Question 958497: I am having a problem setting up an equation for this problem. I keep getting squared variables and I know that's wrong. The problem reads: John recently drove to visit his parents who live 420 miles away on his way there his average speed was 9 miles per hour faster than on his way home if John spent a total of 21 hours driving find the two rates.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! let R1 be his rate going to his parents home and
R2 be his rate going back to his home, also
T1 is the time he took going to his parents home and
T2 is the time he took going back to his home, we are given
T1 + T2 = 21 hours and
R1 = R2 + 9
we use rate times time = distance
(R2 +9) * T1 = 420 and
R2 * T2 = 420
therefore T1 = 420 / (R2+9) and T2 = 420/R2 and we know that
420/(R2+9) + 420/R2 = 21
note that the LCD is R2*(R2+9) and we get the following quadratic
21R2^2 -651R2 -3780 = 0
we use the quadratic formula to solve but first note that the discriminant is
square root(b^2 -4*a*c) = square root ((-651)^2 -(4*21*(-3780))) = 861, then
R2 = (-(-651) + 861) / (2*21)) = 36
R2 = (-(-651) - 861) / (2*21)) = -5
we want the positive value for R2, therefore
R1 = R2 +9 = 36 +9 = 45
John's rate to his parent's house is 45 mph and his rate back to his home is 36 mph
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