Solve the equation $\displaystyle {\frac{{{1}}}{{{x}+{3}}}}-{\frac{{{1}}}{{{x}+{4}}}}={\frac{{{1}}}{{{6}}}}$.

The solutions are $\displaystyle {x}_{{1}}=$ Preview Question 6 Part 1 of 2   and $\displaystyle {x}_{{2}}=$ Preview Question 6 Part 2 of 2  
where $\displaystyle {x}_{{1}}\leq{x}_{{2}}$.