document.write( "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 ???? \n" ); document.write( "
Algebra.Com's Answer #97055 by checkley71(8403)\"\" \"About 
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.
\n" ); document.write( "[300(X+6)+300(X-6)]/(X+6)(X-6)=12
\n" ); document.write( "(300X+1800+300X-1800)/(X^2-36)=12
\n" ); document.write( "600X/(X^2-36)=12 NOW CROSS MULTIPLY.
\n" ); document.write( "12(X^2-36)=600X
\n" ); document.write( "12X^2-432-600X=0
\n" ); document.write( "12X^2-600X-432=0
\n" ); document.write( "12(X^2-50X-36)=0
\n" ); document.write( "USING THE QUADRATIC EQUATUION:\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\" WE GET:
\n" ); document.write( "X=(50+-SQRT[-50^2-4*1*-36])/2*1
\n" ); document.write( "X=(5)=-SQRT[2500+144])/2
\n" ); document.write( "X=(50+-SQRT2644)/2
\n" ); document.write( "X=(50+-51.42)/2
\n" ); document.write( "X=(50+51.42)/2
\n" ); document.write( "X=101.42/2
\n" ); document.write( "X=50.71 ANSWER FOR THE SPEED OF THE MOTOR BOAT.
\n" ); document.write( "PROOF:
\n" ); document.write( "300/(50.71+6)+300/(50.71-6)=12
\n" ); document.write( "300/56.71+300/44.71=12
\n" ); document.write( "5.29+6.71=12
\n" ); document.write( "12=12\r
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