Answer:
Step-by-step explanation:
Let v be her average speed from her parents house
Let t₁ be the time traveled to parents house
Let t₂ be the time traveled from parents house
t₁ + t₂ = 19
t₁ = 19 - t₂
as the distance does not change, we can equate two velocity•time statements
vt₂ = (v + 25)t₁
vt₂ = (v + 25)(19 - t₂)
vt₂ = 19v - vt₂ + 475 - 25t₂
2vt₂ = 19v + 475 - 25t₂
vt₂ = 570 and t₂ = 570/v
2(570) = 19v + 475 - 25(570/v)
1140 = 19v + 475 - 14,250/v
665 = 19v - 14,250/v
665v = 19v² - 14250
0 = 19v² - 665v - 14250
0 = v² - 35v - 750
quadratic formula
v = (35 ±√(35² - 4(1)(-750))) / 2
v = (35 ± 65) / 2
v = - 15 mph which we ignore as she did not spend hours driving backwards
or
v = 50 mph is average speed from parents house
v + 25 = 75 mph is average speed to parents house.
the trip to took 570/75 = 7.6 hrs
the trip from took 570/50 = 11.4 hrs. total of 19 hrs.
