1. ## distance time problem

Q: After going a certain distance in an automoble, a driver found that if he had gone 3 miles an hour faster, he would have traveled the distance in 1 hour less time. and that if he had gone 5 miles an hour faster, he would have gone the distance in 1 hour and 30 minutes less.
What was the distance?

my attempt:
let x = distance and y = time

$\frac{x}{60}+\frac{3}{60} = \frac{x}{y -60}$

$\frac{x}{60}+\frac{5}{60} = \frac{x}{y -90}$

$(x+3)(y-60)=60x$

$(x+5)(y-90)=60x$

$-150x+5y=450-xy$

$-120x+3y=180-xy$

Im stuck at this point i dont think i got the right sim equations

2. ## Re: distance time problem

Q: After going a certain distance in an automoble, a driver found that if he had gone 3 miles an hour faster, he would have traveled the distance in 1 hour less time. and that if he had gone 5 miles an hour faster, he would have gone the distance in 1 hour and 30 minutes less.
What was the distance?
$v = \text{ original speed in mph}$

$t = \text{ original time in hrs}$

$vt = d$

$(v+3)(t-1) = d$

$(v+5)(t-1.5) = d$

3 equations, 3 unknowns ... see what you can do.

3. ## Re: distance time problem

Originally Posted by skeeter
$v = \text{ original speed in mph}$

$t = \text{ original time in hrs}$

$vt = d$

$(v+3)(t-1) = d$

$(v+5)(t-1.5) = d$

3 equations, 3 unknowns ... see what you can do.
thank you very much
im sorry im not allowed to add to ur reputation