Simplifying Algebraic Fractions

What an algebraic fraction is, when it is undefined, how to find its value, how to make equivalent fractions, and how to simplify a fraction by cancelling the factors its numerator and denominator have in common.

Before you start

A fraction is in its simplest form when its numerator and denominator have no common factor except $1$ . To get there, divide both by their highest common factor (HCF): $\dfrac{12}{18} = \dfrac{12 \div 6}{18 \div 6} = \dfrac{2}{3}\text{.}$

In algebra, a power is a product: $x^3 = x \times x \times x$ and $5y^2 = 5 \times y \times y$ . This lesson cancels letters by writing them out like this.

What Is an Algebraic Fraction?

You know fractions of numbers, such as $\dfrac{3}{5}$ and $\dfrac{12}{7}$ . An algebraic fraction has a letter in its numerator, its denominator or both:

$\dfrac{2y}{5}$ (a letter in the numerator), $\dfrac{16}{4h + k}$ (letters in the denominator), $\dfrac{7x}{(2x + 1)(3x - 1)}$ and $\dfrac{3f^2}{f^2 - 1}$ (letters in both).

An algebraic fraction follows the same rules as a fraction of numbers, because the letters stand for numbers. When you replace each letter by a number, the fraction becomes an ordinary fraction with a value.

Example 1: Finding the Value of an Algebraic Fraction

(a) Find the value of $\dfrac{2x + 3}{x - 1}$ when $x = 3$ and when $x = -2$ . (b) Find the value of $\dfrac{a^2 - b}{2ab}$ when $a = -2$ and $b = 3$ .

Solution

(a) When $x = 3$ : $\dfrac{2(3) + 3}{3 - 1} = \dfrac{9}{2} = 4\tfrac{1}{2}\text{.}$

When $x = -2$ : $\dfrac{2(-2) + 3}{-2 - 1} = \dfrac{-1}{-3} = \dfrac{1}{3}\text{.}$ A negative divided by a negative is positive.

(b) Put brackets round each negative number before you square it: $a^2 = {(-2)}^2 = 4$ .

$\dfrac{a^2 - b}{2ab} = \dfrac{4 - 3}{2 \times (-2) \times 3} = \dfrac{1}{-12} = -\dfrac{1}{12}\text{.}$

(a) $4\tfrac{1}{2}$ when $x = 3$ , and $\dfrac{1}{3}$ when $x = -2$ . (b) $-\dfrac{1}{12}$

When Is an Algebraic Fraction Undefined?

You cannot divide by zero: $\dfrac{5}{0}$ has no value. So an algebraic fraction has no value for any number that makes its denominator zero. We say the fraction is undefined for that value.

For example, $\dfrac{5}{x - 3}$ is undefined when $x = 3$ , because then the denominator is $3 - 3 = 0$ .

Undefined values

An algebraic fraction is undefined when its denominator is equal to $0$ .

To find these values, write the equation denominator $= 0$ and solve it.

If the denominator is a product, it is $0$ when any one of its factors is $0$ .

Example 2: Finding Where a Fraction Is Undefined

For what value of the letter is each fraction undefined? (a) $\dfrac{2x + 5}{3x - 9}$ (b) $\dfrac{5}{2x + 3}$ (c) $\dfrac{4}{3y}$

Solution

(a) Solve $3x - 9 = 0$ : $3x = 9$ , so $x = 3$ . Check: $3(3) - 9 = 0$ .

(b) Solve $2x + 3 = 0$ : $2x = -3$ , so $x = -\dfrac{3}{2}$ . The answer need not be a whole number. Check: twice $-\dfrac{3}{2}$ is $-3$ , and $-3 + 3 = 0$ .

The numerator does not matter: only the denominator can make a fraction undefined.

(a) $x = 3$ (b) $x = -1\tfrac{1}{2}$ (c) $y = 0$

Example 3: A Denominator That Is a Product

For what values of $x$ is $\dfrac{7x}{(2x + 1)(3x - 1)}$ undefined?

Solution

The denominator is ${(2x + 1)} \times {(3x - 1)}$ . A product of two numbers is $0$ when one of them is $0$ .

$2x + 1 = 0$ gives $2x = -1$ , so $x = -\dfrac{1}{2}$ .

$3x - 1 = 0$ gives $3x = 1$ , so $x = \dfrac{1}{3}$ .

Check $x = \dfrac{1}{3}$ : the first bracket is $\dfrac{2}{3} + 1 = \dfrac{5}{3}$ and the second is $1 - 1 = 0$ , so the product is $\dfrac{5}{3} \times 0 = 0$ .

The fraction is undefined when $x = -\dfrac{1}{2}$ or $x = \dfrac{1}{3}$ .

Zero is not the same as undefined

A fraction is equal to zero when its numerator is $0$ and its denominator is not: $\dfrac{0}{5} = 0$ .

A fraction is undefined when its denominator is $0$ : $\dfrac{5}{0}$ has no value.

Example 4: Zero or Undefined?

Consider $\dfrac{x - 2}{x + 3}$ . (a) For what value of $x$ is it equal to $0$ ? (b) For what value of $x$ is it undefined?

Solution

(a) The numerator is $0$ when $x = 2$ . Then the denominator is $2 + 3 = 5$ , which is not $0$ . So the fraction is $\dfrac{0}{5} = 0$ .

(b) The denominator is $0$ when $x + 3 = 0$ , that is, when $x = -3$ . Then the fraction is $\dfrac{-5}{0}$ , which has no value.

(a) It is $0$ when $x = 2$ . (b) It is undefined when $x = -3$ .

Equivalent Algebraic Fractions

With numbers, $\dfrac{2}{3} = \dfrac{2 \times 4}{3 \times 4} = \dfrac{8}{12}\text{.}$ These are equivalent fractions: they have the same value. Algebraic fractions work the same way, and you may also multiply or divide by a letter or an expression, as long as it is not zero.

Equivalent fractions

The value of a fraction does not change when its numerator and denominator are both multiplied, or both divided, by the same expression $C$ :

$\dfrac{A}{B} = \dfrac{A \times C}{B \times C}$ and $\dfrac{A}{B} = \dfrac{A \div C}{B \div C}$ , where $B \neq 0$ and $C \neq 0$ .

Multiplying: $\dfrac{x}{3} = \dfrac{x \times 5}{3 \times 5} = \dfrac{5x}{15}\text{.}$ Dividing: $\dfrac{6x}{9} = \dfrac{6x \div 3}{9 \div 3} = \dfrac{2x}{3}\text{.}$

Example 5: Finding the Missing Numerator or Denominator

Find the missing expression in each pair of equivalent fractions. (a) $\dfrac{x}{3} = \dfrac{?}{15y}$ (b) $\dfrac{2}{a} = \dfrac{6ab}{?}$ (c) $\dfrac{x + 1}{4} = \dfrac{?}{8x}$

Solution

(a) Ask what the denominator $3$ was multiplied by to give $15y$ : $3 \times 5y = 15y$ . Multiply the numerator by $5y$ too: $x \times 5y = 5xy$ .

(b) This time the numerators are given: $2 \times 3ab = 6ab$ . Multiply the denominator by $3ab$ too: $a \times 3ab = 3a^2b$ .

(c) $4 \times 2x = 8x$ , so multiply the numerator by $2x$ . The whole numerator $x + 1$ is multiplied, so it keeps its brackets: $2x{(x + 1)}$ .

Check (c) with $x = 1$ : $\dfrac{1 + 1}{4} = \dfrac{1}{2}$ and $\dfrac{2(1)(2)}{8} = \dfrac{4}{8} = \dfrac{1}{2}\text{.}$

(a) $5xy$ (b) $3a^2b$ (c) $2x{(x + 1)}$

Simplifying by Cancelling Common Factors

To simplify an algebraic fraction, divide the numerator and the denominator by a factor they have in common. This is called cancelling the common factor. Keep going until the only common factor left is $1$ : the fraction is then in its simplest form.

To see the common letters, write each power as a product. A factor that is cancelled leaves $1$ behind, so if every factor of the numerator cancels, the numerator is $1$ .

Simplifying an algebraic fraction

1. Write the numerator and the denominator as products of numbers and letters.

2. Cancel each factor that appears in both: the HCF of the numbers, and each letter as many times as it appears in both.

3. Multiply what is left on top, and what is left below.

Example 6: Cancelling a Number and a Letter

Simplify (a) $\dfrac{6x}{9}$ (b) $\dfrac{10a}{15ab}$

Solution

(a) The HCF of $6$ and $9$ is $3$ : $\dfrac{6x}{9} = \dfrac{3 \times 2x}{3 \times 3} = \dfrac{2x}{3}\text{.}$

(b) The HCF of $10$ and $15$ is $5$ , and $a$ is a factor of both: $\dfrac{10a}{15ab} = \dfrac{\cancel{5} \times 2 \times \cancel{a}}{\cancel{5} \times 3 \times \cancel{a} \times b} = \dfrac{2}{3b}\text{.}$

Check (b) with $a = 1$ , $b = 1$ : $\dfrac{10}{15} = \dfrac{2}{3}$ , and $\dfrac{2}{3 \times 1} = \dfrac{2}{3}$ .

(a) $\dfrac{2x}{3}$ (b) $\dfrac{2}{3b}$

Example 7: Cancelling Powers of a Letter

Simplify (a) $\dfrac{x}{x^3}$ (b) $\dfrac{y^3}{y^2}$ (c) $\dfrac{3xy^3}{9x^3y^2}$

Solution

(a) $\dfrac{x}{x^3} = \dfrac{\cancel{x}}{\cancel{x} \times x \times x} = \dfrac{1}{x \times x} = \dfrac{1}{x^2}\text{.}$ The $x$ on top cancels one of the three below and leaves $1$ on top.

(b) $\dfrac{y^3}{y^2} = \dfrac{\cancel{y} \times \cancel{y} \times y}{\cancel{y} \times \cancel{y}} = \dfrac{y}{1} = y\text{.}$

$\dfrac{3xy^3}{9x^3y^2} = \dfrac{\cancel{3} \times \cancel{x} \times y \times \cancel{y} \times \cancel{y}}{3 \times \cancel{3} \times \cancel{x} \times x \times x \times \cancel{y} \times \cancel{y}}$

Left on top: $y$ . Left below: $3 \times x \times x = 3x^2$ . So the fraction is $\dfrac{y}{3x^2}$ .

Check (c) with $x = 1$ , $y = 2$ : $\dfrac{3 \times 1 \times 8}{9 \times 1 \times 4} = \dfrac{24}{36} = \dfrac{2}{3}\text{,}$ and $\dfrac{2}{3 \times 1} = \dfrac{2}{3}$ . With $x = 2$ , $y = 1$ : $\dfrac{6}{72} = \dfrac{1}{12}$ , and $\dfrac{1}{3 \times 4} = \dfrac{1}{12}$ .

(a) $\dfrac{1}{x^2}$ (b) $y$ (c) $\dfrac{y}{3x^2}$

Example 8: Two Letters with Powers

Simplify $\dfrac{8x^2y}{12x^4y^4}$ .

Solution

Take each part in turn.

Numbers: the HCF of $8$ and $12$ is $4$ , and $\dfrac{8}{12} = \dfrac{2}{3}$ .

The letter $x$ : $\dfrac{x \times x}{x \times x \times x \times x}$ . Two factors $x$ cancel, leaving $x \times x = x^2$ below.

The letter $y$ : $\dfrac{y}{y \times y \times y \times y}$ . One $y$ cancels, leaving $y^3$ below.

So $\dfrac{8x^2y}{12x^4y^4} = \dfrac{2}{3x^2y^3}\text{.}$

Check with $x = 2$ , $y = 1$ : $\dfrac{8 \times 4}{12 \times 16} = \dfrac{32}{192} = \dfrac{1}{6}\text{,}$ and $\dfrac{2}{3 \times 4} = \dfrac{1}{6}$ .

$\dfrac{8x^2y}{12x^4y^4} = \dfrac{2}{3x^2y^3}$

A minus sign is a factor of $-1$ : $-a = {(-1)} \times a$ . It may be written on top, below or in front of the fraction: $\dfrac{-a}{b} = \dfrac{a}{-b} = -\dfrac{a}{b}\text{,}$ and $\dfrac{-a}{-b} = \dfrac{a}{b}$ . We usually write the answer with the sign in front, and no minus in the denominator.

Example 9: Negative Coefficients

Simplify (a) $\dfrac{-6a^2b}{9ab^3}$ (b) $\dfrac{4x}{-2xy}$ (c) $\dfrac{-15m}{-5mn}$

Solution

(a) Numbers: $\dfrac{-6}{9} = -\dfrac{2}{3}$ . Letters: $\dfrac{a \times a \times b}{a \times b \times b \times b}$ leaves $a$ on top and $b^2$ below. So the answer is $-\dfrac{2a}{3b^2}$ .

(b) $\dfrac{4}{-2} = -2$ , and the $x$ cancels, leaving $y$ below: $-\dfrac{2}{y}$ .

Check (a) with $a = 2$ , $b = 1$ : $\dfrac{-6 \times 4 \times 1}{9 \times 2 \times 1} = \dfrac{-24}{18} = -\dfrac{4}{3}\text{,}$ and $-\dfrac{2 \times 2}{3 \times 1} = -\dfrac{4}{3}\text{.}$

(a) $-\dfrac{2a}{3b^2}$ (b) $-\dfrac{2}{y}$ (c) $\dfrac{3}{n}$

A Bracket as One Factor

A bracket such as ${(x + 2)}$ is one factor, just like a letter. If the same bracket is a factor of the whole numerator and of the whole denominator, it cancels: ${(x + 2)}^2 = {(x + 2)} \times {(x + 2)}\text{,}$ so one ${(x + 2)}$ can cancel with one below.

When a letter or a bracket cancels, the simplified fraction may have a value where the first one did not: $\dfrac{10a}{15ab}$ has no value when $a = 0$ , but $\dfrac{2}{3b}$ has. The two are equal only for the values where the original fraction is defined, so from here on we state those values with the answer.

Example 10: Cancelling a Common Bracket

Simplify (a) $\dfrac{5(x - 1)}{x - 1}$ (b) $\dfrac{4(x + 2)^2}{6x(x + 2)}\text{,}$ stating the values of $x$ for which each original fraction is undefined.

Solution

(a) $\dfrac{5\cancel{(x - 1)}}{\cancel{(x - 1)}} = 5\text{.}$ The original is undefined when $x = 1$ , so the answer is $5$ for every $x$ except $1$ .

(b) Write the square as a product: $\dfrac{4(x + 2)(x + 2)}{6x(x + 2)}\text{.}$

Cancel the HCF $2$ of $4$ and $6$ , and one ${(x + 2)}$ : $\dfrac{2(x + 2)}{3x}$ .

The original denominator $6x{(x + 2)}$ is $0$ when $x = 0$ or $x = -2$ .

Check (b) with $x = 1$ : $\dfrac{4 \times 9}{6 \times 3} = 2$ , and $\dfrac{2 \times 3}{3} = 2$ . With $x = 2$ : $\dfrac{4 \times 16}{12 \times 4} = \dfrac{4}{3}\text{,}$ and $\dfrac{2 \times 4}{6} = \dfrac{4}{3}$ .

Notice that $\dfrac{2(x + 2)}{3x}$ equals $0$ at $x = -2$ , while the original has no value there. That is why the excluded values come from the original fraction.

(a) $5$ , for $x \neq 1$ (b) $\dfrac{2(x + 2)}{3x}$ , for $x \neq 0$ and $x \neq -2$

Cancel Factors, Never Terms

You may cancel only a factor of the whole numerator and of the whole denominator. In $x + 3$ , the $3$ is a term (something added), not a factor of $x + 3$ . So the $3$ s in $\dfrac{x + 3}{3}$ do not cancel.

What you may do is divide each term of the numerator: $\dfrac{x + 3}{3} = \dfrac{x}{3} + \dfrac{3}{3} = \dfrac{x}{3} + 1\text{.}$

If you are not sure whether a step is allowed, put a number in for the letter on both sides. If the two sides give different values, the step is wrong.

Example 11: Is the Cancelling Right?

Test each statement with a number. (a) $\dfrac{x + 3}{3} = x + 1$ (b) $\dfrac{2x}{2 + x} = x$

Solution

(a) Take $x = 3$ . Left side: $\dfrac{3 + 3}{3} = 2$ . Right side: $3 + 1 = 4$ . They differ, so the statement is false. The $3$ is not a factor of $x + 3$ .

(b) Take $x = 2$ . Left side: $\dfrac{4}{4} = 1$ . Right side: $2$ . False. The $2$ in $2 + x$ is a term, not a factor, so nothing cancels. $\dfrac{2x}{2 + x}$ is already in its simplest form.

One value that agrees does not prove a statement true (in (b), $x = 0$ gives $0$ on both sides), but one value that disagrees proves it false.

Both are false. $\dfrac{x + 3}{3} = \dfrac{x}{3} + 1$ , and $\dfrac{2x}{2 + x}$ cannot be simplified.

Common Mistakes

Cancelling a term instead of a factor: $\dfrac{x + 3}{3} = x + 1$ is wrong. Cancel only what multiplies the whole numerator and the whole denominator.
Losing the $1$ : $\dfrac{5x}{10x^2} = \dfrac{1}{2x}$ , not $2x$ . When every factor on top cancels, $1$ is left on top.
Cancelling the wrong number of letters: $\dfrac{x}{x^3}$ is $\dfrac{1}{x^2}$ , not $x^2$ or $\dfrac{1}{x^3}$ . Write the letters out and count.
Dropping a minus sign: $\dfrac{4x}{-2xy}$ is $-\dfrac{2}{y}$ , not $\dfrac{2}{y}$ .
Thinking a fraction is undefined when its numerator is $0$ . $\dfrac{0}{5} = 0$ ; only a zero denominator makes it undefined.
Reading the excluded values from the simplified fraction. They come from the original: $\dfrac{4(x + 2)^2}{6x(x + 2)}$ is undefined at $x = -2$ although $\dfrac{2(x + 2)}{3x}$ is not.

Practice

Try each question before you open its solution. Check each simplification by putting numbers in for the letters on both sides.

Question 1

For what values is each fraction undefined? (a) $\dfrac{7}{2x + 8}$ (b) $\dfrac{x + 1}{5 - 2x}$ (c) $\dfrac{9}{4ab}$

ShowHide solution

(a) $2x + 8 = 0$ , so $2x = -8$ and $x = -4$ .

(b) $5 - 2x = 0$ , so $5 = 2x$ and $x = \dfrac{5}{2} = 2\tfrac{1}{2}$ .

(a) $x = -4$ (b) $x = 2\tfrac{1}{2}$ (c) $a = 0$ or $b = 0$

Question 2

For the fraction $\dfrac{x + 4}{x(x - 5)}$ , find (a) the values of $x$ for which it is undefined, (b) the value of $x$ for which it is equal to $0$ .

ShowHide solution

(a) The denominator $x{(x - 5)}$ is $0$ when $x = 0$ or $x - 5 = 0$ , that is, $x = 0$ or $x = 5$ .

(b) The numerator is $0$ when $x = -4$ . The denominator is then ${(-4)}{(-9)} = 36$ , not $0$ , so the fraction is $\dfrac{0}{36} = 0$ .

(a) $x = 0$ or $x = 5$ (b) $x = -4$

Question 3

Find the value of (a) $\dfrac{3x - 1}{x + 2}$ when $x = -3$ (b) $\dfrac{3x - 1}{x + 2}$ when $x = \dfrac{1}{2}$ (c) $\dfrac{p^2}{p - q}$ when $p = -3$ and $q = 6$ .

ShowHide solution

(a) $\dfrac{3(-3) - 1}{-3 + 2} = \dfrac{-10}{-1} = 10$

(b) $\dfrac{\frac{3}{2} - 1}{\frac{1}{2} + 2} = \dfrac{\frac{1}{2}}{\frac{5}{2}} = \dfrac{1}{2} \times \dfrac{2}{5} = \dfrac{1}{5}$

(a) $10$ (b) $\dfrac{1}{5}$ (c) $-1$

Question 4

Find the missing expression. (a) $\dfrac{3}{4b} = \dfrac{?}{12ab^2}$ (b) $\dfrac{5x}{2y} = \dfrac{15x^2}{?}$ (c) $\dfrac{a - 2}{3} = \dfrac{?}{6a}$

ShowHide solution

(a) $4b \times 3ab = 12ab^2$ , so the numerator is $3 \times 3ab = 9ab$ .

(b) $5x \times 3x = 15x^2$ , so the denominator is $2y \times 3x = 6xy$ .

(a) $9ab$ (b) $6xy$ (c) $2a{(a - 2)}$

Question 5

Simplify (a) $\dfrac{12p}{18}$ (b) $\dfrac{14m}{21mn}$ (c) $\dfrac{4xy}{10y}$

ShowHide solution

(a) HCF $6$ : $\dfrac{2p}{3}$ .

(b) HCF of $14$ and $21$ is $7$ , and $m$ cancels: $\dfrac{2}{3n}$ .

(a) $\dfrac{2p}{3}$ (b) $\dfrac{2}{3n}$ (c) $\dfrac{2x}{5}$

Question 6

Simplify (a) $\dfrac{15p^2q}{25pq^3}$ (b) $\dfrac{18a^3b^2}{24a^2b^4}$ (c) $\dfrac{7x^2y}{21x^2y^3}$

ShowHide solution

(a) $\dfrac{15}{25} = \dfrac{3}{5}$ . $\dfrac{p \times p}{p}$ leaves $p$ on top. $\dfrac{q}{q \times q \times q}$ leaves $q^2$ below. Answer $\dfrac{3p}{5q^2}$ . Check $p = 2$ , $q = 1$ : $\dfrac{60}{50} = \dfrac{6}{5}$ and $\dfrac{6}{5}$ .

(b) $\dfrac{18}{24} = \dfrac{3}{4}$ . $\dfrac{a \times a \times a}{a \times a}$ leaves $a$ on top. $\dfrac{b \times b}{b \times b \times b \times b}$ leaves $b^2$ below. Answer $\dfrac{3a}{4b^2}$ . Check $a = 2$ , $b = 1$ : $\dfrac{144}{96} = \dfrac{3}{2}$ and $\dfrac{6}{4} = \dfrac{3}{2}$ .

(c) $\dfrac{7}{21} = \dfrac{1}{3}$ . $x^2$ cancels completely. $\dfrac{y}{y \times y \times y}$ leaves $y^2$ below. Every factor on top has cancelled, so $1$ is left: $\dfrac{1}{3y^2}$ .

(a) $\dfrac{3p}{5q^2}$ (b) $\dfrac{3a}{4b^2}$ (c) $\dfrac{1}{3y^2}$

Question 7

Simplify (a) $\dfrac{-12x^2y}{16xy^2}$ (b) $\dfrac{10ab}{-15a^2}$ (c) $\dfrac{-8k^2}{-2k}$

ShowHide solution

(a) $\dfrac{-12}{16} = -\dfrac{3}{4}$ ; one $x$ and one $y$ cancel, leaving $x$ on top and $y$ below: $-\dfrac{3x}{4y}$ .

(b) $\dfrac{10}{-15} = -\dfrac{2}{3}$ ; one $a$ cancels, leaving $b$ on top and $a$ below: $-\dfrac{2b}{3a}$ .

Check (a) with $x = 2$ , $y = 1$ : $\dfrac{-48}{32} = -\dfrac{3}{2}$ , and $-\dfrac{6}{4} = -\dfrac{3}{2}$ .

(a) $-\dfrac{3x}{4y}$ (b) $-\dfrac{2b}{3a}$ (c) $4k$

Question 8

Simplify, stating the value for which the original fraction is undefined. (a) $\dfrac{8x(x + 5)}{2(x + 5)}$ (b) $\dfrac{6(a - 3)}{9(a - 3)^2}$

ShowHide solution

(a) Cancel $2$ and ${(x + 5)}$ : $\dfrac{4x}{1} = 4x$ . The original is undefined when $x = -5$ .

(b) $\dfrac{6(a - 3)}{9(a - 3)(a - 3)}\text{.}$ Cancel $3$ and one ${(a - 3)}$ : $\dfrac{2}{3(a - 3)}$ . Undefined when $a = 3$ .

Check (b) with $a = 4$ : $\dfrac{6}{9} = \dfrac{2}{3}$ , and $\dfrac{2}{3 \times 1} = \dfrac{2}{3}$ .

(a) $4x$ , for $x \neq -5$ (b) $\dfrac{2}{3(a - 3)}$ , for $a \neq 3$

Question 9

Simplify $\dfrac{9x^4(x - y)^3}{27x^2y^3(x - y)}\text{,}$ stating the values for which it is undefined.

ShowHide solution

Numbers: $\dfrac{9}{27} = \dfrac{1}{3}$ .

$\dfrac{x \times x \times x \times x}{x \times x}$ leaves $x^2$ on top. $y^3$ stays below.

$\dfrac{(x - y)(x - y)(x - y)}{(x - y)}$ leaves ${(x - y)}^2$ on top.

So the fraction is $\dfrac{x^2(x - y)^2}{3y^3}$ . The original denominator is $0$ when $x = 0$ , $y = 0$ or $x = y$ .

Check with $x = 2$ , $y = 1$ : $\dfrac{9 \times 16 \times 1}{27 \times 4 \times 1 \times 1} = \dfrac{144}{108} = \dfrac{4}{3}\text{,}$ and $\dfrac{4 \times 1}{3} = \dfrac{4}{3}$ .

$\dfrac{x^2(x - y)^2}{3y^3}$ , for $x \neq 0$ , $y \neq 0$ and $x \neq y$

Question 10

Three pupils simplified a fraction. Test each with a number and say whether it is right. (a) Ali: $\dfrac{2x + 6}{2} = x + 6$ (b) Mei: $\dfrac{x + 5}{x + 2} = \dfrac{5}{2}$ (c) Ravi: $\dfrac{3ab}{3a} = b$

ShowHide solution

(a) $x = 1$ : $\dfrac{8}{2} = 4$ , but $1 + 6 = 7$ . Wrong: Ali divided only the first term by $2$ . Dividing each term: $\dfrac{2x}{2} + \dfrac{6}{2} = x + 3\text{.}$

(b) $x = 1$ : $\dfrac{6}{3} = 2$ , but $\dfrac{5}{2}$ is not $2$ . Wrong: $x$ is a term on top and below, not a factor. The fraction cannot be simplified.

(c) $3a$ is a factor of both $3ab$ and $3a$ : $\dfrac{3a \times b}{3a \times 1} = b$ , for $a \neq 0$ . Right. ( $a = 2$ , $b = 5$ : $\dfrac{30}{6} = 5$ .)

(a) Wrong; it is $x + 3$ . (b) Wrong; it cannot be simplified. (c) Right.

Question 11

A rectangular garden plot has an area of $6x^2y$ m² and a length of $4xy$ m. (a) Find its breadth in terms of $x$ , in its simplest form. (b) Find the breadth when $x = 4$ .

ShowHide solution

(a) Area $=$ length $\times$ breadth, so breadth $=$ area $\div$ length $= \dfrac{6x^2y}{4xy}$ .

Cancel $2$ , one $x$ and $y$ : $\dfrac{3x}{2}$ .

(b) $\dfrac{3 \times 4}{2} = 6$ .

(a) $\dfrac{3x}{2}$ m (b) $6$ m

Question 12

A coach travels $18x^2$ km in $3x$ hours. (a) Find its average speed in terms of $x$ . (b) For what value of $x$ is the fraction you wrote undefined, and why does that value not arise here?

ShowHide solution

(a) Speed $= \dfrac{\text{distance}}{\text{time}} = \dfrac{18x^2}{3x}\text{.}$ Cancel $3$ and one $x$ : $\dfrac{6x}{1} = 6x$ .

(b) $\dfrac{18x^2}{3x}$ is undefined when $x = 0$ . A journey takes some time, so $3x$ is more than $0$ and $x$ cannot be $0$ .

(a) $6x$ km/h (b) $x = 0$ ; the time $3x$ hours is not zero.

Key Points

An algebraic fraction has a letter in its numerator, its denominator or both.
It is undefined when its denominator is $0$ : solve denominator $= 0$ . It is $0$ when its numerator is $0$ and its denominator is not.
Equivalent fractions: $\dfrac{A}{B} = \dfrac{A \times C}{B \times C}$ , where $B \neq 0$ and $C \neq 0$ .
To simplify, write powers as products, such as $x^3 = x \times x \times x$ , and cancel the factors common to the whole numerator and the whole denominator.
A bracket is one factor and cancels like a letter. Cancel factors, never terms.
The simplified fraction equals the original only where the original is defined.
Check a simplification by putting numbers in for the letters on both sides.