Three workers

Three workers, A, B, and C, must work on a specific task. Workers A and B completed the task in 19 days, B with C for 19 days, and A with C for 11 days.

How long would it take to complete the task for everyone alone?
How long would it take to complete the task if all three worked together?

Final Answer:

A =  22 d
B =  139.33 d
C =  22 d
x =  10.1951 d

Step-by-step explanation:

$$\begin{gathered} 19(\frac{1}{A}+\frac{1}{B})=1 \\ 19(\frac{1}{B}+\frac{1}{C})=1 \\ 11(\frac{1}{A}+\frac{1}{C})=1 \\ 19(a+b)=1 \\ 19(b+c)=1 \\ 11(a+c)=1 \\ 19\cdot (a+b)=1 \\ 19\cdot (b+c)=1 \\ 11\cdot (a+c)=1 \\ 19a+19b=1 \\ 19b+19c=1 \\ 11a+11c=1 \\ Row3-\frac{11}{19}\cdot Row1\to Row3 \\ -11b+11c=0.421 \\ Row3-\frac{-11}{19}\cdot Row2\to Row3 \\ 22c=1 \\ c=\frac{1}{22}=0.04545455 \\ b=\frac{1-19c}{19}=\frac{1-19\cdot 0.04545455}{19}=0.00717703 \\ a=\frac{1-19b}{19}=\frac{1-19\cdot 0.00717703}{19}=0.04545455 \\ a=\frac{1}{22}\doteq 0.045455 \\ b=\frac{3}{418}\doteq 0.007177 \\ c=\frac{1}{22}\doteq 0.045455 \\ A=1/a=1/0.0455=22\text{d} \end{gathered}$$

$B=1/b=1/0.0072=139.33\text{d}$

$C=1/c=1/0.0455=22\text{d}$

$x=\frac{1}{a+b+c}=\frac{1}{0.0455+0.0072+0.0455}=10.1951\text{d}$

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