Operations on algebraic fractions

Operations on Algebraic Fractions: Operations on algebraic fractions involve addition, subtraction, multiplication, and division of algebraic expressions in fraction form. Here are the steps to perform each operation:

Addition and Subtraction: The addition and subtraction of algebraic expressions can be calculated using the following steps. Consider the following example:

$\frac{x}{x^2+3x+2} -\frac{x-1}{x^2-1}$

Step 1: Find the LCM of all denominators.

Notice, $x^2+3x+2 = (x+1)(x+2)$ $x^2-1 = (x+1)(x-1)$

For the LCM of $x^2+3x+2$ and $x^2-1$, we have,

L.C.M = Product of Common Factors $\times$ Product of Non Common Factors

The common factor is (x+1) only, and the factors that are not common are $(x-1)$ and $(x-2)$.

$\implies \text{L.C.M} = (x+1)\times(x-1)(x+2)$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= (x+1)(x-1)(x+2)$

Step 2: Now divide the LCM by each denominator and multiply the quotient with the corresponding numerator, i.e.,

$\frac{\text{L.C.M}}{x^2+3x+2} = \frac{(x+1)(x-1)(x+2)}{x^2+3x+2} = (x-1)$

and,

$\frac{\text{L.C.M}}{x^2-1} = \frac{(x+1)(x-1)(x+2)}{x^2-1} = (x+2)$,

therefore,

$\frac{x}{x^2+3x+2} -\frac{x-1}{x^2-1} = \frac{x(x-1) - (x-1)(x+2)}{(x+1)(x-1)(x+2)}$

Step 3: Simply the expression.

$\frac{x}{x^2+3x+2} -\frac{x-1}{x^2-1} = \frac{x(x-1) - (x-1)(x+2)}{(x+1)(x-1)(x+2)}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{x^2-x - (x^2+x-2)}{(x+1)(x-1)(x+2)}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{x^2-x -x^2-x+2}{(x+1)(x-1)(x+2)}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{-2x+2}{(x+1)(x-1)(x+2)}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{-2(x-1)}{(x+1)(x-1)(x+2)}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{-2}{(x+1)(x+2)}$

Hence the sum is $\frac{-2}{(x+1)(x+2)}$.

Multiplication of Algebraic Fractions:

Multiplication of algebraic expressions involves multiplying each term in one expression by each term in the other expression, but before canceling the common factors from the numerator and denominator.

Example: Simplify: $\frac{ab^2+2a}{ab-6+2b-3a}\times \frac{b^2-6b+9}{b^3+2b}$

Solution:

$\frac{ab^2+2a}{ab-6+2b-3a}\times \frac{b^2-6b+9}{b^3+2b} = \frac{ab^2+2a}{ab+2b-3a-6}\times \frac{b^2-6b+9}{b^3+2b}$

Factorize the denominators of both fractions, i.e.,

$\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{a(b^2+2)}{b(a+2)-3(a+2)}\times \frac{b^2-2(b)(3)+3^2}{b(b^2+2)}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{a(b^2+2)}{(b-3)(a+2)}\times \frac{(b-3)^2}{b(b^2+2)}$

Canceling $b^2+2$ and $b-3$ from the numerator and denominator, we get,

$\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{a}{a+2}\times \frac{b-3}{b}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{a(b-3)}{b(a+2)}$

Thus, $\frac{ab^2+2a}{ab-6+2b-3a}\times \frac{b^2-6b+9}{b^3+2b}= \frac{a(b-3)}{b(a+2)}$

Division of Algebraic Expression: In order to divide one expression with another, first change the division to multiplication by taking the reciprocal of the left-hand side term of division, and simply the multiplication.

Example: Simplify $\frac{1}{x^2-9} \div \frac{1}{x+3}$

Solution:

$\frac{1}{x^2-9} \div \frac{1}{x+3} = \frac{1}{x^2-9} \times \frac{x+3}{1}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{1}{(x-3)(x+3)} \times (x+3)$

Cancelling $(x+3)$ in numerator and denominator, we get,

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{1}{x-3}$

Thus, $\frac{1}{x^2-9} \div \frac{1}{x+3}= \frac{1}{x-3}$.