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[" Solve: "(1)/(2(2x+3y))+(12)/(7(3x-2y)...

[" Solve: "(1)/(2(2x+3y))+(12)/(7(3x-2y))=(1)/(2)(7)/(2x+3y)+(4)/(3x-2y)=2," where "2x+3y!=0" and "3x-2y!=0],[" In trapezium "ABCD,AB|DC" and "DC=2AB" .EF drawn parallel to AB cuts "AD" in "F" and "BC" in Esuch that "],[(BE)/(EC)=(3)/(4)" .Diagonal DB intersects EF at "G" .Prove that "7" FE "=10AB" ."]

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In a trapezium ABCD, AB||DC and DC=2AB. EF drawn parallel to AB cuts AD in F and BC in E such that (BE)/(BC)=3/4 .Diagonals DB intersects EF at G. Prove that 7EF=10AB.

In trapezium ABCD. AB||DC and DC = 2AB. A line segment EF drawn parallel to AB cuts AD in F and BC in E such that (BE)/(EC)=3/4 . Diagonal DB intersects EF at G. prove that &EF= 10AB.

In trapezium ABCD. AB||DC and DC = 2AB. A line segment EF drawn parallel to AB cuts AD in F and BC in E such that (BE)/(EC)=3/4 . Diagonal DB intersects EF at G. prove that &EF= 10AB.

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In trapezium ABCD, AB||DC and CD = 2AB, EF drawn parallel to AB cuts AO in F and BC in F such that (BE)/(EC)=(3)/(4) , diagonal DB intersects EF at Q. The find the relation between FE and AB?

Sol v e :1/(2(2x+3y))+(12)/(7(3x-2y))=1/2 7/(2x+3y)+4/(3x-2y)=2 where 2x+3y!=0a n d3x-2y!=0.

In a trapezium ABCD, AB || AD and DC = 2 AB . EF || AB , cuts AD in F and BC in E such that (BE)/(EC) = 3/4 . Diagonal DB intersects EF at G. Prove that, 7EF = 10AB .

Of the trapezium ABCD, AB||DC and DC = 2AB. If the line segment EF drawn parallel to AB meets AD and BC at the points Fand E respectively so that (BE)/(EC)=3/4 and the diagonal DB meets EF at the points G, prove that 7FE=11AB

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