`3x-2y=7` and `(2x)/(3y)=2`

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

The equation of straight line belonging to both the families of lines (x-y+1)+lambda_1(2x-y-2)=0 and (5x+3y-2)+lambda_2(3x-y-4)=0 where lambda_1, lambda_2 are arbitrary numbers is (A) 5x -2y -7=0 (B) 2x+ 5y - 7= 0 (C) 5x + 2y -7 =0 (D) 2x- 5y- 7= 0

Find the solution of x + y = 7 and 2x - 3y = 11 .

Show that the following set of curves intersect orthogonally: y=x^(3) and 6y=7-x^(2)x^(3)-3xy^(2)=-2 and 3x^(2)y-y=2x^(2)+4y^(2)=8 and x^(2)-2y^(2)=4

x+2y+z=7 x+3z=11 2x-3y=1

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 4x - 3y = 7xy and 3x + 2y = 18xy , then (x,y) =