(b)*x-y=8,quad 1x-3y=1

3x+2y=8 2x-3y=1

Solve for x and y: x+y=8" and "2x-3y=1 .

" (2) "3x+y=-1,2x-3y=-8

If (-2,\ 1) is the centroid of the triangle having its vertices at (x ,\ 0),\ (5,\ -2),\ (-8,\ y), then x ,\ y satisfy the relation 3x+8y=0 (b) 3x-8y=0 (c) 8x+3y=0 (d) 8x=3y

If (-2,1) is the centroid of the triangle having its vertices at (x,0),(5,-2),(-8,y), then x,y satisfy the relation 3x+8y=0 (b) 3x-8y=0( c) 8x+3y=0 (d) 8x=3y

Solve the following system of equations: (1)/(3x+y)+(1)/(3x-y)=(3)/(4),quad (1)/(2(3x+y))-(1)/(2(3x-y))=-(1)/(8)

The equation of the normal to the curve 3x^2-y^2=8 which is parallel to x+3y=8 is x-3y=8 (b) x-3y+8=0 (c) x+3y+-8=0 (d) x+3y=0

The line common to the given families of line a(x +y+1)+b(2x-3y-8)=0 and lambda(x-y-1)+mu (2x-y-1)=0 is : x+y+1=0 x+y-1=0 x+y-1=0 None of these