Question 30904
This can be solved algebrically. You have two equations.
1- {{{2X=Y}}} where x is the speed of the first car, and y is the speed of the second (faster) car.
2- {{{3x+3y=96}}} 96 is their distance away from each other, and the 3's stand for the hours. car X has traveled for three hours, and so has car Y.
 Next work out the equation. You can use substitution.
{{{ 3x+3(2Y)=96}}} distribute
then
{{{3x+6x=96}}} add like terms
then
{{{9x=96}}} divide both sides by 9 {{{9x/9=96/9}}}
{{{x=10+2/3}}} so car x (slow car) is traveling at 10 and 2/3 miles an hour.
Now, plug the X value into equation 1 {{{2X=Y}}}
The fast car is travelling at 21 and 1/3 miles an hour.
{{{Y=10+2/3}}}
 You can also check your answer.  Plug in both numbers into equation 2. You get
{{{3(10+2/3)* (21+1/3)=96)}}} if this equation is true, then your answers are correct.
(it  is true)
HOPE THIS HELPS!