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

Sagot :

xy=6
x=6/y......1
3x+2y=12
from1
3*(6/y)+2y=12
{after solving}2y^2-12y+18=0
y^2-6y+9=0      {taking y common}
y^2-3y-3y+9=0
y(y-3)-3(y-3)=0
y=3,3............2
putting 2 in 1
x=2.........3

9x^2+4y^2=
from 2 and 3
9*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]