`x-3y=1,3x-2y+4=0`

Find the point of intersection of the straight lines : (i) 2x+3y-6=0 , 3x-2y-6=0 (ii) x=0 , 2x-y+3=0 (iii) (x)/(3)-(y)/(4)=0 , (x)/(2)+(y)/(3)=1

Solve:(1)/(2(2x+3y))+(12)/(7(3x-2y))=(1)/(2)(7)/(2x+3y)+(4)/(3x-2y)=2 where 2x+3y!=0 and 3x-2y!=0

If y-3=0,x-3y+9=0 and 3x-2y+1=0x+y-3=0,x-3y+9=0 and 3x-2y+1=0 are sides of triangle then area of triangle.

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

The equation of the transvers axis of the hyperbola (x-3)^2+(y=1)^2+(4x+3y)^2 is x+3y=0 (b) 4x+3y=9 3x-4y=13 (d) 4x+3y=0