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

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

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

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

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

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

Show that x = 3 and y = -2 is the solution of the system of linear equations x+2y=-1" and "2x-3y=12 .

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

y^(2)=12x

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

The centres of 3 circles x^2+y^2 =1, x^2+y^2 +6x-2y=1, x^2+y^2-12x+4y=1 are :