just like the previous one
if the ship travelled a total of 149 miles, then let's say it travelled East "d" miles and thus it travelled North the slack of 149 and "d", thus " 149 - d".
now, if the rate travelling East is say "r", then the faster rate going North is " r + 3".
[tex]\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
East&d&r&4\\
North&149-d&r+3&3
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=4r\\
149-d=(r+3)3\\
----------\\
149-\boxed{4r}=3(r+3)
\end{cases}[/tex]
solve for "r".