x+2y=-1;2x-3y=12

2x+3y=12;x-y=1

Solve each pair of equation by using the substitution method. x+2y=-1 and 2x-3y=12.

,x-y=-1, 3x-2y=12

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

Solve the following system of equations by using the method of substitution: x+2y=-1,quad 2x-3y=12

The system of equations x+2y=3,2x-3y=5 and x -12y =1 has

The circles x^2+y^2-12x-12y=0 and x^2+y^2+6x+6y=0 :

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 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