Question 191957


{{{((n-6)/(n^2+11n+24))/((n+1)/(n+3))}}} Start with the given expression.



{{{((n-6)/(n^2+11n+24))((n+3)/(n+1))}}} Multiply the first fraction {{{(n-6)/(n^2+11n+24)}}} by the reciprocal of the second fraction {{{(n+1)/(n+3)}}}.



{{{(((n-6))/((n+8)(n+3)))((n+3)/(n+1))}}} Factor {{{n^2+11n+24}}} to get {{{(n+8)(n+3)}}}.



{{{((n-6)(n+3))/((n+8)(n+3)(n+1))}}} Combine the fractions. 



{{{((n-6)highlight((n+3)))/((n+8)highlight((n+3))(n+1))}}} Highlight the common terms. 



{{{((n-6)cross((n+3)))/((n+8)cross((n+3))(n+1))}}} Cancel out the common terms. 



{{{((n-6))/((n+8)(n+1))}}} Simplify. 



{{{(n-6)/(n^2+9n+8)}}} FOIL the denominator



So {{{((n-6)/(n^2+11n+24))/((n+1)/(n+3))}}} simplifies to {{{(n-6)/(n^2+9n+8)}}}.



In other words, {{{((n-6)/(n^2+11n+24))/((n+1)/(n+3))=(n-6)/(n^2+9n+8)}}} where {{{n<>-8}}}, {{{n<>-3}}}, or {{{n<>-1}}}