Simple factorization.

Both equations could be simplified to, with the first becomes

$(x+y)^{2} + (x+y)(xy+1)=40$

$\Leftrightarrow (x+1)(y+1)(x+y) = 40$

while the second changes to

$(x+1)(y+1) = 8$

So, all one needs to do to find the $xy$ is finding out the value of $x+y$ through substitution of the RHS part of second equation to LHS of the other. By doing that, one would come up with

$x+y = 5$

At this point, one don't have to find the individual value of both $x$ and $y$ as one could use the last equation one came up with to simplify the second equation. Voila,

$xy + 5 + 1 = 8$ $\leftrightarrow xy = 2$

By a little rearrangement in the first equation,we can show that: $x^2y+xy^2+x^2+y^2+2xy+x+y=x^2+2xy+y^2+x^2y+xy^2+x+y$ Factoring,we get: $(x+y)^2+xy(x+y)+1(x+y)=40$ $(x+y)(x+y+xy+1)=40$ Replacing the value of the second equation in the first,we see that: $(x+y)(x+y+xy+1)=(x+y)(8)=40\rightarrow\;x+y=\frac{40}{8}=\boxed{5}$ Replacing the value of $x+y=5$ in the second equation, we get: $xy+5+1=8\rightarrow\;xy=8-5-1=\boxed{2}$

xy (x+y)+(x+y)^2+(x+y)=40, (x+y)(xy+x+y+1)=40

X+y=5, From, xy+x+y+1=8

>>xy=2

Thee equation factorizes to: $(x+y)^2 + xy(x+y) +(x+y) = (x+y)(xy+x+y+1) = 8(x+y) = 40 \Rightarrow\ x+y = 5$ Substituting our result into the 2nd given equation: $xy = 8 - (x+y) - 1 = \boxed{2}$

• We can rearrange the first equation to: $x²y + x² + xy + x) + (xy² + xy + y² +y) = 40$ .
• Factoring the first equation, we have: $x(xy + x + y + 1) + y(xy + x + y + 1) = 40 => (x + y)(xy + x + y + 1) = 40$
• Replacing the second equation in the first one, we have: $(x + y)(8) = 40 => (x + y) = 5$ .
• Replacing (x + y) = 5) in the second equation, we conclude: $xy + x + y + 1 = 8 => xy + 5 + 1 = 8$ .
• Finally, we have xy = 2 .

First Equation:

$x^2y$ + x $y^2$ + $x^2$ + $y^2$ + 2xy + x + y = 40

$\Rightarrow$ $x^2y$ + x + x $y^2$ + y + $x^2$ + 2xy + $y^2$ = 40

$\Rightarrow$ $x^2y$ + x + x $y^2$ + y + $(x+y)^2$ = 40

$\Rightarrow$ x(xy+1) + y(xy+1) + y + $(x+y)^2$ = 40

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Second Equation:

xy + x + y + 1 = 8

$\Rightarrow$ xy + 1 = 8 - x - y

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Substituting into First Equation:

x(xy+1) + y(xy+1) + y + $(x+y)^2$ = 40

$\Rightarrow$ x(8 - x - y) + y(8 - x - y) + $(x+y)^2$ = 40

$\Rightarrow$ 8x - $x^2$ -xy + y(8 - x - y) + $(x+y)^2$ = 40

$\Rightarrow$ 8x - $x^2$ -xy + 8y -xy - $y^2$ + $(x+y)^2$ = 40

$\Rightarrow$ 8x - $x^2$ -xy + 8y -xy - $y^2$ + $x^2$ + 2xy + $y^2$ = 40

Simplify

$\Rightarrow$ 8x + 8y = 40

$\Rightarrow$ 8(x+y) = 40

$\Rightarrow$ (x+y) = 5

Substitute Back into Second Equation:

xy + x + y + 1 = 8

$\Rightarrow$ xy + (x + y) + 1 = 8

$\Rightarrow$ xy + 5 + 1 = 8

$\Rightarrow$ xy + 6 = 8

$\Rightarrow$ xy = 2

Final Answer: xy = 2

x^2y+xy^2+x^2+y^2+2xy+x+y = 40 xy(x+y)+(x+y)^2+(x+y) = 40 (x+y) (xy+x+y+1) = 40 (x+y)*8 = 40 (x+y) = 5

by putting the value of (x+y)=5 in the equation 2, we get

xy=(8-6)=2

Using the Python sympy library: Divide the longer polynomial by the shorter to show that the shorter one is a factor (since the remainder is zero). Then conclude that $x+y$ is 5, the quotient of 40 and 8. Finally, solve the pair of equations that includes this and the one with 8 on the rhs.

Image of interactive session.