jmc

We first find the equations of the lines $l$ and $m.$ Since $l$ passes through $(0,5)$ and $(3,0),$ its slope is $$\frac{0-5}{3-0}=-\frac{5}{3}.$$Since it passes through $(0,5),$ its $y$-intercept is $(0,5)$ so $l$'s equation is $y=-\frac{5}{3}x+5.$

Since $m$ passes through $(0,2)$ and $(7,0)$, it has slope $$\frac{0-2}{7-0}=-\frac{2}{7}.$$Since $m\text{’s}$ $y$-intercept is $(0,2),$ its equation is $y=-\frac{2}{7}x+2.$

Now we find the $x\text{’s}$ at which $m$ and $l$ reach $y=15.$ Setting $y=15$ in both equations, we solve: $$y=15=-\frac{5}{3}x+5.$$Subtracting $5$ and multiplying by $-\frac{3}{5}$ for both sides, we get $x=-6.$ Thus, $l$ reaches $y=15$ when $x=-6.$ Now we solve $$y=15=-\frac{2}{7}x+2.$$Subtracting $2$ and multiplying by $-\frac{7}{2}$ for both sides, we get $x=-\mathrm{45.5.}$ Thus, $m$ reaches $y=15$ when $x=-\mathrm{45.5.}$

Thus, $(-6)-(-45.5)=\boxed{39.5},$ which is our answer.