Question

DEATH STAR Question: find x and y when

Answer 1

\[x \sqrt{y}+y \sqrt{x}=30,x \sqrt{x}+y \sqrt{y}=35\]

Answer 2

what do you get when you square both equation?

Answer 3

i.e., \((x \sqrt{y} + y \sqrt {x})^{2}=30^{2}\) and \((x \sqrt{x} + y \sqrt {y})^{2}=35^{2}\)

Answer 4

\[x^{2}y+2x\sqrt{x}y\sqrt{y}+y^{2}x=900\]

Answer 5

and square the other equation as well.

Answer 6

\[xx^{2}+2xy\sqrt{xy}+yy^{2}=1225\]

Answer 7

do you see that the middle item is actually the same? so you can subtract the first squared equation from the second, and you will get an equation which equals to \(x^3-x^2y-y^2x+y^3=325\)
Let me know if I lost you.

Answer 8

it's actually \((x-y)^2(x+y)=325\)

Answer 9

\[xx^2-yx^2+yy^2-xy^2\]\[=(x-y)x^2+(y-x)y^2\]\[=(x-y)x^2-(x-y)y^2\]\[=(x-y)(x^2-y^2)\]\[=(x-y)(x+y)(x-y)\]\[=(x-y)^2(x+y)\]

Answer 10

okay, thanks what and then

Answer 11

and then i got stuck as well -_-

Answer 12

okay that fine, thanks a lot :)
for help me

Answer 13

x = 4 and y = 9

Answer 14

you guess ??? or ...

Answer 15

i got it by factoring 325 into 25*13, and set \[25=(x-y)^2\], hence \[x-y=5 or -5\] and \[13=(x+y)\]
then \[x=4 or 9\] and \[y=9 or 4\] so basically the two answers are 4 and 9

Answer 16

thanks

Answer 17

no problem! and best response please :)