Dividing Algebraic Fractions
The quotient of two algebraic fractions is found by multiplying the first fraction by the reciprocal of the second. $$ \frac{A}{B} \ : \ \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C} $$
How to divide two algebraic fractions
To divide two algebraic fractions:
- Rewrite the division as a multiplication
- Replace the second fraction with its reciprocal
- Carry out the multiplication
The result is the quotient expressed as an algebraic fraction.
Note. This is the quickest and most straightforward method for dividing algebraic fractions.
A worked example
Let’s look at this division:
$$ \frac{2x}{3y} : \frac{4y}{3z} $$
which can be written as
$$ \frac{ \frac{2x}{3y} }{ \frac{4y}{3z} } $$
I now turn the division into a multiplication by taking the reciprocal of the second fraction:
$$ \frac{2x}{3y} \cdot \frac{3z}{4y} $$
Next, I simplify:
$$ \frac{2x}{\require{cancel} \cancel{3}y} \cdot \frac{\cancel{3}z}{4y} $$
$$ \frac{\cancel{2}x}{y} \cdot \frac{z}{\cancel{4}y} $$
$$ \frac{x}{y} \cdot \frac{z}{2y} $$
Now I multiply the fractions:
$$ \frac{x \cdot z}{y \cdot 2y} $$
$$ \frac{xz}{2y^2} $$
So the quotient of the division is:
$$ \frac{2x}{3y} : \frac{4y}{3z} = \frac{xz}{2y^2} $$
The proof
Consider the general case of dividing two algebraic fractions:
$$ \frac{A}{B} \ : \ \frac{C}{D} $$
where A, B, C, and D are polynomials.
First, I rewrite the division as a single fraction:
$$ \frac{ \frac{A}{B} }{ \ \frac{C}{D} } $$
By the invariance property of fractions, I multiply numerator and denominator by the reciprocal of the denominator, that is by D/C:
$$ \frac{ \frac{A}{B} \cdot \frac{D}{C} }{ \ \frac{C}{D} \cdot \frac{D}{C} } $$
Now I simplify the denominator:
$$ \frac{ \frac{A}{B} \cdot \frac{D}{C} }{ \ \frac{\cancel{C}}{D} \cdot \frac{D}{\cancel{C}} } $$
$$ \frac{ \frac{A}{B} \cdot \frac{D}{C} }{ \ \frac{1}{\cancel{D}} \cdot \frac{\cancel{D}}{1} } $$
$$ \frac{ \frac{A}{B} \cdot \frac{D}{C} }{ \ \frac{1}{1} \cdot \frac{1}{1} } $$
$$ \frac{ \frac{A}{B} \cdot \frac{D}{C} }{ 1 } $$
$$ \frac{A}{B} \cdot \frac{D}{C} $$
This establishes the equivalence between dividing two algebraic fractions and multiplying the first by the reciprocal of the second.
And so on.
