Fairly complex distance time questions

Question

Hi I am trying to solve this distance time speed question with many variables but I am not sure how to work backwards.

Betty and Carl live in the same apartment building work at the same office, and set off for work in the morning at the same times. They must each travel 65km, and they arrive at work at the same time. Betty travels by car at 45km/h, parks, the walks at 5km/h the rest of the way to work. Carl takes the bus, which travels at 40km/h, to the same car park as Betty uses, and then rides his bike the rest of the way at 6km/h. If they both leave home at 6am, what time do they arrive at work.

Answer 1

45 * tCar + 5 * tWalk = 65 <--- first official equation

40 * tBus + 6* tBike = 65 <--- 2nd official equation

tCar + tWalk = tBus + tBike <--- 3rd official equation

45*tCar = 40 * tBus <--- 4th official equation : distance from apartment to parking lot

Solves 4th equation for tCar

tCar = (8/9) * tBus

Substitutes that into equations #1 and equations #3

45 * (8/9) * tBus + 5 * tWalk = 65

40 * tBus + 6 * tBike = 65

(8/9) tBus + tWalk = tBus + tBike

writes the system in standard form:

40 * tBus + 5 * tWalk = 65 <--- equation ALPHA

40 * tBus + 6 * tBike = 65 <--- equation BETA

(1/9) * tBus + tBike - tWalk = 0 <--- equation DELTA

next subtracts equation BETA minus equation ALPHA to eliminates tBus from first two of these:

6 * tBike - 5 * tWalk = 0 ----> tBike = (5/6)*tWalk <--- equation OMEGA

Then to eliminate tBus from BETA and DELTA does -360* DELTA + BETA:

-354*tBike + 360* tWalk = 65 <--- equation EPSILON

Substitutes OMEGA into EPSILON

-354( (5/6) * tWalk ) + 360 * tWalk = 65

-1770 * tWalk + 2160* tWalk = 390 <--- multiplies both sides by 6

390 * tWalk = 390

tWalk = 1

per OMEGA, tBike = 5/6

Per the original first equation:

45 * tCar + 5 * tWalk = 65

45 * tCar + 5 * 1 = 65

45 * tCar + 5 = 65

45 * tCar = 60

tCar = 60/45 = 4/3

Per the original 2nd equation:

40 * tBus + 6 * (5/6) = 65

40 * Tbus + 5 = 65

40 * tBus = 60

tBus = 60/40 = 3/2

CHECK:

(3/2)*40 = 60 = (4/3)*45 <--- apartment to parking lot

1*5 = 5 = (5/6)*6 <--- walking or biking to work from parking lot

3/2 + 5/6 = 9/6 + 5/6 = 14/6 = 7/3 <-- tBus + tBike

4/3 + 1 = 4/3 + 3/3 = 7/3 <-- tCar + tWalk

Their total travel time is 7/3 hours = 2 hours and 1/3 hour = 2 hours and 20 minutes

The arrive at 8:20 am

Thank you for this challenging problem