Solve!

Let $a=x+y$ and $b=xy$ . Then, with the use of the identities we have the system: $a^3-3ab=7\\ a^2-b+a=4$ Solve for $b$ from the first equation and plug in the second:

$b=\dfrac{a^3-7}{3a} \\ a^2-\dfrac{a^3-7}{3a}+a=4$

We arrive to a cubic equation:

$2a^3+3a^2-12a+7=0$

Factor ir using the rational root test:

$(a-1)^2(2a+7)=0$

So we have the solutions $(a,b)=(1,-2),(-7/2,19/4)$ .

Finally, for $x$ and $y$ to be real, we need that $a^2-4b\geq 0$ . Only the first solution is valid, leading us to $\boxed{2}$ different pairs such that $x+y=1$ and $xy=-2$ , giving us $(x,y)=(2,-1),(-1,2)$ .