4x-1}{2x+3}$
swap $ x,y $
$x=\dfrac{4y-1}{2y+3}$
Solve back $y$ in terms of $x$
$y=\dfrac{3x+1}{-2x+4}, $ done.
EDIT1:
It is an interesting bi-linear or fractional linear function. Notice that coefficients in the left diagonal got swapped and signs of right diagonal elements changed,
$$ \dfrac{a x + b }{c x + d} \rightarrow \dfrac{d x - b }{-c x + a} $$
leaving $ (a d - b c) $ unaltered.
Best Answer
So you can write $$x(1-a)=a$$ If $a=1$ we have $$x\times 0=1$$, which is impossible. For $a\neq 1$ we get $$x=\frac{a}{1-a}$$