I need help with applications involving quadratic equation word problems!!

Question

stuart bikes 36 mi, averaging a certain speed. The return trip is made at an average speed that is 3mph slower. The total time for the round trip is 7hr. Find stuarts average speed on each part of the trip.

Answer 1

 

Let X = rate for first half of the trip

Let X + 3 = rate for return trip

 

The total distance for the trip 72 miles (36 miles to and 36 miles on return trip)

 

The total time is 7 hours

 

Let T₁ = time for first leg of trip

Let T₂ = time for return trip

 

T₁ + T₂ = 7

 

FOR FIRST LEG OF THE TRIP:

   

   Rate • Time = Distance

 

   X • (T₁) = 36

 

FOR RETURN TRIP:

 

   

   Rate • Time = Distance

 

   (X -3) • T₂ = 36

 

We have the following  equations in our system:

 

    T₁ + T₂ = 7

    X•T₁ = 36

    (X - 3) •T₂ = 36

 

We can solve for T₂ in the first equation

 

    T₂ = 7 - T₁

 

We can substitute for T₂ in the other 3rd equation ... that gives us

 

    X•T₁ = 36

    (X - 3)(7 - T₁)= 36

 

Solving for X in the first equation,

 

    X = 36/T₁

 

Substituting that into equation 2, gives us a 1 variable equation

 

    ( 36/T₁ - 3 )(7 - T₁ ) = 36

   

Addition of fractions in first binomial

 

 

    (36/T₁ - 3T₁/T₁)(7 - T₁) = 36

   

    ((36 - 3T₁)/T₁)(7 - T₁) = 36

 

Clear the fraction by multiplying both sides of equation by T₁

 

     (36 - 3T₁)(7 - T₁) = 36T₁

 

Factor out 3 and divide both sides by 3

 

     3(12 - T₁)(7 - T₁) = 36T₁

 

     (12 - T₁)(7 - T₁) = 12T₁

 

Simplify, combine like terms

 

     84 - 12T₁ - 7T₁ + (T₁)² = 12T₁

 

     (T₁)²  -19T₁ + 84 = 12T₁

 

     (T₁)² - 31T₁ + 84 = 0

 

Factor and solve using the zero product property

 

      (T₁ - 28)(T₁ - 3) = 0

 

      T₁ - 28 = 0 ... T₁ = 28   ... we can discard this value because total time was 7 hours

   

      OR

 

      T₁ - 3 = 0 ... T₁ = 3  

 

So the time for leg 1 of the trip is 3 hours ... 4 hours for the return trip

 

Using an earlier equation, we can compute the rates ... x and x - 3

 

     x • T₁ = 36

 

     x(3) = 36

 

     x = 12

 

So the rate was 12 mi/hr for first part of trip,

and the rate was 12 - 3, or 9 mi/hr for return trip