[2x+3y+5=0],[3x-2y-12=0]

Solve the following system of equations graphically; 2x+3y+5=0;quad 3x-2y-12=0

Solve for x and y: 2x - 3y-13 =0 , 3x -2y+12 =0

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

find x and y : 2x+3y-8=0, 3x+y-5=0

find x and y : 2x+y-5=0 3x+2y-8=0

If the lines x + 3y - 5 = 0 , 5x +2y-12 = 0 and 3x -ky-1=0 do not form a triangle, then a value of k is

Solve by cross multiplication x+2y+1=0 ,2x-3y-12=0