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}