[2x-(3)/(y)=9],[3x+(7)/(y)=2,y!=0]

Solve the following system of equations: 2x-(3)/(y)=9,quad 3x+(7)/(y)=2,y!=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 : (x, y ne 0) (10)/(2x+y)+(3)/(2x-y)=3, (15)/(2x+y)+(9)/(2x-y)=6

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

(x)/(2) - ( y )/(9) = 6, ( x) /( 7) + ( y ) /(3) = 5 .

3x-4y=-7,5x-2y=0

Solve the system: x + y - 2z = 0, 2x - 3y + z = 0, 3x - 7y + 10z = 0, 6x - 9y + 10z = 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))

Equation of the circle cutting orthogonal these circles x^(2)+y^(2)-2x-3y-7=0x^(2)+y^(2)+5x-5y+9=0 and x^(2)+y^(2)+7x-9y+29=0 is: