Fish, Cabbages, & Soaps

$\text { 4 soaps = 5 cabbages } \Rightarrow \text {2 soaps = 2.5 cabbages }$

$\text {2 cabbages + 2 soaps = 4.5 cabbages = }$

$\text { = 1.5 × 3 cabbages = 1.5 × 2 fish = } \boxed { \text { 3 fish} }$

In order to answer this question, we need to evaluate how much one cabbage and one soap is worth in the rate of fish.

First, since $3$ cabbages equalizes $2$ fish, $1$ cabbage would equalize $\dfrac{2}{3}$ fish.

Then since $4$ cabbages equalizes $5$ cabbages, $1$ cabbage would equalize $\dfrac{5}{4}$ cabbages and, in turn, equalizes $\dfrac{5}{4}\cdot \dfrac{2}{3} = \dfrac{5}{6}$ fish.

Therefore, $2$ cabbages plus $2$ soaps will equalize $2\cdot \dfrac{2}{3}+2\cdot \dfrac{5}{6} = \dfrac{4+5}{3} = \boxed{3}$ fish.

Let the value of a fish, cabbage and soap be $f$ , $c$ and $s$ respectively, then we have:

$\begin{cases} 2f = 3c & \implies c = \frac 23f \\ 5c = 4s & \implies s = \frac 54 c = \frac 54 \times \frac 23 f = \frac 56 f \end{cases}$

$\implies 2c + 2s = 2 \left(\frac 23 + \frac 56\right) f = 2 \times \frac 96 f = \boxed{3} f$

Fish = $f$

Cabbage = $c$

Soap = $s$

Given

$2f = 3c\\ 5c = 4s \\ \implies \\ c = \dfrac{2}{3}f \\ s = \dfrac{5}{4}c = \dfrac{10}{12}f$

Now the question

$2c + 2s = ?f \\ = 2(c+s) \\ = 2 ( \dfrac{2}{3}f + \dfrac{10}{12}f) \\ = 4(\dfrac{1}{3}f + \dfrac{5}{12}f) \\ = 4(\dfrac{4f + 5f}{12}) = \dfrac{4 \times 9 f}{12} \\ = 3f$