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at 9:00 a.m Kruti started from home going 60 miles per hour. At 11:00 a.m Courtney started after her on the same road going 70 miles per hour. At what time did Courtney catch up to Kruti?

Hi, I will be glad to help you with your problem

The best way to solve distance-rate-time problems involving systems of equations is to create a drawing, and then a table to help you form the equations.

I'm just going to help with the table.

         D.          R.          T.  

Because Courtney is going to catch up with Kruti, they are going the same distance.  We don't know what that distance is, so we use a variable, d. In the table both Kruti and Courtney will have a d in the Distance column (D)

         D.          R.          T.  
Kruti          d
Courtney      d

Kruti is going 60 miles an hour whereas Courtney is going 70.  These values are their rates and go in the R column

         D.          R.          T.  
Kruti          d          60
Courtney      d          70

Time is probably the trickiest part in these problems.
Again, we don't know the exact time they meet, that is what we want to find out, we again use a variable, t.  Because Kruti has a 2 hour head start, instead of t we use t + 2 for Kruti and t for Courtney.

         D.          R.          T.  
Kruti          d          60          t + 2
Courtney      d          70          t

Now that the table is complete, equations can be formed. Distance = rate x time

Kruti's equation
d = 60(t + 2)

Courtney's equation
d = 70t

Since their distances are the same (they meet) the equations can be set equal to eachother

60(t + 2) = 70t

Solve for t to determine how many hours past 11, the time Courtney starts) that they meet.

I hope this helps.


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Anne Losch


Solving equations, graphing, evaluatin equations, factoring, functions, systems of equations, rational equations, exponent, complex numbers, word problems, logarithms, polynomials, and all topics in an Algebra 1, Algebra 2, or College Algebra class


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