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):}

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

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

1 + 5x + 3y = 9 2x 3y = 12 X = (x, y) - 4