Question 5684
Your answer looks fine actually, well done. However, being a mathematician, I have my own tried and tested method, shown here.


As you say, speed = distance/time.


Let x=speed of the boat in still water.
Let t = time taken to travel upstream


<b>Upstream</b>:
speed = x-9
distance = 12
time = t


<b>Downstream</b>:
speed = x+9
distance = 12
time = 3-t


so, upstream, we get x-9 = 12/t and downstream, we get x+9 = 12/(3-t).


so, re-arrange for t, since we are not interested in that, and we wish to get rid of it:


t = 12/(x-9) and 3-t = 12/(x+9) --> t = 3 - 12/(x+9).


Equating these 2 gives us 12/(x-9) = 3 - 12/(x+9), which is 12/(x-9) + 12/(x+9) = 3 --> your equation :-).


So, multiply every term by (x+9)(x-9) -->


{{{(12(x+9)(x-9))/(x-9) + (12(x+9)(x-9))/(x+9) = 3(x+9)(x-9)}}}. we can cancel some terms on the left hand side:


{{{12(x+9) + 12(x-9) = 3(x+9)(x-9)}}} --> divide everything by 3
{{{4(x+9) + 4(x-9) = (x+9)(x-9)}}}
{{{4x+36 + 4x-36 = (x^2-81)}}}
{{{8x = x^2 - 81}}}


--> {{{x^2 - 8x - 81 = 0}}}


So, factorise (not sure it does easily) or plug into the quadratic formula. I shall leave this for you to do. Good luck. Oh yes, check my working because i am not happy that the equation doesn't factorise...perhaps the answer is a nive fraction...work it out :-)



jon.