Let v be the rate that Ezekiel paddles in still water and s be the rate of the current.
When Ezekiel paddles upstream, the rate of the movement is v-s, and when he paddles downstream the rate of the movement is v+s.
Since he paddles upstream 12 miles in 4 hours, then:
[tex]v-s=\frac{12}{4}[/tex]
Since he paddles downstream the same 12 miles in 1.5 hours, then:
[tex]v+s=\frac{12}{1.5}[/tex]
We obtained a 2x2 system of equations. Simplify the fractions:
[tex]\begin{gathered} v-s=3 \\ v+s=8 \end{gathered}[/tex]
Solve the system using the elimination method. To do so, add both equations:
[tex]\begin{gathered} \Rightarrow v-s+v+s=3+8 \\ \Rightarrow v+v+s-s=11 \\ \Rightarrow2v+0=11 \\ \Rightarrow2v=11 \\ \Rightarrow v=\frac{11}{2} \\ \\ \therefore v=5.5 \end{gathered}[/tex]
Replace v=5.5 in any of the equations and solve for s:
[tex]\begin{gathered} v+s=8 \\ \Rightarrow5.5+s=8 \\ \Rightarrow s=8-5.5 \\ \\ \therefore s=2.5 \end{gathered}[/tex]
Therefore, Ezekiel paddles at 5.5 mi/h in still water. The rate of the current is 2.5 mi/h.
