SOLUTION: A motor boat travels 300km downstream and returns 300km upstream in a total of 12 hours. The speed of the current is 6km/h. Write the equation that could be used to determine the s

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A motor boat travels 300km downstream and returns 300km upstream in a total of 12 hours. The speed of the current is 6km/h. Write the equation that could be used to determine the s      Log On


   

Question 132748: A motor boat travels 300km downstream and returns 300km upstream in a total of 12 hours. The speed of the current is 6km/h. Write the equation that could be used to determine the speed of the motorboat in still water? I think it is (300/x )+(300/x+6)=12 Can someone please confirm this for me ????
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
300/(X-6)+300/(X+6)=12 THE FIRST FRACTION IS THE SPEED UP STREAM AGAINST THE TIDE & THE SECOND FRACTION IS THE SPEED WITH THE CURRENT.
[300(X+6)+300(X-6)]/(X+6)(X-6)=12
(300X+1800+300X-1800)/(X^2-36)=12
600X/(X^2-36)=12 NOW CROSS MULTIPLY.
12(X^2-36)=600X
12X^2-432-600X=0
12X^2-600X-432=0
12(X^2-50X-36)=0
USING THE QUADRATIC EQUATUION:x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ WE GET:
X=(50+-SQRT[-50^2-4*1*-36])/2*1
X=(5)=-SQRT[2500+144])/2
X=(50+-SQRT2644)/2
X=(50+-51.42)/2
X=(50+51.42)/2
X=101.42/2
X=50.71 ANSWER FOR THE SPEED OF THE MOTOR BOAT.
PROOF:
300/(50.71+6)+300/(50.71-6)=12
300/56.71+300/44.71=12
5.29+6.71=12
12=12