SOLUTION: A computer encoder A can finish a job in as many days as encoder B. Encoder C can finish the same job in 2 days less than encoder B. Together, the three encoders worked for 2 days;

Question 908967: A computer encoder A can finish a job in as many days as encoder B. Encoder C can finish the same job in 2 days less than encoder B. Together, the three encoders worked for 2 days; after which encoder, A and B got sick and were forced to stop working leaving encoder C alone to finish the job in three more days. Find the rate of each encoder.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
a=b
c=b-2
2/a+2/b+2/c+3/c=1
2/b+2/b+2/(b-2)+3/(b-2)=1
4/b+5/(b-2)=1
4*(b-2)+5*b/(b*(b-2))=1
4*b-8+5*b=b^2-2b
9b-8=b^2-2b
b^2-11b+8=0
a=10.217, b=10.217, c=8.21699
check
2/10.217+2/10.217+2/8.21699+3/8.21699=1
ok

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