Question 190453
{{{2/(n+2) + 1/n + (n+6)/(n^2+2n)}}} Start with the given expression.



{{{2/(n+2) + 1/n + (n+6)/(n(n+2))}}} Factor the last denominator



Take note that the LCD is {{{n(n+2)}}}



{{{(2*n)/(n*(n+2)) + 1/n + (n+6)/(n(n+2))}}} Multiply the first fraction by {{{n/n}}} to get the denominator equal to the LCD



{{{(2*n)/(n*(n+2)) + (1*(n+2))/(n*(n+2)) + (n+6)/(n(n+2))}}} Multiply the second fraction by {{{(n+2)/(n+2)}}} to get the denominator equal to the LCD



{{{(2n)/(n(n+2)) + (n+2)/(n(n+2)) + (n+6)/(n(n+2))}}} Distribute



{{{(2n+n+2+n+6)/(n(n+2))}}} Combine the fractions



{{{(4n+8)/(n(n+2))}}} Combine like terms.



{{{(4(n+2))/(n(n+2))}}} Factor out the GCF 



{{{(4*cross((n+2)))/(n*cross((n+2)))}}} Cancel out like terms



{{{4/n}}} Simplify



So {{{2/(n+2) + 1/n + (n+6)/(n^2+2n)=4/n}}} where {{{n<>-2}}} or {{{n<>0}}}