J3 RevEx2f Q27

Meihua - t = (x + 2) hr; d = 50 km; m = (a - 5) km/h [let m = Meihua's speed and a = Ailin's speed]
Ailin - t = x hr; d = 40 km; a km/h [a = Ailin's speed]

speed = distance / time

your working as I remembered;
50/(x + 2) + 5 = 40/x
your equation is correct but was based on speed where your variable x represents time. However the question wanted you to calculate the speed. So you should use the following equation that is based on time with variable a that represents Ailin's speed.

50/(a - 5) - 2 = 40/a

you will reach this quadratic equation

-2a(square) + 20a + 200 = 0

you need to use the quadratic formula to solve the above equation

a = {-20 +- squareroot[20square - 4(-2)(200)]} / -4

the squareroot is 44.72 - you need calculator or squareroot table to get this.

finally you will arrive with the following answers

a = -6.18 or 16.18

-6.18 is not acceptable for speed as it was negative valued

So Ailin's speed is 16.18 km/h
Meihua's speed = Ailin's speed - 5 therefore 11.18 km/h

It's easy isn't it... The other 3 questions you try first... If you still have problem with it, then please let me know...

By the way, have a meaningful rabbit year and an unforgetable CNY celebrations that is spiced with quadratic math.

Take care and take rest...