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