At time [tex]t[/tex] (in hours), train W travels a distance of
[tex](45\,\mathrm{mph})t[/tex]
away from Abbington. Meanwhile, train X starts 150 mi away from Abbington and is getting closer, so its distance from Abbington is
[tex]150\,\mathrm{mi} - (55\,\mathrm{mph})t[/tex]
When the two trains meet, we have
[tex]45t = 150 - 55t[/tex]
Solve for [tex]t[/tex].
[tex]45t = 150 - 55t \implies 100t = 150 \implies t = \dfrac{150}{100} = 1.5[/tex]
The trains pass each other after 1.5 hours, at which point train W will have traveled a distance of
[tex](45\,\mathrm{mph})(1.5\,\mathrm h) = \boxed{67.5\,\rm mi}[/tex]
