Question 32318
Let the rate for Moe be x
Let the rate for Bob be 1.4+x
Bob's partial time = 2 hours and 20 minuts (1/3 hours) or 7/3 hours:
Equation:
{{{30/(1.4+x)-30/x=7/3}}}--->Multiply the whole equation by 3 to remove the fraction:
{{{90/(1.4+x)-90/x=7}}}
90[(1.4+x)-(x)]=7[(1.4+x)(x)]}}}
{{{126=7(1.4x+x^2)}}}
{{{7x^2+9.8x-126=0}}}
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
a=7, b=9.8, c=-126
*[tex x=\frac{-9.8\frac{+}{-}sqrt(9.8^2-4*7*-126)}{2*7}]
SImplfy:
x=3.6
3.6+1.4=5
Total time for each: -->30/(5)=6 and 30/3.6=8.33
<font color = red>Hence, Bob's rate is 5kmph with 6 hours of total, and Moe's rate is 3.6kmph with 8.33 hours of total.
</font>Paul.