y=3x

Solve the following equations : 3x-y=3 x-y=4

Two straight lines u=0a n dv=0 pass through the origin and the angle between them is tan^(-1)(7/9) . If the ratio of the slope of v=0 and u=0 is 9/2 , then their equations are (a) y+3x=0 and 3y+2x=0 (b) 2y+3x=0 and 3y+x=0 (c) 2y=3x and 3y=x (d) y=3x and 3y=2x

If y=x^(1/3)-x^(-1/3) , prove that y^(3)+3y=x-(1)/(x)

Solve graphically y=[3-x]

Find the value of x and y when 3x+y=3 9x-3y=9

The value of (x-y)^3+(x+y)^3+3(x-y)^2(x+y)+3(x+y)^2(x-y) is

If one of the diagonals of a square is along the line x=2y and one of its vertices is (3, 0), then its sides through this vertex are given by the equations (A) y-3x+9=0, 3y+x-3=0 (B) y+3x+9=0, 3y+x-3=0 (C) y-3x+9=0, 3y-x+3=0 (D) y-3x+9=0, 3y+x+9=0

If one of the diagonals of a square is along the line x=2y and one of its vertices is (3, 0), then its sides through this vertex are given by the equations (A) y-3x+9=0, 3y+x-3=0 (B) y+3x+9=0, 3y+x-3=0 (C) y-3x+9=0, 3y-x+3=0 (D) y-3x+9=0, 3y+x+9=0

1/(3x+y)+1/(3x-y)=3/4 and 1/(3x+y)-1/(3x-y)= -1/4