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Given that
\(\begin{align*} \frac{1}{x}+\frac{1}{y}&=3,\\ xy+x+y&=4, \end{align*}\)
compute \(x^2y+xy^2\).

 Jan 26, 2021
 #1
avatar+116126 
+2

1/x  + 1/y  =  3

(x + y)   / xy   =  3

x + y =   3xy

 

 

So   

 xy  + x + y =  4

xy + 3xy =  4

4xy  =  4

xy   =1

 

x^2y + xy^2  =

xy (y + x)  =

xy ( x + y)  

1 ( 3xy)  =

1 ( 3(1)) =

3

 

 

cool cool cool

 Jan 26, 2021
 #2
avatar+311 
+2

The first equation becomes
\(\frac{x+y}{xy}=3\Rightarrow x+y=3xy\)
Substituting into the second equation,
\(4xy=4\Rightarrow xy=1\)
Thus x+y=3.

The quantity we desire factors as xy(x+y), so it is equal to \(1(3)=\boxed{3}\).

 

 

-hihihi

 

😎😎😎

 Jan 26, 2021

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