1. Introduction
Algebraic fractions have properties which are the same as those for numerical fractions, the only difference being that the numerator (top) and denominator (bottom) are both algebraic expressions.
Example 1: Simplify the following fractions:
Solution:
$$ \frac{2b}{7b^2} = \frac {2 \redcancel b}{7b \redcancel b} = \frac{2}{7b} $$Solution:
$$ \begin{eqnarray} \frac{3x+x^2}{6x^2} & = & \frac{ x \times (3+x)^\ }{x \times 6x} \\ & = & \frac{ \redcancel x \times (3+x)}{ \redcancel x \times 6x} = \frac{3+x}{6x} \end{eqnarray} $$Note that the cancellation is allowed since $x$ is a common factor of the numerator and denominator.
Sometimes a little more work is necessary before an algebraic fraction can be reduced to a simpler form.
Example 2: Simplify the algebraic fraction:
$$ \frac{x^2 - 2x + 1}{x^2 + 2x - 3} $$Solution:
In this case, the numerator and denominator can be factored into two terms (see the module on Simplifying Expressions):
$$ x^2-2x+1 = (x-1)^2 ~~ \text{and} ~~ x^2+2x-3 = (x-1)(x+3) $$With this established the simplification proceeds as follows:
$$ \begin{eqnarray} \frac{x^2-2x+1}{x^2+2x-3} & = & \frac{(x-1) \times \redcancel{(x-1)}}{(x+3) \times \redcancel{(x-1)}} \\ & = & \frac{(x-1)}{(x+3)} \end{eqnarray} $$Exercise 1: Simplify each of the following algebraic fractions.
Solution:
$$ \begin{eqnarray} \frac{8y}{2y^3} & = & \frac{4 \times 2y}{y^2 \times 2y} \\ & = & \frac{4 \times \redcancel 2 \times \redcancel y}{y^2 \times \redcancel 2 \times \redcancel y} \\ & = & \frac{4}{y^2} \end{eqnarray} $$Solution:
$$ \begin{eqnarray} \frac{2y}{4x} & = & \frac{2 \times y}{2 \times 2x} \\ & = & \frac {\redcancel 2 \times y}{\redcancel 2 \times 2x} = \frac{y}{2x} \end{eqnarray} $$Solution: This time, instead of expanding the factors, it is easier to use the rule for powers (see the module on Powers):
$$ \frac{a^m}{a^n} = a^{m-n} $$Thus:
$$ \begin{eqnarray} \frac{7a^6b^3}{14a^5b^4 } & = & \frac{7}{14} \times \frac{a^6}{a^5} \times \frac{b^3}{b^4} \\ & = & \frac{1}{2} \times a^{6-5} \times b^{3-4} \\ & = & \frac{1}{2} \times a^1 \times b^{-1} = \frac{a}{2b} \end{eqnarray} $$Solution:
$$ \begin{eqnarray} \frac {(2x)^2}{4x} & = & \frac {2 \times x \times 2 \times x}{2 \times 2 \times x} \\ & = & \frac {\redcancel 2 \times \redcancel x \times \redcancel 2 \times x}{\redcancel 2 \times \redcancel 2 \times x} = x \end{eqnarray} $$Solution:
$$ \begin{eqnarray} \frac{5y+2y^2}{7y} & = & \frac{y \times (5+2y)}{7 \times y} \\ & = & \frac{\redcancel y \times (5+2y)}{7 \times \redcancel y} = \frac{5+2y}{7} \end{eqnarray} $$Solution:
$$ \begin{eqnarray} \frac {5ax}{15a+10a^2} & = & \frac {5 \times a \times x}{5 \times a \times (3+2a)} \\ & = & \frac {\redcancel 5 \times \redcancel a \times x }{ \redcancel 5 \times \redcancel a \times (3 + 2a)} = \frac{x}{3+2a} \end{eqnarray} $$Solution:
$$ \begin{eqnarray} \frac{2z^2−4z}{2z^2 −10z} & = & \frac {2 \times z \times (z-2)}{2 \times z \times (z-5)} \\ & = & \frac {\redcancel 2 \times \redcancel z \times (z-2)}{\redcancel 2 \times \redcancel z \times (z-5)} = \frac{z-2}{z-5} \end{eqnarray} $$Solution: In this case some initial factorisation is needed (see the module on Factorising Expressions).
$$ y^2+7y+10 = (y+5)(y+2) ~~ \text{and} ~~ y^2−25=(y+5)(y−5) $$Thus:
$$ \begin{eqnarray} \frac{y^2 +7y+10}{y^2 −25} & = & \frac {\redcancel{(y+5)}(y+2)}{\redcancel{(y+5)}(y-5)} = \frac {y+2}{y-5} \\ \end{eqnarray} $$Solution: Again, in this case, some initial factorisation is needed (see the module on Factorising Expressions).
$$ w^2−5w−14 = (w−7)(w+2) ~~ \text{and} ~~ w^2−4w−21 = (w−7)(w+3) $$Thus:
$$ \begin{eqnarray} \frac {w^2−5w−14}{w^2−4w−21} & = & \frac {\redcancel{(w-7)}(w+2)}{\redcancel{(w-7)}(w+3)} = \frac {w+2}{w+3} \\ \end{eqnarray} $$Click on questions to reveal their solutions
Now try this short quiz:
Quiz 1: Which of the following is a simplified version of $ \displaystyle \frac{t^2+3t−4}{t^2−3t+2} $ ?
Explanation: The numerator and denominator respectively factorise as:
$$ t^2+3t−4=(t−1)(t+4) ~~ \text{and} ~~ t^2−3t+2=(t−1)(t−2) $$So that:
$$ \begin{eqnarray} \frac{t^2+3t−4}{t^2−3t+2} & = & \frac {\redcancel{(t-1)}(t+4)}{\redcancel{(t-1)}(t-2)} = \frac {t+4}{t-2} \end{eqnarray} $$So far, simplification has been acheived by cancelling common factors from the numerator and denominator. There are fractions which can be simplified by multiplying the numerator and denominator by an appropriate common factor, thus obtaining an equivalent, simpler expression.
Example 3: Simplify the following fractions:
(a) $ \displaystyle \frac { \frac 1 4 + y}{\frac 1 2} $
Solution: In this case, multiplying both the numerator and denominator by 4 gives:
$$ \frac { \frac 1 4 + y}{\frac 1 2} = \frac { 4(\frac 1 4 +y)}{4(\frac 1 2)} = \frac {1+4y}{2} $$(b) $ \displaystyle \frac { 3x + \frac 1 x}{2} $
Solution: To simplify this expression, multiply the numerator and denominator by $x$. Thus:
$$ \frac { 3x + \frac 1 x}{2} = \frac { x(3x + \frac 1 x)}{2x} = \frac { 3x^2 + 1}{2x} $$Now try this exercise on similar examples:
Exercise 2: Simplify each of the following algebraic fractions.
Solution: The fraction is simplified by multiplying both the numerator and denominator by $ \blue 2 $:
$$ \frac {4y - \frac 3 2}{2} = \frac {\blue 2(4y - \frac 3 2)}{\blue 2 \times 2} = \frac {8y - 3}{4} $$Solution: This fraction is simplified by multiplying both the numerator and denominator by $ \blue 4 $:
$$ \frac {2x + \frac 1 2}{x + \frac 1 4} = \frac {\blue 4(2x + \frac 1 2)}{\blue 4(x + \frac 1 4)} = \frac {8x+2}{4x+1} $$Solution: In this case, since the numerator contains the fraction $ \frac 1 3 $ and the denominator contains the fraction $ \frac 1 2 $, the common factor needed is $ 2 \times 3 = \blue 6 $. Thus:
$$ \frac {z - \frac 1 3}{z - \frac 1 2} = \frac {\blue 6(z - \frac 1 3)}{\blue 6(z - \frac 1 2)} = \frac {6z - 2}{6z - 3} $$Solution: For this fraction the required multiplier is $ \blue x $:
$$ \frac {2 - \frac 1 x}{2} = \frac {\blue x(2 - \frac 1 x)}{2\blue x} = \frac {2x - 1}{2x} $$Solution: Here the numerator contains the fraction $ \frac 2 t $ and the denominator is the fraction $ \frac 1 2 $, so the required multiplier is $\blue{2t}$:
$$ \frac {3t - \frac 2 t}{\frac 1 2} = \frac {\blue{2t}(3t - \frac 2 t)}{\blue{2t}\frac 1 2} = \frac {6t^2 - 4}{t} $$Solution: For the last past of this exercise, since the numerator includes the fraction $ \frac 1 {2z} $ and the denominator includes the fraction $ \frac 1 {3z} $, the common multiplier is $ \blue{6z} $:
$$ \frac {z - \frac 1 {2z}}{z - \frac 1 {3z}} = \frac {\blue {6z}(z - \frac 1 {2z})}{\blue{6z}(z - \frac 1 {3z})} = \frac {6z^2 - 3}{6z^2 - 2} $$Click on questions to reveal their solutions
For the last part of this section, try the following short quiz.
Quiz 2: Which of the following is a simplified version of $ ~ \displaystyle \frac {x - \frac 1 {x+1}}{x-1} $ ?
Explanation: For $ \displaystyle \frac {x - \frac 1 {x+1}}{x-1} $, the common multiplier is $ \blue{x+1} $. Multiplying the numerator and denominator by this gives:
$$ \begin{eqnarray} \frac {x - \frac 1 {x+1}}{x-1} & = & \frac {\blue{(x+1)}(x - \frac 1 {x+1})}{\blue{(x+1)}(x-1)} \\ & = & \frac {(x+1)x - (x + 1)(\frac 1 {x+1})}{(x^2-1)} = \frac {x^2+x-1}{x^2-1} \end{eqnarray} $$2. Addition of Algebraic Fractions
Addition (and subtraction) of algebraic fractions proceeds in exactly the same manner as for numerical fractions.
Example 4: Write the following sum as a single fraction in is simplest form:
$$ \frac 2 {x+1} + \frac 1 {x+2} $$Solution: The least common multiple of the denominators (see the module on Fractions) is $ \blue{(x+1)(x+2)} $. Thus:
$$ \begin{eqnarray} \frac 2 {x+1} + \frac 1 {x+2} & = & \frac {2 \times \blue{(x+2)}}{(x+1) \times \blue{(x+2)}} + \frac {1 \times \blue{(x+1)}}{(x+2) \times \blue {(x+1)}} \\ & = & \frac {2x+4}{(x+1)(x+2)} + \frac {x+1}{(x+2)(x+1)} \\ & = & \frac {(2x+4)+(x+1)}{(x+1)(x+2)} = \frac {3x+5}{(x+1)(x+2)} \end{eqnarray} $$Exercise 3: Evaluate each of the following fractions.
Solution: The least common multiple of the denominators is $ \blue{yz} $. Thus:
$$ \begin{eqnarray} \frac 2 y + \frac 3 z & = & \frac {2 \times \blue{z}}{y \times \blue{z}} + \frac {3 \times \blue{y}}{z \times \blue {y}} \\ & = & \frac {2z} {yz} + \frac {3y} {yz} = \frac {3y+2z}{yz} \end{eqnarray} $$Solution: Here the least common multiple of the denominators is $ \blue{15y} $, so
$$ \begin{eqnarray} \frac 1 {3y} - \frac 2 {5y} & = & \frac {\blue{5} \times 1}{\blue{5} \times 3y} - \frac {\blue{3} \times 2}{\blue{3} \times 5y} \\ & = & \frac 5 {15y} - \frac 6 {15y} \\ & = & \frac {5-6}{15y} = - \frac 1 {15y} \end{eqnarray} $$Solution: The least common multiple of the denominators of the two fractions in this case is $ \blue{6} $. Thus:
$$ \begin{eqnarray} \frac {3z+1} 3 - \frac {2z+1} 2 & = & \frac {\blue{2} \times (3z+1)}{\blue{2} \times 3} - \frac {\blue{3} \times (2z+1)}{\blue{3} \times 2} \\ & = & \frac {6z+2} 6 - \frac {6z+3} 6 \\ & = & \frac {(6z+2) - (6z+3)} 6 = - \frac 1 6 \end{eqnarray} $$Simplification in this case has shown that the difference of these two fractions is independent of $z$.
Solution: There least common multiple of the denominators of the two fractions in this case is $ \blue{2t} $. The sum simplifies as follows:
$$ \begin{eqnarray} \frac {3t+1} 2 + \frac 1 t & = & \frac {\blue{t} \times (3t+1)}{\blue{t} \times 2} + \frac {\blue{2} \times 1}{\blue{2} \times t} \\ & = & \frac {3t^2+t} {2t} + \frac 2 {2t} = \frac {3t^2+t+2} {2t} \end{eqnarray} $$Solution: Here, the required least common multiple of the denominators is the factor $ \blue{2(x-1)} $. Proceeding as before:
$$ \begin{eqnarray} \frac {x+1} 2 + \frac 1 {x-1} & = & \frac {\blue{(x-1)} \times (x+1)}{\blue{(x-1)} \times 2} + \frac {\blue{2} \times 1}{\blue{2} \times (x-1)} \\ & = & \frac {(x^2-1)} {2(x-1)} + \frac 2 {2(x-1)} \\ & = & \frac {(x^2-1)+2} {2(x-1)} = \frac {x^2+1} {2(x-1)} \end{eqnarray} $$Solution: Here the least common multiple of the denominators is $ \blue{(w+3)(w−1) = w^2+2w−3} $:
$$ \begin{eqnarray} \frac 2 {w+3} - \frac 5 {w-1} & = & \frac {\blue{(w-1)} \times 2}{\blue{(w-1)} \times (w+3)} - \frac {\blue{(w+3)} \times 5}{\blue{(w+3)} \times (w-1)} \\ & = & \frac {2w-2} {(w+3)(w-1)} - \frac {5w+15} {(w+3)(w-1)} \\ & = & \frac {(2w-2)-(5w+15)} {(w+3)(w-1)} \\ & = & \frac {-3w-17} {(w+3)(w-1)} = - \left( \frac {(3w+17)} {(w+3)(w-1)} \right) \end{eqnarray} $$Click on questions to reveal their solutions
Quiz 3: Which of the following values of $ \blue{a} $ is needed if:
$$ \frac {\blue{a}} {2x+1} + \frac 1 {x+2} = \frac {4x+5} {(2x+1)(x+2)} ~ \text{?} $$Explanation: Writing all the fractions with a common denominator:
$$ \begin{eqnarray} \frac {4x+5} {(2x+1)(x+2)} & = & \frac {\blue a} {2x+1} + \frac 1 {x+2} \\ & = & \frac {\blue a(x+2)} {(2x+1)(x+2)} + \frac {(2x+1)} {(2x+1)(x+2)} \\ & = & \frac {\blue ax +2 \blue a +2x+1} {(2x+1)(x+2)} \\ & = & \frac {(\blue a+2)x +(2\blue a +1)} {(2x+1)(x+2)} \end{eqnarray} $$so that $ (\blue a+2)x+(2 \blue a+1) = 4x+5 $. This gives two equations
$$ \begin{eqnarray} \blue a+2 & = & 4 & ~~~\text{(coefficients of } x \text{)}\\ 2\blue a+1 & = & 5 & ~~~\text{(constant coefficients)} \end{eqnarray} $$The solution to both equations is $ \blue a = 2 $.
3. Simple Partial Fractions
The last quiz was an example of partial fractions, i.e. the technique of decomposing a fractions as a sum of simpler fractions. This section will consider the simpler forms of this technique.
Example 5: Find the partial fraction decomposition of $ 4 / (x^2-4) $.
Solution: The denominator factorises as $ x^2-4 = (x-2)(x+2) $ (see the module on Quadratics). The partial fractions will, therefore, be of the form $ a / (x-2) $ and $ b / (x+2) $. Thus:
$$ \begin{eqnarray} \frac a {x-2} + \frac b {x+2} & = & \frac 4 {(x-2)(x+2)} \\ \frac {a(x+2)} {(x−2)(x+2)} + \frac {b(x−2)} {(x+2)(x−2)} & = & \frac 4 {(x-2)(x+2)} \\ \frac {(a+b)x+2(a−b)} {(x−2)(x+2)} & = & \frac 4 {(x-2)(x+2)} \\ \text{so that: } (a+b)x+2(a−b) & = & 4 \end{eqnarray} $$This enables $a$ and $b$ to be found. For the equation to be true for all values of $x$ the coefficients must match, i.e.
$$ \begin{eqnarray} a+b & = & 0 & ~~~\text{(coefficients of } x \text{)}\\ 2a-2b & = & 4 & ~~~\text{(constant coefficients)} \end{eqnarray} $$- where the first equation holds since there is no $x$ term in $ 4 / (x^2-4) $. This set of simultaneous equations may be solved to give $ a = 1 $ and $ b = -1 $ (see the module on Simultaneous Equations for a method of finding these equations). Thus:
$$ \frac 4 {(x-2)(x+2)} = \frac 1 {x-2} - \frac 1 {x+2} $$Exercise 4: For each of the following, find $a$ and $b$:
Solution: Taking common denominators:
$$ \begin{eqnarray} \frac a {x-2} + \frac b {x+2} & = & \frac {4x}{(x-2)(x+2)} \\ \frac {a(x+2)} {(x-2)(x+2)} + \frac {b(x-2)} {(x-2)(x+2)} & = & \frac {4x}{(x-2)(x+2)} \\ \frac {(a+b)x + (2a-2b)} {(x-2)(x+2)} & = & \frac {4x}{(x-2)(x+2)} \\ \text{so that } (a + b)x + (2a − 2b) & = & 4x \end{eqnarray} $$Equating coefficients in this case gives:
$$ \begin{eqnarray} a+b & = & 4 & ~~~\text{(coefficients of } x \text{)}\\ 2a-2b & = & 0 & ~~~\text{(constant coefficients)} \end{eqnarray} $$Solving this set of equations gives $ a = 2$, $ b = 2 $. Hence:
$$ \frac 2 {x-2} + \frac 2 {x+2} = \frac {4x} {(x-2)(x+2)} $$Solution: Taking common denominators:
$$ \begin{eqnarray} \frac a {z-3} + \frac b {z+2} & = & \frac {z+7}{(z-3)(z+2)} \\ \frac {a(z+2)} {(z-3)(z+2)} + \frac {b(z-3)} {(z-3)(z+2)} & = & \frac {z+7}{(z-3)(z+2)} \\ \frac {(a+b)z + (2a-3b)} {(z-3)(z+2)} & = & \frac {z+7}{(z-3)(z+2)} \\ \text{so that } (a + b)z + (2a − 3b) & = & z+7 \end{eqnarray} $$Equating coefficients in this case gives:
$$ \begin{eqnarray} a+b & = & 1 & ~~~\text{(coefficients of } x \text{)}\\ 2a-3b & = & 7 & ~~~\text{(constant coefficients)} \end{eqnarray} $$Solving this set of equations gives $ a = 2$, $ b = -1 $. Hence:
$$ \frac 2 {z-3} - \frac 1 {z+2} = \frac {z+7} {(z-3)(z+2)} $$Solution: Taking common denominators:
$$ \begin{eqnarray} \frac a {w-4} + \frac b {w+1} & = & \frac {3w-2}{(w-4)(w+1)} \\ \frac {a(w+1)} {(w-4)(w+1)} + \frac {b(w-4)} {(w-4)(w+1)} & = & \frac {3w-2}{(w-4)(w+1)} \\ \frac {(a+b)w + (a-4b)} {(w-4)(w+1)} & = & \frac {3w-2}{(w-4)(w+1)} \\ \text{so that } (a + b)w + (a − 4b) & = & 3w-2 \end{eqnarray} $$Equating coefficients in this case gives:
$$ \begin{eqnarray} a+b & = & 3 & ~~~\text{(coefficients of } x \text{)}\\ a-4b & = & -2 & ~~~\text{(constant coefficients)} \end{eqnarray} $$Solving this set of equations gives $ a = 2$, $ b = 1 $. Hence:
$$ \frac 2 {w-4} + \frac 1 {w+1} = \frac {3w-2} {(w-4)(w+1)} $$Click on questions to reveal their solutions
Now try this short quiz:
Quiz 4: If $ \displaystyle \frac a {2x-3} + \frac b {3x+4} = \frac {x+7} {(2x-3)(3x+4)} $, which of the following is the solution to the equation?
Explanation: Writing all the fractions with a common denominator:
$$ \begin{eqnarray} \frac a {2x-3} + \frac b {3x+4} & = & \frac {x+7} {(2x-3)(3x+4)} \\ \frac {a(3x+4)} {(2x-3)(3x+4)} + \frac {b(2x-3)} {(2x-3)(3x+4)} & = & \frac {x+7} {(2x-3)(3x+4)} \\ \frac {(3a+2b)x+(4a-3b)} {(2x-3)(3x+4)} & = & \frac {x+7} {(2x-3)(3x+4)} \\ \text{so that } (3a+2b)x + (4a−3b) & = & x+7 \end{eqnarray} $$This gives two equations:
$$ \begin{eqnarray} 3a+2b & = & 1 & ~~~\text{(coefficients of } x \text{)}\\ 4a-3b & = & 7 & ~~~\text{(constant terms)} \end{eqnarray} $$Solving these gives: $ a = 1$, $ b = -1 $.
4. Quiz on Algebraic Fractions
In each of the following choose...