Algebraic Fraction

A fraction of numbers, like $\dfrac34$, is one number divided by another. An algebraic fraction is the same thing with letters in it: a polynomial divided by a polynomial, such as

$$\frac{x}{y}, \qquad \frac{a + 3}{a^2 + 5a + 6}, \qquad \frac{d}{30} + \frac{d}{60}$$

It follows every rule that number fractions follow, because the letters are numbers, just numbers we haven’t been told yet. Put a = 2 in $\dfrac{a + 3}{a^2 + 5a + 6}$ and you get $\dfrac{5}{20}$, an ordinary fraction.

One rule is new: a fraction means nothing when its denominator is 0. $\dfrac{1}{x - 3}$ has no value at x = 3, so whatever we do with it, we keep in mind that x ≠ 3.

Lowest terms: cancel factors

A fraction is in its lowest terms when its numerator and denominator have no common factor. With numbers you divide both by their highest common factor: $\dfrac{12}{18} = \dfrac{2 \times 6}{3 \times 6} = \dfrac23$. With letters, factorize both first, then cancel the common factors:

$$\frac{a + 3}{a^2 + 5a + 6} = \frac{a + 3}{(a + 2)(a + 3)} = \frac{1}{a + 2}$$

Check with a = 2: $\dfrac{5}{20} = \dfrac14$ and $\dfrac{1}{2 + 2} = \dfrac14$ ✓.

To simplify a sum or difference of algebraic fractions is to write it as one fraction in its lowest terms.

First look at the denominators. Fractions with the same denominator are like fractions: add or subtract the numerators and keep the one denominator. Fractions with different denominators are unlike fractions: first rewrite them over a common denominator, the LCM of the denominators.

a. Like fractions:

$$\frac{x}{x - y} + \frac{y}{x - y} = \frac{x + y}{x - y}$$

b. Unlike fractions. The denominators (x − y) and (x + y) have no common factor, so their LCM is their product. Multiply the top and bottom of each fraction by the other denominator:

$$\begin{aligned} \frac{x}{x - y} - \frac{y}{x + y} &= \frac{x(x + y)}{(x - y)(x + y)} - \frac{y(x - y)}{(x + y)(x - y)} \\ &= \frac{x^2 + xy - xy + y^2}{(x + y)(x - y)} \\ &= \frac{x^2 + y^2}{x^2 - y^2} \end{aligned}$$

c. Factorize first: $a^2 - b^2 = (a - b)(a + b)$. The first denominator (a − b) is already inside the second, so the LCM is (a − b)(a + b), and only the first fraction needs to be multiplied, by (a + b):

$$\begin{aligned} \frac{1}{a - b} - \frac{b}{(a - b)(a + b)} &= \frac{(a + b) - b}{(a - b)(a + b)} \\ &= \frac{a}{(a - b)(a + b)} = \frac{a}{a^2 - b^2} \end{aligned}$$

The steps, the same as for number fractions:

1. Factorize every denominator (and numerator, if it helps).
2. Cancel any fraction that can be reduced on its own; it makes the rest shorter.
3. Find the LCM of the denominators (Figure 8.1): every factor that appears, each as many times as it appears in the denominator that has it most often.
4. Multiply each numerator by the factors its denominator is missing; write everything over the LCM.
5. Multiply out the top, collect like terms, and factorize the result to see if anything cancels.

The same LCM you used for $\dfrac14 + \dfrac16$: 4 is 2 × 2 and 6 is 2 × 3, so the LCM is 2 × 2 × 3 = 12, not 4 × 6 = 24. Multiplying the denominators always gives a common denominator, but often a bigger one than you need, and the extra factor must be canceled at the end.

Worked examples

How can you be sure an answer like that is right? Put in a number. If the working is right, the first expression and the answer give the same value for every allowed a; if there is a slip, almost any number shows it. With a = 4: $\dfrac{3}{3} + \dfrac{2}{-4} + \dfrac{-1}{-1} = 1.5$, and $\dfrac{3 \times (-1)}{1 \times (-2)} = 1.5$ ✓.

Back to the bus

The bus’s average speed is 40 km/h, whatever the length of the road.

Here is why. Let the road be d km long. Time is distance ÷ speed, so the climb takes $\dfrac{d}{30}$ hours and the run down $\dfrac{d}{60}$ hours. The whole trip is 2d km, so

$$\begin{aligned} \text{average speed} &= \frac{2d}{\dfrac{d}{30} + \dfrac{d}{60}} \\ &= \frac{2d}{\dfrac{2d + d}{60}} = \frac{2d}{\dfrac{3d}{60}} \\ &= 2d \times \frac{60}{3d} = 40 \text{ km/h} \end{aligned}$$

The d cancels, so the length of the road doesn’t matter. With any two speeds u and v the same steps give

$$\text{average speed} = \frac{2d}{\dfrac{d}{u} + \dfrac{d}{v}} = \frac{2uv}{u + v}$$

What it means: the bus spends twice as long climbing as coming down, so the slow speed counts twice as much in the average, which is why it is nearer 30 than 60. A round trip never averages the two speeds simply unless they are equal. Mathematicians call $\dfrac{2uv}{u + v}$ the harmonic mean of the two speeds.

In the SEE’s Compulsory Mathematics paper, Algebra carries 15 marks in three questions (CDC’s specification grid). In CDC’s model paper this unit shares a 5-mark question with Unit 9:

Part	Level	Marks	What it asks
a.	application	3	solve an exponential equation (Unit 9)
b.	understanding	2	simplify a sum and difference of three algebraic fractions

The 2 marks of part b are usually one for the LCM step and one for the simplified answer. Here is a question in that shape.

Question 1

Change into the lowest terms.

Question 2

Simplify.

Question 3

Simplify.

Questions 4 and 5