How do you simplify #(2p^3q^2)/(8p^4q) div (4pq^2)/(16p^4)#?

2 Answers
Jul 16, 2015

Answer:

#((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))=p^2/q#

Explanation:

We start off with what's given:

#((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))#

Recall: #(a/b)/(c/d)=(a/b)/(c/d)*(d/c)/(d/c)=a/b*d/c#

So,

#((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))=((2p^3q^2)/(8p^4q))*((16p^4)/(4pq^2))#

Recall: #a^5/a^3=a^5*a^-3=a^(5-3)=a^2#

So,

#((2p^3q^2)/(8p^4q))((16p^4)/(4pq^2))##=##((p^3q^2)/(4p^4q))((4p^4)/(pq^2))#

The #((p^3q^2)/(4p^4q))# part becomes:

#((p^3q^2)/(4p^4q))=(p^3q^2)(4^(-1)p^(-4)q^(-1))=4^(-1)p^(3-4)q^(2-1)=4^(-1)p^(-1)q^1#

#=q/(4p)#

The #((4p^4)/(pq^2))# part becomes:

#((4p^4)/(pq^2))=(4p^4)(p^(-1)q^(-2))=4p^(4-1)q^(-2)=4p^3q^(-2)#

#=(4p^3)/q^2#

So, our original setup was:

#((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))=((2p^3q^2)/(8p^4q))*((16p^4)/(4pq^2))=((p^3q^2)/(4p^4q))((4p^4)/(pq^2))#

#=(q/(4p))((4p^3)/q^2)=(4p^3q)/(4pq^2)=p^2/q#

Jul 17, 2015

Answer:

The answer is #p^2/q#.

Explanation:

#(2p^3q^2)/(8p^4q)##-:##(4pq^2)/(16p^4)#

Reduce the numerical part of each fraction.

#(cancel2^1p^3q^2)/(cancel8^4p^4q)##-:##(cancel4^1pq^2)/(cancel16^4p^4)# =

#(p^3q^2)/(4p^4q)##-:##(pq^2)/(4p^4)# =

Simplify each fraction.

Use the exponent rule #x^m/x^n=x^(m-n)# for each fraction.

#(p^3q^2)/(4p^4q)=(p^(3-4)q^(2-1))/(4)=(p^(-1)q)/4#

#(pq^2)/(4p^4)=(p^(1-4)q^2)/(4)=(p^-3q^2)/4#

Use exponent rule #x^(-m)=1/(x^m)# for each fraction.

#(q)/(4p)-:(q^2)/(4p^3)#

In order to divide fractions, invert the second fraction to get its reciprocal, and then multiply.

#(q)/(4p)xx(4p^3)/(q^2)# =

#(4p^3q)/(4pq^2)# =

Cancel the #4#.

#(cancel4^1p^3q)/(cancel4^1pq^2)#

Apply the exponent rules #x^m/x^n=x^(m-n)# and #x^(-m)=1/(x^m)#.

#(p^(3-1)q^(1-2))=p^2q^(-1)=(p^2)/q#