Question 202359
Let r = speed of slower car


First, let's set up the equation for the slower car


{{{d=rt}}} Start with the distance-rate-time equation.



{{{352=rt}}} Plug in {{{d=352}}} (since the slower car goes 352 miles)



{{{352/r=t}}} Divide both sides by "r" to isolate "t".



{{{t=352/r}}} Rearrange the equation



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Now let's set up the equation for the faster car



{{{d=rt}}} Start with the distance-rate-time equation.



{{{456=rt}}} Plug in {{{d=456}}} (since the faster car went 456 miles)



{{{456=(r+13)*t}}} Replace "r" with "r+13" (since the faster car goes 13 mph faster



{{{456=(r+13)*(352/r)}}} Plug in {{{t=352/r}}} (the previous equation we isolated)



{{{456=(352/r)(r+13)}}} Rearrange the terms.



{{{456r=352(r+13)}}} Multiply both sides by "r".



{{{456r=352r+4576}}} Distribute.



{{{456r-352r=4576}}} Subtract {{{352r}}} from both sides.



{{{104r=4576}}} Combine like terms on the left side.



{{{r=(4576)/(104)}}} Divide both sides by {{{104}}} to isolate {{{r}}}.



{{{r=44}}} Reduce.



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Answer:


So the solution is {{{r=44}}} 



This means that the slower car is going 44 mph and the faster car is going 44+13=57 mph