Problem: Three machines P,Q, and R, working together, can do a job in x hours. When working alone P needs an additional 6 hours to do the job; Q, one additional hour; and R​,x additional hours. The value of x​ is:
Answer Choices:
A. 32​
B. 1211​
C. 23​
D. 2
E. 3
Solution:
Since x1​ represents the fractional part of the job done in 1 hour when the three machines operate together, and so forth,
x+61​+x+11​+x+x1​=x1​ or, more simply,
x+61​+x+11​−2x1​=0
∴2x(x+1)+2x(x+6)−(x+6)(x+1)=0
\therefore 3 x^{2}+7 x-6=0, x=\dfrac{2}