The combustion can be described by the following reaction:
$$\ce{C_xH_y (g) + ($x + y/4$) O2 (g) ->[\Delta] x CO2(g) + $y/2$ H2O (g)}$$
All products are gaseous at first, but as the system cools down to the room temperature, water condenses attributing to the decrease in $\pu{30 mL}$. Further decrease in $\pu{40 mL}$ is due to reaction of alkali with the remaining gaseous product, $\ce{CO2}$:
$$\ce{CO2 (g) + 2 M^IOH (aq) -> M^I2CO3 (aq) + H2O}$$
From the ideal gas law it's obvious that amounts of gaseous substances are proportional to their volumes:
$$n = \frac{p}{RT}V \quad\to\quad n\propto V$$
Let's rewrite the first equation with assigned volumes:
$$\ce{\underset{\pu{20 mL}}{C_xH_y (g)} + ($x + y/4$) O2 (g) ->[\Delta] \underset{\pu{40 mL}}{x CO2(g)} + \underset{\pu{30 mL}}{$y/2$ H2O (g)}}$$
Since $n\propto V$, and for the balanced chemical equation amounts of reactants and products are equal, so are their volumes. Hence:
Q: Do they mean that the combined volumes of the reactants and the resultants are the same? That's an incorrect inference, isn't it?
A: Yes; no, this is correct for the given reaction.
That "strange line"
$$\pu{20 mL} + 20(x + y/4)~\pu{mL} - \pu{20 mL} = \pu{30mL}$$
is a direct consequence of that as it can also be rewritten as
$$\underset{\ce{C_xH_y}}{\pu{20 mL}} + \underset{\ce{O2}}{20(x + y/4)~\pu{mL}} = \underset{\ce{CO2}}{\pu{20 mL}} + \underset{\ce{H2O}}{\pu{30mL}}.$$
However, as this is a single equation with two unknowns, it only illustrates the preservation of the total amounts of products and reactants. In order to find the formula of unknown hydrocarbon, you just have to equate corresponding coefficients with the ratio between $\ce{C_xH_y}$, $\ce{CO2}$ and $\ce{H2O}$ ($20:40:30 = 1:2:1.5$):
$$x = 2$$
$$y/2 = 1.5 \to y = 3$$
which results in empirical formula $\ce{C2H3}$. In order to find out the molecular formula you would also need auxiliary data such as molecular weight of the hydrocarbon.
