`3x-y=2`
`2x-y=3`

Solve the following system of the linear equations by cramer's rule 3x+y+z=2,2x-4y+3z=-1 and 4x+y-3z=-11

3x+2y=8 2x-3y=8

3x+2y=8 2x-3y=1

3x+2y=5 2x-3y=7

If the straight lines 3x+4y+2=0,2x-y+3=0 and 2x+ay-6=0 are concurrent,then a=

If the equation of circle circumscribing the quadrilateral formed by the lines in order are 2x+3y=2,3x-2y=3,x+2y=3and2x-y=1 is given by x^(2)+y^(2)+lamdax+muy+v=0 . Then the value of |lamda+2mu+v| is

3x+2y=5 2x+3y=7

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

In the formula p=(3x-2y)/.(2x-3y) , make x as the subject of the formula.