Rational

\(\frac{5n}{n-4}+3=\frac{3n+8}{n-4}\)

\(\frac{2n-12}{n-4}=\frac{3n+8}{n-4}\)

\((2n-12)(n-4)=(3n+8)(n-4)\)

notice that I don't divide both sides by n-4. (because n-4 can equal 0)

\(2n^2-20n+48=3n^2-4n-32\)

\(n^2+16n-80=0\)

\((n+20)(n-4)=0\)

\(n=-20, n=4\)

 

JP

\(N=25- \sqrt(561)/4 \)

 

\(N=(25+ \sqrt(561))/4\)

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