Is this algebra allowed Given
$$\frac{\frac{a}{a+b}}{\frac{c}{c+d}} = 1.8 \text{ (1)}$$
and 
$$\frac{a}{a+b}-\frac{c}{c+d} = \frac{2}{15} \text{ (2)}$$
and
$a = 30$ 
and 
$a+b+c+d=220 \text{ (3)}$
Solve the remaining

This is what I did 
According to $(1)$ $$\frac{a}{a+b} = 1.8 \frac{c}{c+d}$$
$$\Leftrightarrow \frac{5}{9} \cdot \frac{a}{a+b} = \frac{c}{c+d}$$
then using $(2)$
$$\frac{a}{a+b} - \frac{c}{c+d} = \frac{2}{15}$$
$$\frac{30}{30+b} - \frac{5 \times 30}{9(30+b)} = \frac{2}{15} \implies b = 70$$
using $(3)$
$(30+70) + (c+d) = 220 \implies (a+c) = 120$ 
going back to $(2)$
$$\frac{a}{a+b} - \frac{c}{c+d} = \frac{2}{15}$$ 
$$\frac{3}{10} - \frac{c}{120} = \frac{2}{15} \implies c = 20, d = 100$$
Therefore : $a = 30, b = 70, c = 20, d = 100$
Confused if some of these algebra isn't allowed like rearranging then using it in another equation like:
$$\frac{c}{c+d} = \frac{5}{9} \times  \frac{a}{a+b}$$
in the initial step. 

Your answer is perfectly fine. Giving fractional algebraic values to another variable fraction is as good as substituting non-fractional variables such as $p=q$. You can always do this unless the denominator equals $0$.
