Assembly sick worker

Two assemblers with the same daily output were to complete the assembly in 12 days. After 4 days, one of them got sick, and the other finished the assembly in another 16 days. How long would it take everyone to do the assembly by themselves?

Final Answer:

a =  24 d
b =  24 d

Step-by-step explanation:

$$\begin{gathered} 12/a+12/b=1 \\ 4/a+(16+4)/b=1 \\ x=1/a;y=1/b \\ 12\cdot x+12\cdot y=1 \\ 4\cdot x+(16+4)\cdot y=1 \\ 12x+12y=1 \\ 4x+20y=1 \\ Row2-\frac{4}{12}\cdot Row1\to Row2 \\ 16y=0.67 \\ y=\frac{0.66666667}{16}=0.04166667 \\ x=\frac{1-12y}{12}=\frac{1-12\cdot 0.04166667}{12}=0.04166667 \\ x=\frac{1}{24}\doteq 0.041667 \\ y=\frac{1}{24}\doteq 0.041667 \\ a=1/x=1/0.0417=24\text{d} \end{gathered}$$

$b=1/y=1/0.0417=24\text{d}$

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