Question 203094
Let's make t = time (in hours) that the first car travels.



First, let's set up the equation for the first car to leave. Recall that the distance "d" equals the speed "r" multiplied by the time "t". So this means that 



{{{d=rt}}}



{{{d=60t}}} Plug in {{{r=60}}} (the speed of the first car)



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Now let's set up the second equation (for the second car). Since the first car travels "t" hours, and it has a 2 hour head start, this means that the second car is going to travel "t-2" hours (eg, the first car travels t=6 hours and the second travels t-2=6-2=4 hours). 



So this means that we start with {{{d=rt}}} and plug in {{{r=75}}} and replace "t" with "t-2" to get {{{d=75(t-2)}}}



Note: the "d"s of both equations are the same "d" since the cars will have gone the distance when they meet.



So we have the two equations {{{d=60t}}} and {{{d=75(t-2)}}}



{{{d=60t}}} Start with the first equation.



{{{75(t-2)=60t}}} Plug in {{{d=75(t-2)}}}



{{{75t-150=60t}}} Distribute.



{{{75t=60t+150}}} Add {{{150}}} to both sides.



{{{75t-60t=150}}} Subtract {{{60t}}} from both sides.



{{{15t=150}}} Combine like terms on the left side.



{{{t=(150)/(15)}}} Divide both sides by {{{15}}} to isolate {{{t}}}.



{{{t=10}}} Reduce.



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Answer:


So the solution is {{{t=10}}} which means that the first car traveled for 10 hours



Subtract 2 from this answer to get {{{10-2=8}}}. So this tells us that the second car traveled 8 hours.



So it takes 8 hours for the second car to catch up and meet the first car.