solving simultaneous: tom, julie leave towns A, B to meet...

[yumm]

question1 -tom leaves town A and travels at a constant speend of u km/h towards town B. at the same time julie leaves town B and travels towards town A at a constant speend of vkm/h. Town B is dkm from town A.

what is their distance from ton A when they meet ?

attempt
sum of their distance is d km

q2-a man walks 70km. he walks x km at 8 km/h and y km at 10km/h
a)if the man walks at 10km/h for the time he was walking at 8km/h and at 8km/h for the time he was walking at 10km/h, he walks 72 km. find x and y.

attempt
8x+10y=70
10x+8y=72 ???

 

[Denis]

question1 -tom leaves town A and travels at a constant speend of u km/h towards town B. at the same time julie leaves town B and travels towards town A at a constant speend of vkm/h. Town B is d km from town A.
what is their distance from town A when they meet ?
attempt
sum of their distance is d km

Yes...but that's no attempt yumm: you're GIVEN that!

Try again. Hint: let t = time ; so they both travelled for t hours before meeting.
And remember that speed = distance / time

 

[Denis]

q2-a man walks 70km. he walks x km at 8 km/h and y km at 10km/h
a)if the man walks at 10km/h for the time he was walking at 8km/h and at 8km/h for the time he was walking at 10km/h, he walks 72 km. find x and y.
attempt
8x+10y=70
10x+8y=72 ???

NO. You are multiplying speed by distance to get distance :shock:

speed = distance / time ; so distance = speed * time

What grade are you in, yumm?

 

[yumm]

nevermind about the questions, soroban helped me so thanks anyway

 

[Deleted member 4993]

nevermind about the questions, soroban helped me so thanks anyway

You mean Soroban spoon-fed you. That's okay for one problem - but did you learn from it?

You have posted similar problems (2 equations and 2 unknowns) before:

viewtopic.php?f=16&t=29505&p=112945#p112945

those were solved for you.

You did not show any sign of learning from those solved examples.