Answer:
speed of the Jon visiting parents = 56 mph
speed of the Jon when returning from home = 56 - 14 = 42 mph
Step-by-step explanation:
given,
distance of Jon parent's house = 648 mile
avg speed when he was visiting his parent's house be 'x' mph
avg speed when he is returning from his parent's house be 'x-14' mph
total time taken = 27 hours
total distance = speed × time
648 = x × t₁
![t_1 = (648)/(x)](https://img.qammunity.org/2020/formulas/physics/high-school/z4cnf2sn9jrr3v8ohq3l72k5btwucc6602.png)
648 = ( x - 14 ) × t₂
![t_2 = (648)/(x-14)](https://img.qammunity.org/2020/formulas/physics/high-school/wg4ma45mmbrbahwkz4z1kce5egrgqs9bkm.png)
t = t₁ + t₂
![27 = (648)/(x) + (648)/(x-14)](https://img.qammunity.org/2020/formulas/physics/high-school/ytuzgzf6qckdsx86v6g7fmimhphcawicmy.png)
![1 = 24 ((1)/(x) + (1)/(x-14))](https://img.qammunity.org/2020/formulas/physics/high-school/cnzc5zrm1b4o0ar10lhm83iqa5jy4zvaf8.png)
x² - 62 x + 336 = 0
x² - 56 x - 6 x + 336 = 0
(x - 56 )(x - 6)=0
on solving
x = 56 ,6
hence, speed of the Jon visiting parents = 56 mph
speed of the Jon when returning from home = 56 - 14 = 42 mph