<PRE><B>Question 1123073</B>
rate of car is x and rate of coach is x-30
:
we use rate * time = distance
:
let t1 be the time for the car to travel the 700 miles and t2 be the time it takes the coach to travel 700 miles
:
we know that
:
t1+t2 = 20 hours, then
: 
t2 = 20-t1
:
we have two equations in two unknowns
:
1) x * t1 = 700
:
2) (x-30) * (20-t1) = 700
:
equation 2 becomes the following after multiplying the left side of =
:
20x-600-xt1+30t1 = 700
:
simplifying we have
:
30t1-xt1 = 1300-20x, and
:
t1 = (1300-20x)/(30-x)
:
substitute for t1 in equation 1
:
x * (1300-20x)/(30-x) = 700
:
1300x-20x^2 = 21000-700x
:
20x^2 -2000x +21000 = 0
:
******************************
x^2 -100x +1050 = 0
******************************
:
use the quadratic formula to solve for x
:
x = (-(-100) +square root((-100)^2 -4*1*1050))/(2*1) = 88.0789 Km/hr
:
x = (-(-100) -square root((-100)^2 -4*1*1050))/(2*1) = 11.9211 Km/hr
:
remember that x is the rate for the car and x-30 is the rate for the coach
:
therefore 11.9211 Km/hr can not be the rate for the car and 
:
the rate for the car is 88.0789 Km/hr
:
88.0789 * t1 = 700
:
*********************************
t1 = 700/88.0789 = 7.9474
:
return trip took 7.9474 hours
*********************************
:
</PRE>