Question

X2+Y2=100 when xy=18 what is the value of x and y

Answer 1

\[\large (x+y)^2 = (x^2 + y^2) + 2(xy)\]

Answer 2

Find x + y first..

Answer 3

i wonder what water's plan is -_- guess i'll check it out

Answer 4

x = -100, y = -1, Y = ±8
x = -10, y = -10, Y = ±8

Answer 5

This is by wolfram alpha..

Answer 6

http://www.wolframalpha.com/input/?i=xy%3D18+find+X2%2BY2%3D100+

Answer 7

wolfram doesnt really give right answers...if you dont know how to input it right...

Answer 8

u can see here

Answer 9

\[(x+y)^2 = 136 \implies (x+y)= \pm \sqrt{136}\]

Answer 10

here comes a monologue solution >:))

Answer 11

Im so lost right now -_-

Answer 12

Now :

\[x = \sqrt{136} - y\]

\[(\sqrt{136} - y)y = 18\]

Answer 13

i agree @Seamonkey0912 i think @waterineyes should go slow dont you agree? ;)

Answer 14

And now I am also lost..

Answer 15

Let me try first..

Answer 16

maybe you should've done it the traditional way...

Answer 17

\[y^2 - \sqrt{136}y + 18 = 0 \implies y = \frac{\sqrt{136} \pm 8}{2}\]

Answer 18

x^2 + y^2 = 100

xy = 18

^from there i can say x = 18/y

now substitute this value of x into x^2 + y^2 =100

giving you (18/y)^2 + y^2 = 100

18^2/y^2 + y^2 = 100

324/y^2 + y^2 = 100

now multiply all terms by y^2

324 + y^4 = 100y^2

y^4 - 100y^2 + 324 = 0

use quadratic formula to get the value of y

Answer 19

x^2+y^2 is not equal to (x+y)^2

Answer 20

How do you know and where I have written this ??

Answer 21

Ahhh o_O still lost I'm sorry

Answer 22

I used:

\[(x+y)^2 = x^2 + y^2 + 2xy \]

Answer 23

Are you having answers with you @Seamonkey0912

Answer 24

I thought it was +8, -8 but I can't get it to equal to 100

Answer 25

http://www.wolframalpha.com/input/?i=xy+%3D+18+and+x%5E2+%2B+y%5E2+%3D+100

Answer 26

I got:

\[y = 4 \pm \sqrt{34}\]

Answer 27

Now find x :

\[x = 2 \sqrt{34} - 4 \pm \sqrt{34} \implies x = 3\sqrt{34} - 4 \quad Or \quad \sqrt{34} -4\]

Answer 28

Similarly:

\[x = -2 \sqrt{34} - y\]

\[y^2 + 2\sqrt{34}y + 18 = 0 \implies y = -\sqrt{34} \pm 4\]

And now the x again..

Answer 29

using the idea
\[ (x+y)^2 = x^2+2xy+y^2 \]
and the given values
\[ x^2+y^2= 100 \text{ and } 2xy=36 \]
\[ (x+y)^2= 136 \]
or
\[ x+y= 2\sqrt{34} \]
also \[ xy= 18 \]
one way to solve is replace y= 18/x in the first equation to get
\[ x+\frac{18}{x}-2\sqrt{34} =0 \]
\[ x^2 -2\sqrt{34}x + 18= 0 \]
use the quadratic formula to get
\[ x= \sqrt{34} \pm 4 \]
that means
\[ x= \sqrt{34}+4 \text{ and } y= \sqrt{34}-4 \]

check:
\[ xy= ( \sqrt{34}+4)( \sqrt{34}-4)= 34-16= 18 \]
and we will see that
\[ x^2+y^2= 100 \]