Question

I need explanation with this problem been struggling and it's practice for my exam coming up this semester

Answer 1

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;

`\begin{gathered} f\text{ + l = 2} \\ 90\text{ f + 60l = 160} \end{gathered}`

So, we proceed to solve these equations simultaneously

`\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{ = }(20)/(30) \\ \\ l\text{ = }(2)/(3)\text{ hours} \end{gathered}`

To get f, we simply substitute l into the first part of the equations;

`\begin{gathered} \text{from; f = 2-l} \\ \\ f\text{ = 2-}(2)/(3) \\ f\text{ = }(4)/(3)\text{ hours} \end{gathered}`

Since an hour is 60 minutes;

`\begin{gathered} l\text{ = }(2)/(3)*\text{ 60 = 40 minutes} \\ \\ f\text{ = }(4)/(3)*60\text{ = 80 minutes} \end{gathered}`