[+3y=14," (iv) "(5)/(x-1)+(1)/(y-2)=2],[-4y=23,(6)/(x-1)-(3)/(y-2)=1]

If (5)/(x-1)+(1)/(y-2)=2,(6)/(x-1)+(-3)/(y-2)=1 then x=……….

Find the S.D. between the lines : (i) (x)/(2) = (y)/(-3) = (z)/(1) and (x -2)/(3) = (y - 1)/(-5) = (z + 4)/(2) (ii) (x -1)/(2) = (y - 2)/(3) = (z - 3)/(2) and (x + 1)/(3) = (y - 1)/(2) = (z - 1)/(5) (iii) (x + 1)/(7) = (y + 1)/(-6) = (z + 1)/(1) and (x -3)/(1) = (y -5)/(-2) = (z - 7)/(1) (iv) (x - 3)/(3) = (y - 8)/(-1) = (z-3)/(1) and (x + 3)/(-3) = (y +7)/(2) = (z -6)/(4) .

(1)/(2(3x+4y))=(1)/(5(2x-3y))=(1)/(4),

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))

Solve : (1)/(2x)+(1)/(4y)-(1)/(3z)=(1)/(4),(1)/(x)=(1)/(3y),(1)/(x)-(1)/(5y)+(4)/(z)=2 (2)/(15)