In trapezoid $ABCD$, $\overline{AB}$ is parallel to $\overline{CD}$, $AB = 7$ units, and $CD = 10$ units. Segment $EF$ is drawn parallel to $\overline{AB}$ with $E$ lying on $\overline{AD}$ and $F$ lying on $\overline{BC}$. If $BF:FC = 3:4$, what is $EF$? Express your answer as a common fraction.
Extend segments $\overline{DA}$ and $\overline{CB}$ until they intersect at point $G$ as shown.  Define $x=GB$ and $y=BF/3$ .  Then $BF=3y$ and $FC=4y$.  (We chose $y=BF/3$ so we could represent both $BF$ and $FC$ without fractions).  Using the similarity of triangles $GBA$ and $GCD$, we have $\frac{x}{7}=\frac{x+7y}{10}$.  Solving this equation for $x$, we find $x=49y/3$.  Now using the similarity of triangles $GBA$ and $GFE$ and substituting $49y/3$ for $x$ we get \begin{align*}
\frac{x}{7}&=\frac{x+3y}{EF} \\
\frac{\frac{49}{3}y}{7}&=\frac{\frac{49}{3}y+3y}{EF} \\
\frac{49}{21}&=\frac{58}{3\,EF} \\
EF &= \boxed{\frac{58}{7}}\text{ units}.
\end{align*}

(Note: Intuitively, since $F$ is three-sevenths of the way from $B$ to $C$, $EF$ should be three-sevenths of the way from $7$ to $10$.  This intuition is correct and may be proved using the same approach as above.)

[asy]
size(150);
defaultpen(linewidth(.7pt)+fontsize(8pt));
pair A=(0,0), B=(7,0), C=(8,-4), D=(-2,-4), Ep=3/7*D+4/7*A, F=3/7*C+4/7*B, G=(14/3,28/3);
pair[] dots={A, B, C, D, Ep, F, G};
draw(dots);

draw(G--C--D--cycle);
draw(A--B);
draw(Ep--F);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$E$",Ep,W);
label("$F$",F,E);
label("$G$",G,N);
label("$x$",midpoint(G--B),W);
label("$3y$",midpoint(B--F),W);
label("$4y$",midpoint(F--C),W);
label("$7$",midpoint(A--B),N);
label("$10$",midpoint(C--D),N); [/asy]