If [tex]3x+2y =12[/tex] and [tex]xy=6[/tex], then find the value of [tex]9x^{2} + 4y^{2} [/tex]

xy=6x=6/y......13x+2y=12from13*(6/y)+2y=12{after solving}2y^2-12y+18=0y^2-6y+9=0      {taking y common}y^2-3y-3y+9=0y(y-3)-3(y-3)=0y=3,3............2putting 2 in 1x=2.........39x^2+4y^2=from 2 and 39*4+4*9=36+36=72

[tex]3x+2y=12\\
(3x+2y)^2=144\\
9x^2+12xy+4y^2=144\\\\
9x^2+4y^2=144-12xy\\
9x^2+4y^2=144-12\cdot6\\
9x^2+4y^2=144-72\\
\boxed{9x^2+4y^2=72}
[/tex]