kyuhunl wrote:When working together at their constant rates, machines A and B can complete a certain task in 4 hours. How many hours would it take machine B to complete the task, when working alone at its constant rate?
1) When working alone at its constant rate, machine A can finish the task in 6 hours.
2) When working alone at their constant rates, it takes machine B 6 hours longer than machine A to finish the task
$${T_{A \cup B}} = 4\,{\text{h}}$$
$$? = {T_B}\,\,\,\left[ {\text{h}} \right]$$
$$\left( * \right)\,\,\,\frac{1}{{{T_{A \cup B}}}} = \frac{1}{{{T_A}}} + \frac{1}{{{T_B}}}$$
$$\left( 1 \right)\,\,{T_A} = 6\,{\text{h}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\frac{1}{4} = \frac{1}{6} + \frac{1}{{{T_B}}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{T_B}\,\,{\text{unique}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{SUFF}}.$$
$$\left( 2 \right)\,\,{T_B} = {T_A} + 6\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\frac{1}{4} = \frac{1}{{{T_A}}} + \frac{1}{{{T_A} + 6}}\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,{T_A} > 0\,\,{\text{unique}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{T_B}\,\,{\text{unique}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{SUFF}}.$$
$$\left( {**} \right)\,\,\,\,\frac{1}{4} = \frac{{2{T_A} + 6}}{{{T_A}\left( {{T_A} + 6} \right)}}\,\,\,\,\, \Rightarrow \,\,\,\, \ldots \,\,\,\, \Rightarrow \,\,\,\,{T_A}^2 - 2{T_A} - 24 = 0\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{roots}}\,\,{\text{product}}} \,\,\,\,\left( {\frac{c}{a} = } \right)\frac{{ - 24}}{1} < 0\,\,\,\, \Rightarrow \,\,\,\,{T_A} > 0\,\,{\text{unique}}$$
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
