Question 117491
{{{((n^2+5n+4)/(2n^2+2n))((n^2-n-12)/(n^2-16))}}} Start with the given expression


{{{(((n+4)(n+1))/(2n^2+2n))((n^2-n-12)/(n^2-16))}}}   Factor {{{n^2+5n+4}}} to get {{{(n+4)(n+1)}}} 


{{{(((n+4)(n+1))/(2n(n+1)))((n^2-n-12)/(n^2-16))}}}   Factor {{{2n^2+2n}}} to get {{{2n(n+1)}}} 


{{{(((n+4)(n+1))/(2n(n+1)))(((n-4)(n+3))/(n^2-16))}}}   Factor {{{n^2-n-12}}} to get {{{(n-4)(n+3)}}} 


{{{(((n+4)(n+1))/(2n(n+1)))(((n-4)(n+3))/((n+4)(n-4)))}}}   Factor {{{n^2-16}}} to get {{{(n+4)(n-4)}}} 



{{{(n+4)(n+1)(n-4)(n+3)/2n(n+1)(n+4)(n-4)}}} Combine the fractions




{{{cross((n+4))cross((n+1))cross((n+4))(n+3)/(2n)cross((n+1))cross((n+4))cross((n-4))}}} Cancel like terms




{{{(n+3)/(2n)}}} Simplify



So {{{((n^2+5n+4)/(2n^2+2n))((n^2-n-12)/(n^2-16))}}} simplifies to {{{(n+3)/(2n)}}}



In other words, {{{((n^2+5n+4)/(2n^2+2n))((n^2-n-12)/(n^2-16))=(n+3)/(2n)}}}



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{{{(3pq^3/5)/(9p^2q^2/10)}}} Start with the given expression




{{{(30pq^3/45p^2q^2)}}} Combine the fractions by multiplying


{{{p^(1-2)q^(3-2)}}} Remember when you divide monomials, you subtract their corresponding exponents. For instance {{{x^2/x^3=x^(2-3)=x^-1}}}



{{{(2/3)p^-1q^1}}} Simplify.  Remember to reduce the coefficients (which are numbers in front of the variables) to get {{{30/45=2/3}}}.



{{{(2/3)(1/p)q}}} Rewrite {{{p^-1}}} as {{{1/p}}}



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Answer:


So {{{(3pq^3/5)/(9p^2q^2/10)}}} simplifies to {{{(2/3)(1/p)q}}}.


In other words, {{{(3pq^3/5)/(9p^2q^2/10)=(2/3)(1/p)q}}}.