The first part of the journey took 4/3 hours (80 minutes)
The last part of the journey too 2/3 hours (40 minutes)
Here, we want to set-up equations to solve
We start by filling the table
Line 1
The rate for the first part of the race is 90 mph
The time for the first part of the race is F
Line 2
The time for the second part of the race is L
The distance (product of the rate and time) is 60L
So, adding these up give us the following equations to be added and solved below;
[tex]\begin{gathered} f\text{ + l = 2} \\ 90\text{ f + 60l = 160} \end{gathered}[/tex]So, we proceed to solve these equations simultaneously
[tex]\begin{gathered} \text{from equation i, f = 2-l} \\ \text{substitute this into i}i \\ 90(2-l)\text{ + 60l = 160} \\ 180-90l\text{ + 60l = 160} \\ 30l\text{ = 180-160} \\ 30l\text{ = 20} \\ l\text{ = }\frac{20}{30} \\ \\ l\text{ = }\frac{2}{3}\text{ hours} \end{gathered}[/tex]To get f, we simply substitute l into the first part of the equations;
[tex]\begin{gathered} \text{from; f = 2-l} \\ \\ f\text{ = 2-}\frac{2}{3} \\ f\text{ = }\frac{4}{3}\text{ hours} \end{gathered}[/tex]Since an hour is 60 minutes;
[tex]\begin{gathered} l\text{ = }\frac{2}{3}\times\text{ 60 = 40 minutes} \\ \\ f\text{ = }\frac{4}{3}\times60\text{ = 80 minutes} \end{gathered}[/tex]