f'(6)/(x+y)-(7)/(x-y)=3,(1)/(2(x+y))-(1)/(3(x-y))=0

Solve for x and y : (1)/( 7x ) + ( 1) /( 6y) = 3, (1)/( 2x) - (1)/( 3y ) = 5 (x ne 0, y ne 0)

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

Solve by cross-multiplication method. (i) 8x-3y=12,5x=2y +7 (ii) 6x+ 7y -11 =0, 5x+ 2y =13 (iii) (2)/(x)+(3)/(y) =5, (3)/(x) -(1)/(y) +9=0

Solve for x and y (2)/(x)+(2)/(3y)=(1)/(6),(3)/(x)+(2)/(y)=0

( 1 ) /(x) + (1 ) /( y) = 7 , ( 2)/(x) + ( 3) /( y) = 17 (x ne 0, y ne 0 ) .

Simplify : (i) (5x - 9y) - (-7x + y) (ii) (x^(2) -x) -(1)/(2)(x - 3 + 3x^(2)) (iii) [7 - 2x + 5y - (x -y)]-(5x + 3y -7) (iv) ((1)/(3)y^(2) - (4)/(7)y + 5) - ((2)/(7)y - (2)/(3)y^(2) + 2) - ((1)/(7)y - 3 + 2y^(2))

If the points (x_(1),y_(1)),(x_(2),y_(2)), and (x_(3),y_(3)) are collinear show that (y_(2)-y_(3))/(x_(2)x_(3))+(y_(3)-y_(1))/(x_(3)x_(1))+(y_(1)-y_(2))/(x_(1)x_(2))=0

A triangle has vertices A_(i) (x_(i),y_(i)) for i= 1,2,3,. If the orthocenter of triangle is (0,0) then prove that |{:(x_(2)-x_(3),,y_(2)-y_(3),,y_(1)(y_(2)-y_(3))+x_(1)(x_(2)-x_(3))),(x_(3)-x_(1) ,,y_(3)-y_(1),,y_(2)(y_(3)-y_(1))+x_(2)(x_(3)-x_(1))),( x_(1)-x_(2),,y_(1)-y_(2),,y_(3)(y_(1)-y_(2))+x_(3)(x_(1)-x_(2))):}|=0