jmc

We can find two equations from the given information: $$xy=56$$ $$\frac{7}{x}+\frac{14}{y}=4$$ We can solve for $y$ in the first equation: $$y=56/x$$ Now, substitute into the second equation: $$\begin{array}{rl}\frac{7}{x}+\frac{14}{56/x}& =4\\ \Rightarrow \frac{7}{x}+\frac{x}{4}& =4\end{array}$$ Multiplying through by $4x$, we can clear all fractions: $$\begin{array}{rl}28+x^2& =16x\\ \Rightarrow x^2-16x+28& =0\end{array}$$ Factoring, we find: $$(x-2)(x-14)=0$$ $$x=2\text{ or }x=14$$ If we solve for $y$ using our original equation, we find either $y=28$ or $y=4$, giving two ordered solution pairs, $(2,28)$ and $(14,4)$. However, only the first satisfies the requirement that $x<y$. Thus, the value of $x$ is $\boxed{2}$.