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

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

(4)/(3)x+2y=8;2x+3y=12

" (2) "3x+y=-1,2x-3y=-8

Solve: 2x+3y=2; x-2y=8

Solve by Cramer's rule: 3x+2y=4 8x+3y=13

The equation of the normal to the curve 3x^2-y^2=8 which is parallel to x+3y=8 is x-3y=8 (b) x-3y+8=0 (c) x+3y+-8=0 (d) x+3y=0

Determine, algebraically, the vertices of the triangle formed by the lines 3x-y=3 2x-3y=2 and " " x+2y=8

Solve : 2x+3y=7 3x-2y=8

slove the following equation by elimation method 2x+3y=8, x-2y=-3

3x+2y=4 ; & 8x+5y=9