



 
Jalisa, I can see two ways to approach this problem. Both depend on the fact that distance = time × rate. The first method is algebraic. Let d be the distance the slow car travels until they meet and t the time they travel until they meet. Thus for the slow car
The fast car has to travel an additional 15 miles so for the fast car
You now have two equations in two unknowns which you can solve for t. The second method uses the difference in their rates. The fast car travels at 10 miles per hour faster than the slow car and has to make up the 15 mile head start that the slow car has. At 10 miles per hour how long does it take to travel 15 miles? I hope this helps,  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 