Question 11852
Let x = distance from a to b
x+20 = distance from b to c

18 mph = rate from a to b
10 mph = rate from b to c


{{{Time = (Distance)/(Rate)}}}


Total time = {{{t[1] + t[2] = 8}}}
{{{t[1] = x/18}}}   
{{{t[2] = (x+20)/10}}}


{{{x/18 + (x+20)/10 = 8}}}


For simplicity sake, multiply both sides of the equation by the product of 18 times 10 or 180.  (The fact is that 90 is the LCD, and it would work even better, but 180 is easier to come up with and it works nearly as well.)


{{{180*(x/18) + 180*((x+20)/10 ) = 180*8}}}


Divide out the denominators:
{{{10x + 18(x+ 20) = 1440}}}
{{{ 10x + 18x + 360 = 1440}}}
{{{28x + 360 - 360 = 1440 - 360}}}
{{{28x = 1080}}}
{{{x = 1080/28 = 38 4/7 }}} miles


Time at 18 mph = {{{D/R= D * (1/R)=  (1080/28) *( 1/18) = 60/28 = 15/7 }}}hours.


Time at 10 mph = {{{ (1080/28 + 20)* (1/10) = ((1080+560)/ 28) *(1/10)= (1640/28) *(1/10) = 164/28 = 41/7}}} hours.


Strange answer, but it does check (with a LOT of work!)


{{{t[1] + t[2] = 8}}}
{{{(1080/28)*(1/18) + ((1080/28)+2)* (1/10) = 8}}} hours.
{{{ 1080/(28*18) + 1640/(28*10) = 8}}}
{{{15/7 + 41/7 = 56/7 = 8}}}

Believe it or not, this checks.


R^2 at SCC