Question 246091
How did you get x=2 & y=3?
-----------------------------
8x^2-18xy+9y^2=0
(4x-3)(2x-3)=0
4x-3=0
4x=3
x=3/4 ans. y=1 & y=.5
2x-3=0
2x=3
x=3/2 ans. y=2 & y=1.
8(3/4)^2-18(3/4y)+9y^2=0
8*9/16-54/4y+9y^2=0
72/16-27y/2+9y^2=0
9y^2-27y/2+9/2=0
9(y^2-3y/2+1/2)=0
9(y^2-1.5y+.5)=0
9(y-1)(y-.5)=0
y-1=0
y=1 ans
y-.5=0
y=.5 ans.
8(3/2)^2-18(3/2y)+9y^2=0
8*9/4-27y+9y^2=0
18-27y+9y^2=0
9y^2-27y+18
9(y^2-3y+2)=0
(y-2)(y-1)=0
y-2=0
y=2 ans
y-1=0
t=1 ans