Steam on the saturation line is rarely all vapour. Its quality $x$ is the fraction of each kilogram that has actually boiled: $x = 0$ is saturated water, $x = 1$ is dry saturated steam, and a working header lives somewhere near 0.95. Quality is a pure fraction — no units, ever. If your $x$ arrives wearing kJ/kg, a rearrangement upstream is lying to you.

Read aloud: x equals h minus h-f, over h-f-g — $x = \dfrac{h - h_f}{h_{fg}}$. Every letter has a job. $h$ is the measured enthalpy of the mixture in kJ/kg; $h_f$ is the saturated liquid enthalpy at that same pressure; $h_{fg}$ is the latent heat at that same pressure. The numerator is the latent heat this kilogram actually took up; the denominator is how much was on offer. Whichever of the four the question leaves blank is the one you solve for.

The same physics has a second face. Written from the $h_f$ and $h_g$ columns it is the lever rule: $h = (1 - x)\,h_f + x\,h_g$ — h equals one minus x, times h-f, plus x times h-g. The wet share $(1-x)$ carries liquid enthalpy, the dry share $x$ carries vapour enthalpy, and the two shares add to one whole kilogram. Use whichever form matches the columns your table gives you; because $h_g = h_f + h_{fg}$, they can never disagree.

And the named mistake this lesson exists to prevent: dividing by $h_g$ instead of $h_{fg}$. They differ by the liquid enthalpy, they sit next to each other in the table, and the wrong one gives a plausible-looking answer every single time. Latent heat is the one with the journey in its subscript.