[2x+3y=2],[x-2y=8]

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

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

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

2x+3y=8 4x-5y=-6

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

(a) 2x+y=5 ; 3x+2y=8

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

Factorize the following : (i) 20 x^3-40 x^2+80 x (ii) 2x^3y^2-4x^2y^3+8x y^4 (iii) 10 m^3n^2+15 m^4n-20 m^2n^3 (iv) 2a^4b^4-3a^3b^5-6a b^3

2x + 3y = 8 , x - 2y + 3 = 0

For hyperbola whose center is at (1, 2) and the asymptotes are parallel to lines 2x+3y=0 and x+2y=1 , the equation of the hyperbola passing through (2, 4) is (a) (2x+3y-5)(x+2y-8)=40 (b) (2x+3y-8)(x+2y-8)=40 (c) (2x+3y-8)(x+2y-5)=30 (d) none of these