DIVISION OF ALGEBRAIC FRACTION

Like common fractions, algebraic fractions are also divisible. The process is the same as common fractions. 

First, you will Invert the fraction. Next, you Change $÷$ to $\times$. Thirdly, You multiply the fraction. And lastly, you reduce the fraction, if there are any common factors.

The process is illustrated below

๐Ÿ‘‰This post extends our posts on algebraic fractions ๐Ÿ‘ˆ

Example 1
Write $\frac{2y}{2y^2+3y-5}÷\frac{y^3}{y^2-y}$ in the simplest form possible.

Solution: 
First, we invert the fraction and change the division sign$(÷)$ to multiplication.

$\frac{2y}{2y^2+3y-5}\times\frac{y^2-y}{y^3}$

Multiply numerators and denominators
$\frac{(2y)(y^2-y)}{(2y^2-3y-5)(y^3)}$

By factorization
$\frac{(2)(y)(y)(y-1)}{(2y+5)(y-1)(y)(y)(y)}$
$\frac{(2)\require{cancel}\bcancel{(y)(y)(y-1)}}{(2y+5)\require{cancel}\bcancel{(y)(y)(y-1)}(y)}$
$\frac{2}{y(2y+5)}$

Example 2
Simplify $\frac{2x^2+9x+4}{x-3}÷\frac{8x+4}{x^2-9}$

Solution:
Invert the fraction
$\frac{2x^2+9x+4}{x-3}\times\frac{x^2-9}{8x+4}$

$x^2-9$ is difference of two squares.
$\frac{2x^2+9x+4}{x-3}÷\frac{(x+3)(x-3)}{8x+4}$
$\frac{(2x^2+9x+4)(x+3)(x-3)}{(x-3)(8x+4)}$
$\frac{(2x+1)(x+4)(x+3)(x-3)}{(x-3)(4)(2x+1)}$
$\frac{\require{cancel}\bcancel{(2x+1)(x-3)}(x+4)(x+3)}{\require{cancel}\bcancel{(2x+1)(x-3)}(4)}$
$\frac{(x+4)(x+3)}{4}$

Example 3
Write $\frac{x^2-7x+12}{x-3}÷\frac{x-4}{x^2}$ in the simplest form possible.

Solution:
Overturn the fraction
$\frac{x^2-7x+12}{x-3}\times\frac{x^2}{x-4}$

$\frac{x^2-7x+12)(x^2)}{(x-3)(x-4)}$
$\frac{(x-3)(x-4)(x^2)}{(x-3)(x-4)}$
$\frac{\require{cancel}\bcancel{(x-3)(x-4)}(x^2)}{\require{cancel}\bcancel{(x-3)(x-4)}}$
$\frac{x^2}{1}$
$x^2$

Example 4
Solve $\frac{x^3-2x^2-13x-10}{x^2}÷\frac{x^2+3x+2}{x^4}$

Solution:
First, we overturn the fraction
$\frac{x^3-2x^2-13x-10}{x^2}\times\frac{x^4}{x^2+3x+2}$

Multiplications of numerators and denominators
$\frac{(x^3-2x^2-13x-10)(x^4)}{(x^2)(x^2+3x+2)}$

By factorizing
$\frac{(x-5)(x+1)(x+2)(x^2)(x^2)}{(x^2)(x+1)(x+2)}$
$\frac{\require{cancel}\bcancel{(x+1)(x+2)(x^2)}(x-5)(x^2)}{\require{cancel}\bcancel{(x+1)(x+2)(x^2)}}$.
$\frac{(x-5)(x^2)}{1}$
$(x-5)(x^2)$

Related post

Example 5
Write $\frac{2x+1}{8x^2}÷\frac{2x+1}{x^2}$ in the simplest form

Solution:
Invert the fractions
$\frac{2x+1}{8x^2}\times\frac{x^2}{2x+1}$

$\frac{(2x+1)(x^2)}{(8)(x^2)(2x+1)}$
$\frac{\require{cancel}\bcancel{(2x+1)(x^2)}}{\require{cancel}\bcancel{(2x+1)(x^2)}(8)}$
$\frac{1}{8}$
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