" 1."3x+5y=12,5x+3y=4

3 x + 5y = 12, 5x + 3y = 4 .

The maximum value of p for which the lines 3x-4y=2, 3x-4y=12, 12x+5y=7 and 12x+5y=p constitute the sides of a rhombous is

The minimum value of p for which the lines 3x-4y=2, 3x-4y=12, 12x+5y=7 and 12x+5y=p constitute the sides of a rhombus is

The minimum value of p for which the lines 3x-4y=2, 3x-4y=12, 12x+5y=7 and 12x+5y=p constitute the sides of a rhombus is

{:(3x + 5y = 12),(3x - 5y = - 18):}

Simplify : (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

Simplify. (1.5x -4y)(1.5x +4y +3) -4.5x + 12y

Solve the following system of equations by matrix method : 5x+7y+2=0 4x+6y+3=0 5x+2y=3 3x+2y=5 3x+4y-5=0 x-y+3=0 3x+y=19 3x-y=23 3x+7y=4 x+2y=-1 3x+y=7 5x+3y=12

4x -5y-11z=12,. x-3y+z=1 2x + 3y-7z = 2