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

2 Answers
Jul 16, 2015

((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

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