Question

Two cars left the city for a suburb, 120 km away, at the same time. The speed of one of the cars was 20 km/hour greater than the speed of the other, and that is why it arrived at the suburb 1 hour before the other car. What is the speed of both cars.?

Answer 1

Slower car: `40"km/h"`
Faster car: `60"km/h"`

Explanation:

Let `x=`hours it took the slower car to reach the suburb
The time it took the faster car to reach the suburb is `x-1`

Since distance`div`time`=`speed,
The speed of the slower car is `120"km"div(x)"h"=120/x"km/h"`
The speed of the faster car is `120"km"div(x+1)"h"=120/(x-1)"km/h"`

Since the faster car travelled `20"km/h"` faster than the slower car,
`120/(x-1)=120/x+20`

Now multiply both sides by `(x-1)` and then `x` to start solving.
`(120(x-1))/(x-1)=(120(x-1))/x+20(x-1)`
`(120x(x-1))/(x-1)=(120x(x-1))/x+20x(x-1)`

Simplify, distribute, factor, disregard, etc. Essentially, solve.
`(120xcolor(red)cancelcolor(black)(x-1))/color(red)cancelcolor(black)(x-1)=(120color(red)cancelcolor(black)x(x-1))/color(red)cancelcolor(black)x+20x(x-1)`
`color(red)cancelcolor(black)(120x)=color(red)cancelcolor(black)(120x)-120+20x^2-20x`
`0=20x^2-20x-120`
`0=x^2-x-6`
`0=x^2+2x-3x-6`
`0=x(x+2)-3(x+2)`
`0=(x+2)(x-3)`
`x+2=0orx-3=0`
`x="-"2orx=3`

Since `x` is the number of hours for the slower car to reach the suburb, a negative solution can be disregarded as extraneous as negative time does not (yet?) exist. `x=3`

We know that the speed of the slower car is `120/x"km/h"` and that of the faster car is `120/(x-1)"km/h"` so plugging in `x`, we get:
Slower car: `120/((3))=120/3=40"km/h"`
Faster car: `120/((3)-1)=120/2=60"km/h"`