[x+y=7-0],[2x-3y=9-square]

Fill in the blanks: x/7=y/3=(7x-9y)/(square)

The area of the parallelogram formed by the lines 2x-3y+a=0, 3x-2y-a=0, 2x-3y+3a=0 and 3x-2y-2a=0 in square units , is

2x-y-5=0; x+3y-9=0

One diagonal of a square is along the line 8x-15y=0 and one of its vertex is (1,2) . Then the equations of the sides of the square passing through this vertex are 23x+7y=9,7x+23y=5323x-7y+9=0,7x+23y+53=023x-7y-9=0,7x+23y-53=0 these

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

Complete the following: 2x+y=8 x: 0, square , 3 y: square , 4, 2 (x,y): square , square , (3,2) x-y=1 x: square , 2, square y: -1, square , 4 (x,y): square , square , square

2x+3y-5=0,2x+3y+15=0,x+y-7=0,x+y+7=0 are sides of a parallelogram.Then the centre of the parallelogram is

Complete the following activity to solve simultaneous equations : 4x+ 3y = 18 " "...(1) 5x - 3y = 9 " " ...(2) Adding equations (1) and (2) , +{:(4x+3y=18" ...(1)"),(5x-3y=9" ...(2)"),(bar(squarex" "=square)):} :. X = (square)/(square ) " " :. X = square Substituting x = 3 in equation (1) , 4 xx square + 3y = 18 " " :. 3y = 18 - square " " :. 3y = square :. y = 2 (x,y ) = (square, square ) is the solution

2x+3y=9 in this equation replace y by (7-4x)/5 and find x.

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