Basic algebra manipulation

Problem: A baseball player travels from his home city, Jasonville, to Giambi City for a baseball game. He drives at m miles an hour. After the game, he flies home instead, traveling at p miles an hour. If the distance from jasonville to Giambi City is v miles, and it took him j hours longer to drive than to fly, what is j?

The answer is vm-vp. I set up a rate distance time chart for the first leg there and the second leg home. Then I made x+j=v/m (rate formula) and x=v/p Then I subsituted v/p for x and set j = to v/m-v/p. That was the answer I got. What did I do wrongly?

Re: Basic algebra manipulation

Time taken to fly = $\displaystyle pv$

Time taken to drive = $\displaystyle mv = pv + j$

Rearrange to get;

$\displaystyle j = mv - pv = v(m - p)$

Re: Basic algebra manipulation

I do not agree with the given "answer", because the units do not work out. vm - vp = (distance)(speed), which is not time.

you had it correct the first time ...

rate(time) = distance

m(t+j) = v

pt = v

m(t+j) = pt

mt + mj = pt

mj = pt - mt

j = pt/m - t

j = v/m - v/p

Re: Basic algebra manipulation

Code:

` |....@m mph.............v miles....................>v/m hours`

v/p hours<................v miles................@p mph.....|

v/m = v/p + j

j = v(p - m) / (pm)

Re: Basic algebra manipulation

It must be a typo in my book... All too common these days. Thank you for clarifying.