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Given that

1/x + 1/y = 1/3

xy + x + y = 4

 

Compute x^2*y + x*y^2.

 Apr 15, 2022
 #1
avatar+326 
+3

G'day mate, 

 

"y" in this case is "a", I accidentally used "a" throughout, sorry!!!!!

 

\(\begin{bmatrix}\frac{1}{x}+\frac{1}{a}=\frac{1}{3}\\ xa+x+a=4\end{bmatrix}\)

 

\(\mathrm{Plug\:the\:solutions\:}x=-\frac{3a}{3-a}\mathrm{\:into\:}xa+x+a=4\)

 

We get a = \(a=\frac{1}{2}+i\frac{\sqrt{11}}{2} .and. a=\frac{1}{2}-i\frac{\sqrt{11}}{2}\)

 

\(\mathrm{Plug\:the\:solutions\:}a=\frac{1}{2}-i\frac{\sqrt{11}}{2},\:a=\frac{1}{2}+i\frac{\sqrt{11}}{2}\mathrm{\:into\:}\frac{1}{x}+\frac{1}{a}=\frac{1}{3}\)

 

That gives \(x=\frac{1}{2}+\frac{\sqrt{11}}{2}i and x=\frac{1}{2}-\frac{\sqrt{11}}{2}i\)

 

So now we have our values for (x, y): \(\begin{pmatrix}x=\frac{1}{2}+\frac{\sqrt{11}}{2}i,\:&y=\frac{1}{2}-i\frac{\sqrt{11}}{2}\\ x=\frac{1}{2}-\frac{\sqrt{11}}{2}i,\:&y=\frac{1}{2}+i\frac{\sqrt{11}}{2}\end{pmatrix}\)

 

Plugging that in: \(x^2\cdot y + x\cdot y^2\):

 

\(\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)^2\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)+\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)^2=3\)

 

\(\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)^2\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)+\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)^2=3\)

 

So the answer is \(3.\)

 

 

-Vinculum 

 

smileysmileysmiley

 Apr 15, 2022
 #2
avatar+122390 
+2

1/x + 1/y  =  1/3     ⇒   (x + y) / xy  =   1/3   ⇒  (x + y)  = (1/3)xy  ⇒ 3(x + y) = xy

 

And

xy + ( x + y) = 4

3xy + 3(x + y) = 12

3xy + xy = 12

4xy = 12

xy  = 3

 

Since (x +y) / xy  =  1/3   this implies that  (x + y)   =1

 

So

 

x^2y  + y^2x   =

 

xy ( x + y)   =

 

3 ( 1)   = 3

 

 

cool cool cool

 Apr 15, 2022

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