Solve
`sqrt3x+y=1`
`x+sqrt3y=1`

Find the angle between the lines sqrt(3)x+y=1 and x+sqrt(3)y=1

Find angles between the lines sqrt(3)x+y=1 and x+sqrt(3)y=1

Solve x^(2)-(sqrt3+1)x+sqrt3=0 .

Prove that the lines sqrt(3)x+y=0,sqrt(3)y+x=0,sqrt(3)x+y=1 and sqrt(3)y+x=1 form a rhombus.

Prove that the diagonals of the parallelogram formed by the lines sqrt(3)x+y=0, sqrt(3) y+x=0, sqrt(3) x+y=1 and sqrt(3) y+x=1 are at right angles.

If x =1 + sqrt2 + sqrt3 and y =1 + sqrt2 - sqrt3, then the value of (x ^(2) + 4xy + y ^(2))/(x + y) is:

The angle between the lines sqrt(3)x-y-2=0 and x-sqrt(3)y+1=0 is

Solve (sqrt3+1)^(2_x)+(sqrt3-1)^(2_x)=2^(3x) .

The equation of tangent to the curve y=int_(x^(2))^(x^(3))(dt)/(sqrt(1+t^(2))) at x=1 is sqrt(3)x+1=y (b) sqrt(2)y+1=x sqrt(3)x+y=1(d)sqrt(2)y=x