" If "1/x+2/y=4" and "3/y-1/x=11" then "

1/(3x+y)+1/(3x-y)=3/4 and 1/(3x+y)-1/(3x-y)= -1/4

3x+y=1,2x=11y+3

The graph of y = x^(2) and y= 2 -x interesects at (1,1) and (-2,4) then the roots of required quadratic equations are:

The tangent to the circle x^(2)+y^(2)-4x+2y+k=0 at (1,1) is x-2y+1=0 then k=

The tangents to the circle x^(2)+y^(2)-4x+2y+k=0 at (1,1) is x-2y+1=0 then k=

The tangent to the circle x^(2)+y^(2)-4x+2y+k=0 at (1,1) is x-2y+1=0 then k=

Solve by matrix method. 1/x+2/y-3/z=-4 2/x+3/y+2/z=2 3/x-1/y-4/z=11