We will start by simplifying the given equation by getting a common denominator. Observe that the denominators will cancel out. Multiplying both sides by the quantity $(n+1)(2-n)$, we have $(2-n)(2-n) + (2n-4)(n+1) = 7(n+1)(2-n)$.
The left hand side simplifies readily: $4-4n+n^2 + (2n-4)(n+1) = 4-4n+n^2 + 2n^2-2=n^2+2n$. Going back to the right side, we have
$(2-n)(2-n) + (2n-4)(n+1) = (2-n)^2 + (2n-4)(n+1) =(2-n)^2 + 2(n-2)(n+1)$.
To compute $(2-n)^2$, we distribute the negative sign first: $(2-n)^2=(-1)(n-2)^2$. Now we can use difference of squares: $(-1)(n-2)^2=-(n^2-4n+4)=-n^2+4n-4$. Distributing $2(n-2)(n+1)$, we have \[2(n-2)(n+1) =2(n^2-n-2n-2) = 2(n^2-3n-2).\]Adding them back up, we have that $n^2 +2n = (-n^2+4n-4) + 2(n^2-3n-2)$.
Simplifying the right side, this means that $n^2+2n=-n^2+4n-4+2n^2-6n-4=3n^2-5n-12$. Rearranging the terms, we get $3n^2-7n-12=0$. Note that $3n^2-7n-12=(3n+4)(n-3)$. Thus $n= -\frac 43, 3$, and we take the solution $n=\boxed{3}$.
Solve for $n$ in the equation $\frac{2-n}{n+1} + \frac{2n-4}{2-n} = 7$.
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