Let $A=A(t,x),B=B(t,x), C=C(t,x)$ be arbitrary smooth function. Let $F=F(u,v)$ and $G=G(u,v)$.
Could anyone show me how the solutions of \begin{align} 2(BF_u-AF_v)-2B^2-B&=0\\ 2(BG_u-AG_v)+2A^2+A&=0 \end{align} are \begin{align} F&=\frac{2b+1}{2}u+f(r)\\ G&=\frac{2a+1}{2}v+g(r) \end{align} where $r=au+bv$, and $f,g$ are arbitrary smooth function of $r$. I can do this in Maple and Mathematica, I will be appreciated if someone can show me how to get the exact expression by hand.
You should use the so-called Method of characteristics. (However, explanation on wiki is not complete enough, but you can find it somewhere else)
Note that your equations are actually independent of each other. Therefore, I will show what to do only for the first equation.
Let us write the first equation in the form $$ B F_u - A F_v = \frac{2 B^2 + B}{2}. $$ Using the Method of characteristics, we write $$ \frac{du}{B} = \frac{dv}{-A} = \frac{dF}{\frac{2 B^2 + B}{2}}. $$ Therefore, we get two independent equations from here: $$ \frac{du}{B} = \frac{dv}{-A} \quad \mbox{and} \quad \frac{d u}{B} = \frac{dF}{\frac{2 B^2 + B}{2}}. $$ Simply integrating them, we get $$ \frac{u}{B} + \frac{v}{A} = C_1 \quad \mbox{and} \quad \frac{2 B+1}{2} u - F = C_2. $$ the Method of characteristics says that any function $\Phi(C_1, C_2) = 0$ will be a solution of your problem. Therefore, $C_2 = \varphi(C_1)$ with an arbitrary function $\varphi$ will be also a solution. Substituting $C_1$ and $C_2$ we get $$ F = \frac{2B + 1}{2} u - \varphi \left(\frac{u}{B} + \frac{v}{A} \right), $$ which is absolutely equivalent to your answer.