(v) `sqrt2 x+sqrt3 y = 0 and sqrt3 x-sqrt8 y = 0`

(v) Solve for x and y if (sqrt2 x+sqrt 3 y)=0 and (sqrt3 x- sqrt8 y)=0

Solve the following pair of linear equations by the substitution method. (i) x+y=14 ;x- y=4 (ii) s-t=3; s/3+t/2=6 (iii) 3x-y=3;9x-3y=9 (iv) 0. 2 x+0. 3 y=1. 3 ;0. 4 x+0. 5 y=2. 3 (v) sqrt(2)x+sqrt(3)y=0;sqrt(3)x-sqrt(8)y=0 (vi) 3x/2-5y/2=-2;x/3+y/2=13/6

The locus of the point of intersection of the lines sqrt3 x- y-4sqrt3 t= 0 & sqrt3 tx +ty-4 sqrt3=0 (where t is a parameter) is a hyperbola whose eccentricity is: (a) sqrt3 (b) 2 (c) 2/sqrt3 (d) 4/3

The locus of the point of intersection of the lines sqrt3 x- y-4sqrt3 t= 0 & sqrt3 tx +ty-4 sqrt3=0 (where t is a parameter) is a hyperbola whose eccentricity is:

Solve the following system of equations: sqrt(2)x-sqrt(3)y=0,\ \ \ \ sqrt(3)x-sqrt(8)y=0

Solve the following system of equations: sqrt(2)x-sqrt(3)y=0,\ \ \ \ sqrt(3)x-sqrt(8)y=0

Prove that the locus of the point of intersection of the lines sqrt3 x -y -4 sqrt3 k =0 and sqrt3 k x+ ky - 4 sqrt3 =0, for different values of k, is a hyperbola whose eccentricity is 2.

The locus of the point of intersection of lines sqrt3x-y-4sqrt(3k) =0 and sqrt3kx+ky-4sqrt3=0 for different value of k is a hyperbola whose eccentricity is 2.

An equilateral triangle whose two vertices are (-2, 0) and (2, 0) and which lies in the first and second quadrants only is circumscribed by a circle whose equation is : (A) sqrt(3)x^2 + sqrt(3)y^2 - 4x +4 sqrt(3)y = 0 (B) sqrt(3)x^2 + sqrt(3)y^2 - 4x - 4 sqrt(3)y = 0 (C) sqrt(3)x^2 + sqrt(3)y^2 - 4y + 4 sqrt(3)y = 0 (D) sqrt(3)x^2 + sqrt(3)y^2 - 4y - 4 sqrt(3) = 0

{:(sqrt(5)x - sqrt(7)y = 0),(sqrt(7)x - sqrt(3)y = 0):}