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Homework Statement
solve ## x^2/y + y^2/x=9, 1/x+1/y=3/4##[/B]
Homework Equations
The Attempt at a Solution
##9x^3+9y^3=81xy##
## 4x+4y-3xy=0##
##x^3+y^3-9xy=0, 12x+12y-9xy=0##
##x^3-12x+y^3-12y=0## [/B]
chwala said:Homework Statement
solve ## x^2/y + y^2/x=9, 1/x+1/y=3/4##[/B]
Homework Equations
The Attempt at a Solution
##9x^3+9y^3=81xy##
## 4x+4y-3xy=0##
##x^3+y^3-9xy=0, 12x+12y-9xy=0##
##x^3-12x+y^3-12y=0## [/B]
ehild said:So you have
##(x+y)(x^2-xy+y^2) = 9xy##
I deleted my post which was edited by the moderator, as it did not give any new hint in that form.stevendaryl said:That was my hint, except in the alternative form:
##(\frac{1}{x}+\frac{1}{y})(x^2-xy+y^2) = 9##
It is easy to get that solution and the other one systematically, by using new variables v=x+y, z=xy.mathwonk said:i just looked at it and said the nicest reciprocals that add up to 3/4 are 1/2 and 1/4. fortunately 2^3 + 4^3 is also divisible by 9.
ehild said:I deleted my post which was edited by the moderator, as it did not give any new hint in that form.
The equation x^{3}+y^{3}=9xy was written in the OP (multiplied by 9). No need to use the reciprocals.
The factorization of x^{3}+y^{3} is well known, taught in the 9th class.
yep this does not lead anywhere.....chwala said:Homework Statement
solve ## x^2/y + y^2/x=9, 1/x+1/y=3/4##[/B]
Homework Equations
The Attempt at a Solution
##9x^3+9y^3=81xy##
## 4x+4y-3xy=0##
##x^3+y^3-9xy=0, 12x+12y-9xy=0##
##x^3-12x+y^3-12y=0## [/B]
There is a mistake in the first line, and you have to indicate what z is, and how did you get the equation ##(9/16)z^2-3z-12=0##.chwala said:I got it though...i won't show all the steps
##(x^3+y^3)/(xy)=1 ##, ##(x^3+y^3)/(xy)=9
##(x+y)(x^2-xy+y^2)=x^3+y^3##
and
##(x+y)/(xy)=3/4##
##(x+y)(x^2-xy+y^2)/9=4/3 (x+y)##
##x^2-xy+y^2=12##
##(9/16)z^2-3z-12=0##
##z=8, z=-8/3##
##xy=8, xy=-8/3##
##x+y=3/4##
##x+y=6##
##x=6-y##
##y=2, x=4##
##y=4, x=2##
##y=-1+(11/3)^{0.5} , x=-1-(11/3)^{0.5}##
##y=-1-(11/3)^{0.5}, x=-1+(11/3)^{0.5}##
yeah i see that mistake..ehild said:There is a mistake in the first line, and you have to indicate what z is, and how did you get the equation ##(9/16)z^2-3z-12=0##.
Otherwise, it is correct.
I see nowchwala said:yeah i see that mistake..
##x^2-xy+y^2=12##
→##(x+y)^2-3xy=12##
we know that ##x+y =(3/4)xy##
you still cannot see?