Algebra Word Problem - See description

Question

Greg drove at a constant speed in a rainstorm for 245 miles. He took a break and the rain stopped. He then drove 104 miles at a speed that was 3 miles per hour faster than his previous speed. If he drove for 7 hours find the cars speed for each part of the trip.

Answer 1

Distance = (rate)(time), so time= (distance)/(rate)

 

Time for first part of trip + time for second part = 7

 

Let x = rate for first part of trip

So, x + 3 = rate for second part of trip

 

245/x + 104/(x + 3) = 7

Multiply through by the LCD: x(x + 3) to get

 

245(x + 3) + 104x = 7x(x + 3)

245x + 735 + 104x = 7x² + 21x

349x + 735 = 7x² + 21x

 

7x² - 328x - 735 = 0

 

 

By the quadratic formula  

 

x = (328 ± sqrt ((-328)² -  4(7)(-735)))/(2(7))

 

   = (328 ± sqrt(128164))/14

 

   = (328 ± 358)/14 

 

   = 686/14, -30/14

 

   = 49, -2.14

 

Since x can't be negative, x = 49 and x + 3 = 52

 

ANSWER:  The speed for the first part of the trip was 49 mph and                  the speed for the rest of the trip was 52 mph. 

Answer 2

distance=rate*time

d=rt

245=rt

104=(r+3)(7-t)

from 245=rt we get 245/t=r

substitute this into the second equation

104=[(245/t)+3](7-t)

multiply both sides by t

104t=t[(245/t)+3](7-t)

104t=(245+3t)(7-t)

104t=1715+21t-245t-3t²

104t=1715-224t-3t²

104t-1715+224t+3t²=0

3t²+328t-1715=0

factor this trinomial using the following method...

3*1715=3*5*343=3*5*7*49=3*5*7*7*7

3*5*7*7*7=(3*5)*(7*7*7) and (7*7*7)-(3*5)=343-15=328, the coefficient of the middle term

(3t     )(t     )=0 so far...

the 3 has to multiply the 5 to get 15

(3t     )(t     5) so far...

the 343 has to multiply the 5 also to get 1715

(3t     343)(t     5)=0 so far...

the 328 is positive

(3t+343)(t-5)=0

t-5=0

t=5 hours

go back to the first equation and substitute

245=5r

r=245/5

r=49 mph for 245 miles

r+3=49+3=52 mph for the next 104 miles

check: 245=49*5, 245=245

          104=52*(7-5)=52*2=104