Avg speed = miles/hours = 315/t
Let t1 be original time to drive there. Let t2 be time to drive back.
t1 + t2 = 9 ------> t2 = 9 - t1
315/t1 = 315/t2 + 24
By substitution you have:
[tex]\frac{315}{t_1} = \frac{315}{9-t_1} + 24[/tex]
Eliminate fractions by multiplying by common denominator. (t)(9-t)
[tex]315(9-t_1) = 315 t_1 + 24 t_1 (9-t_1) \\ \\ 24t_1 ^2 - 846t_1+2835 = 0 \\ \\ 8t_1^2 - 282t_1 + 945 = 0[/tex]
Solve using quadratic formula (Note t1 < 9)
[tex]t_1 = \frac{15}{4} \\ \\ t_2 = 9-t_1 = \frac{36}{4} -\frac{15}{4} = \frac{21}{4}[/tex]
Finally, determine the two speeds.
[tex]\frac{315}{(15/4)} = 315*\frac{4}{15} = 84 \\ \\ \frac{315}{(21/4)} = 315*\frac{4}{21} = 60[/tex]
Final Answer:
Jamie traveled 84 mph on the way there and 60 mph on the way back.
