Solving this system of equations without a CAS: $4=a+c$, $5=d+ac+b$, $4=ad+bc$, $4=bd$ Solving this system of equations : 
$$4=a+c$$ $$5=d+ac+b$$ $$4=ad+bc$$ $$4=bd$$
Anyone knows how to solve this without a CAS? 

We obtain $$c=4-a$$ and $$d=\frac{4}{b},$$ which gives
$$\frac{4}{b}+b+a(4-a)=5$$ and
$$a\cdot\frac{4}{b}+b(4-a)=4.$$
From the last equation we obtain:
$$a=\frac{4b^2-4b}{b^2-4},$$ which gives
$$\frac{4}{b}+b+\frac{4b^2-4b}{b^2-4}\left(4-\frac{4b^2-4b}{b^2-4}\right)=5$$ or
$$\frac{b^2-5b+4}{b}+\frac{16b(b-4)(b-1)}{(b^2-4)^2}=0$$ or
$$(b-1)(b-4)\left(\frac{1}{b}+\frac{16b}{(b^2-4)^2}\right)=0$$ or
$$(b-1)(b-4)=0$$ because $$(b^2-4)^2+16b^2>0.$$
Can you end it now?
