Algebraic fractions are fractional expressions which contain pronumerals in the fractions, e.g. \[ \frac{x}{a^2 - b^2} - \frac{x}{a^2+ab} \ \]
To multiply algebraic fractions, first multiply the numerators. The result is the numerator of the expression. Next multiply the denominators. The result is the denominator of the expression.
Cancel any terms common to both the numerator and denominator.
Example To multiply \[ \begin{align} \frac{x^2-4}{x-2} \quad \text{and} \quad \frac{1}{x+2} \end{align} \ \] means \[ \begin{align} \frac{x^2-4}{x-2}\times \frac{1}{x+2} &= \frac{(x-2)(x+2)}{x-2}\times \frac{1}{x+2}\\ &= 1 \end{align} \ \]
Multiply \(\dfrac{x^2-1}{1} \text{ and } \dfrac{x+1}{x-1}\).
To divide algebraic fractions, invert the divisor and multiply numerators and denominators as before.
Example To divide \[ \begin{align} \frac{x^2-9}{x+1} \quad \text{by} \quad \frac{x+3}{x+1} \end{align} \ \] means \[ \begin{align} \frac{x^2-3}{x+1} \div \frac{x+3}{x+1} &= \frac{(x-3)(x+3)}{x+1}\times \frac{x+1}{x+3}\\ &= x-3 \end{align} \ \]
Divide \(\dfrac{x^2-4}{x+1} \text{ by } \dfrac{x+2}{x+1}\).
Here is a quick review of LCMs and LCDs before dealing with addition and subtraction of algebraic fractions.
LCM and LCD
The Least common Multiple (LCM) of a set of numbers is the product of highest powers of factors of the numbers.
For example, given three integers: 4, 6, 18, they factor as \[ \begin{align} 4 &= 2^2 \\ 6 &= 2 \times 3 \\ 18 &= 2 \times 3^2 \end{align} \ \] so \[ \begin{align} \text{LCM} &= 2^2 \times 3^2 \\ &= 36 \end{align} \ \]
Find the LCM of 8, 9, and 6.
The Least Common Denominator (LCD) of a set of fractions is the LCM of denominators.
If the three factors above were denominators of fractions, then \[ \begin{align} \frac{1}{4} + \frac{1}{6} + \frac{1}{18} &= \frac{9 + 6 + 2}{36} \end{align} \ \]
LCDs of Algebraic Fractions
This is similar to case for integers.
Example Given three algebraic factors \[ \begin{align} &4x^2 \\ &6(x+1)^2 \\ &18(x-2) \end{align} \ \] the LCM is the LCM of the integer factors times the LCM of the algebraic factors \[ \begin{align} \text{LCM} &= 2^2 \times 3^2 \times x^2 \times (x+1)^2 \times (x-2) \\ &= 36 \times x^2 \times (x+1)^2 \times (x-2) \end{align} \ \]
If the three factors are denominators of fractions, then \[ \begin{align} &\frac{1}{4x^2} + \frac{1}{6(x+1)^2} + \frac{1}{18(x-2)} \\ &= \frac{9(x+1)^2(x-2) + 6x^2(x-2) + 2x^2(x+1)^2}{36 \times x^2 \times (x+1)^2 \times (x-2)} \end{align} \ \]
Find the LCM of \(8x^2, 8(x+1) \text{ and } 6(x-2)^2\).
Note - you can use any common denominator, that is, any multiple of the LCM, but the LCD reduces the amount of calculation required.
To add or subtract algebraic fractions:
Guided Examples
O E Qs
Example To add \[ \begin{align} \frac{1}{x-3} \quad \text{and} \quad \frac{2}{x^2-9} \end{align} \ \] factorise and put all terms over a common denominator \[ \begin{align} \frac{1}{x-3} + \frac{2}{(x-3)(x+3)} &= \frac{x+3+2}{(x-3)(x+3)} \\ &= \frac{x+5}{(x-3)(x+3)} \end{align} \ \] after collecting like terms.
Add \(\dfrac{1}{x-2} \text{ and } \dfrac{4x}{x^2-4}\).
Example To subtract \[ \begin{align} \frac{1}{x^2-1} \quad \text{from} \quad \frac{2}{x^2} \end{align} \ \] put all terms over a common denominator \[ \begin{align} \frac{2}{x^2} - \frac{1}{x^2-1} &= \frac{2}{x^2}- \frac{1}{(x-1)(x+1)} \\ &= \frac{2(x-1)(x+1) - x^2}{x^2(x-1)(x+1)} \\ &= \frac{x^2-2}{x^2(x-1)(x+1)} \end{align} \ \] after expanding and collecting like terms in the numerator.
Subtract \(\dfrac{1}{x^2-2} \text{ from } \dfrac{3}{x^2}\).
Example To simplify \[ \begin{align} &\qquad \frac{x}{a^2 - b^2} - \frac{x}{a^2+ab} \\ \end{align} \ \] put all terms over a common denominator \[ \begin{align} &=\frac{a^2x+abx-a^2x+b^2x}{(a^2-b^2)(a^2+ab)} \\ &=\frac{abx+b^2x}{(a^2-b^2)(a^2+ab)} \end{align} \ \]
Factorise numerator and denominator \[ =\frac{bx(a+b)}{(a-b)(a+b)(a^2+ab)} \ \]
Cancel any common factors \[ =\frac{bx}{a(a-b)(a+b)} \quad \]
A compound fraction occurs when either or both numerator and denominator are fractions themselves, eg \[ \frac{x - \dfrac{1}{x}}{x + \dfrac{1}{x}} \ \]
To simplify compound fractions :
Example To simplify the example above, put numerator and denominator over their common denominators \[ \begin{align} &\qquad \frac{x - \dfrac{1}{x}}{x + \dfrac{1}{x}} \qquad \quad \\ &=\frac{\dfrac{x^2 - 1}{x}}{\dfrac{x^2 + 1}{x}} \end{align}\ \]
Divide the numerator by the denominator (multiply the numerator by the reciprocal of the denominator) \[ = \frac{x^2 - 1}{x}.\frac{x}{x^2 + 1} \quad \]
Cancel any common factors \[ = \frac{x^2 - 1}{x^2 + 1} \qquad \quad \ \]
Simplify \(\dfrac{2x+1/x}{x+1/2x}\).