Algebraic Fraction

Unit 11 · Compulsory Mathematics · 9 min read

Algebraic Fraction

What the syllabus asks for 2
  • Algebraic Fraction: concepts, formulas and worked problems
  • Exercises of the SEE type

A fraction such as \frac{3}{4} has numbers on the top and bottom. An algebraic fraction has algebraic expressions there, for example \frac{x^2-9}{x^2+5x+6}. We add, subtract, multiply, divide and simplify algebraic fractions by exactly the same rules as ordinary fractions; the only new skill is factorisation, because we can cancel only factors, never terms. Algebraic fractions appear in formulae for speed, work and electricity, and in the SEE they give a dependable 3- or 4-mark question on simplification.

What the SEE asks from this chapter

Expect: (i) reduce a fraction to its lowest terms by factorising; (ii) add or subtract two or three fractions with different denominators using the LCM; (iii) multiply or divide fractions and simplify to a single expression; (iv) a "simplify" or "prove" question combining all of these, often ending in a surprisingly simple answer like 1 or x+1. Factorise first, cancel second, and write every step.

11.1 Algebraic Fraction: concepts, formulas and worked problems

Factorisation toolkit

Everything in this chapter rests on these factorisations. Learn them until you can see them at a glance.

Factorisation formulas

a^2-b^2=(a+b)(a-b)

a^2+2ab+b^2=(a+b)^2 \qquad a^2-2ab+b^2=(a-b)^2

a^3+b^3=(a+b)(a^2-ab+b^2)

a^3-b^3=(a-b)(a^2+ab+b^2)

Type Method Example
Common factor take out the HCF 6x^2+9x=3x(2x+3)
Difference of squares a^2-b^2 x^2-25=(x+5)(x-5)
Trinomial x^2+px+q two numbers with sum p, product q x^2+5x+6=(x+2)(x+3)
Trinomial ax^2+bx+c split the middle term 2x^2+7x+3=(2x+1)(x+3)
Grouping pair the terms ax+ay+bx+by=(a+b)(x+y)
Sum / difference of cubes the two formulas above x^3-8=(x-2)(x^2+2x+4)

figure
Figure 1. Area model of a² − b². Remove the b × b square from the a × a square and re-arrange the rest into a rectangle (a + b) long and (a − b) wide.

Factorising before simplifying

Problem. Factorise (a) x^2-7x+12, (b) 3x^2-x-2, (c) a^3+27.

Solution. (a) We need two numbers with product 12 and sum -7: -3 and -4. So x^2-7x+12=(x-3)(x-4).

(b) 3\times(-2)=-6; split -x as -3x+2x: 3x^2-3x+2x-2=3x(x-1)+2(x-1)=(x-1)(3x+2).

(c) a^3+27=a^3+3^3=(a+3)(a^2-3a+9).

Reducing to lowest terms

Cancelling rule

\frac{p\times r}{q\times r}=\frac{p}{q}\qquad (r\ne 0)

You may cancel only a common factor of the whole numerator and the whole denominator. Never cancel a term that is added or subtracted: \frac{x+3}{x+5} is already in lowest terms, and it is not \frac{3}{5}.

Reduce to lowest terms

Problem. Simplify \frac{x^2-9}{x^2+5x+6}.

Solution. Factorise both parts: x^2-9=(x+3)(x-3) and x^2+5x+6=(x+2)(x+3). Then

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

Check with x=1. Original: \frac{1-9}{1+5+6}=\frac{-8}{12}=-\frac{2}{3}. Answer: \frac{1-3}{1+2}=-\frac{2}{3}. ✓

Reduce with a trinomial

Problem. Simplify \frac{a^2-a-6}{a^2-9}.

Solution. a^2-a-6=(a-3)(a+2) and a^2-9=(a-3)(a+3). Cancel (a-3):

\frac{a^2-a-6}{a^2-9}=\frac{a+2}{a+3}

Reduce with cubes

Problem. Simplify \frac{x^3-8}{x^2-4}.

Solution. x^3-8=(x-2)(x^2+2x+4) and x^2-4=(x-2)(x+2). So the fraction is \frac{x^2+2x+4}{x+2}.

Common mistakes
  • Cancelling terms: \frac{x+3}{x+5}\ne\frac{3}{5} and \frac{x^2+4}{x+4}\ne x.
  • Forgetting a bracket when subtracting a fraction: \frac{1}{a}-\frac{b-c}{a}=\frac{1-(b-c)}{a}, not \frac{1-b-c}{a}.
  • Cancelling before factorising — if you cannot see a common factor, factorise first.
  • Writing "=0" when numerators cancel: \frac{x}{x}=1, not 0.

HCF and LCM of expressions

To add fractions you need a common denominator, and the best one is the lowest common multiple (LCM) of the denominators. As in Chapter 8, factorise first. The HCF takes each common factor with its smallest power; the LCM takes every factor that appears, with its largest power.

Expressions Factorised HCF LCM
x^2-4 and x^2+4x+4 (x-2)(x+2) and (x+2)^2 x+2 (x-2)(x+2)^2
x^2-x-6 and x^2-9 (x-3)(x+2) and (x-3)(x+3) x-3 (x-3)(x+2)(x+3)
6x^2y and 9xy^2 2\cdot3\cdot x^2y and 3^2\cdot xy^2 3xy 18x^2y^2
LCM before adding

Problem. Add \frac{1}{x^2-4}+\frac{1}{x^2+4x+4}.

Solution. From the table the LCM is (x-2)(x+2)^2. Multiply the first fraction's top and bottom by (x+2) and the second's by (x-2):

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

Check with x=3. \frac{1}{5}+\frac{1}{25}=\frac{6}{25}, and the answer gives \frac{6}{1\times25}=\frac{6}{25} ✓.

Multiplication and division

Rules

\frac{p}{q}\times\frac{r}{s}=\frac{pr}{qs}\qquad \frac{p}{q}\div\frac{r}{s}=\frac{p}{q}\times\frac{s}{r}=\frac{ps}{qr}

To divide, invert the second fraction and multiply. Factorise everything and cancel before multiplying out.

Multiplying

Problem. Simplify \frac{x^2-5x+6}{x^2-4}\times\frac{x+2}{x-3}.

Solution. x^2-5x+6=(x-2)(x-3) and x^2-4=(x-2)(x+2). Thus

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

Every factor cancels, leaving 1.

Dividing

Problem. Simplify \frac{a^2-b^2}{a^2+2ab+b^2}\div\frac{a-b}{a+b}.

Solution. Factorise: \frac{(a+b)(a-b)}{(a+b)^2}\times\frac{a+b}{a-b}. The factors (a+b) and (a-b) cancel completely:

\frac{(a+b)(a-b)(a+b)}{(a+b)(a+b)(a-b)}=1

Three fractions

Problem. Simplify \frac{x^2+3x+2}{x^2+x-2}\times\frac{x^2-1}{x^2+2x}.

Solution. x^2+3x+2=(x+1)(x+2), x^2+x-2=(x+2)(x-1), x^2-1=(x+1)(x-1), x^2+2x=x(x+2). So

\frac{(x+1)(x+2)}{(x+2)(x-1)}\times\frac{(x+1)(x-1)}{x(x+2)}

Cancel (x+2) once and (x-1) once:

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

Addition and subtraction

Steps for \frac{p}{q}\pm\frac{r}{s}
  1. Factorise every denominator.
  2. Find the LCM of the denominators: each different factor once, with its highest power.
  3. Rewrite each fraction with the LCM as denominator (multiply top and bottom by the missing factors).
  4. Add or subtract the numerators — put brackets round numerators that are subtracted.
  5. Simplify the numerator; factorise and cancel if possible.
Same denominator

Problem. Simplify \frac{x}{x^2-y^2}-\frac{y}{x^2-y^2}.

Solution. The denominators are equal, so \frac{x-y}{x^2-y^2}=\frac{x-y}{(x-y)(x+y)}=\frac{1}{x+y}.

Different denominators

Problem. Simplify \frac{1}{x-1}+\frac{1}{x+1}.

Solution. The LCM is (x-1)(x+1).

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

Subtraction

Problem. Simplify \frac{x}{x-2}-\frac{2}{x+2}.

Solution. LCM =(x-2)(x+2).

\frac{x(x+2)-2(x-2)}{(x-2)(x+2)}=\frac{x^2+2x-2x+4}{x^2-4}=\frac{x^2+4}{x^2-4}

The numerator x^2+4 does not factorise, so this is the final answer.

Three fractions

Problem. Simplify \frac{2}{x^2-1}+\frac{1}{x+1}-\frac{1}{x-1}.

Solution. x^2-1=(x+1)(x-1), so the LCM is (x+1)(x-1):

\frac{2+(x-1)-(x+1)}{(x+1)(x-1)}=\frac{2+x-1-x-1}{(x+1)(x-1)}=\frac{0}{x^2-1}=0

Subtraction with a bracket trap

Problem. Simplify \frac{1}{x-2}-\frac{1}{x-1}-\frac{1}{(x-1)(x-2)}.

Solution. The LCM is (x-1)(x-2).

\frac{(x-1)-(x-2)-1}{(x-1)(x-2)}=\frac{x-1-x+2-1}{(x-1)(x-2)}=\frac{0}{(x-1)(x-2)}=0

Unlike denominators with a pair

Problem. Simplify \frac{x+2}{x-2}-\frac{x-2}{x+2}.

Solution. LCM =(x-2)(x+2). Numerator: (x+2)^2-(x-2)^2=(x^2+4x+4)-(x^2-4x+4)=8x.

\frac{8x}{x^2-4}

Fractions inside fractions

Complex fraction

Problem. Simplify \frac{x^2-\frac{1}{x^2}}{x-\frac{1}{x}}.

Solution. Put p=x and q=\frac{1}{x}. Then the numerator is p^2-q^2 and the denominator is p-q. Therefore

\frac{p^2-q^2}{p-q}=p+q=x+\frac{1}{x}=\frac{x^2+1}{x}

Continued fraction

Problem. Simplify \frac{1}{1-\frac{1}{1+\frac{1}{x}}}.

Solution. Work from the inside out.

  • 1+\frac{1}{x}=\frac{x+1}{x}, so its reciprocal is \frac{x}{x+1}.
  • 1-\frac{x}{x+1}=\frac{(x+1)-x}{x+1}=\frac{1}{x+1}.
  • The reciprocal of \frac{1}{x+1} is x+1.

The answer is x+1.

Sum of squares

Problem. Simplify \left(x+\frac{1}{x}\right)^2-\left(x-\frac{1}{x}\right)^2.

Solution. Use a^2-b^2=(a+b)(a-b): here a+b=2x and a-b=\frac{2}{x}. The product is 2x\times\frac{2}{x}=4.

figure
Figure 2. A route map for any algebraic-fraction question: factorise first, then decide whether to cancel (× and ÷) or find the LCM (+ and −).

Restrictions

A fraction is undefined when its denominator is 0. In \frac{x-3}{x+2} we need x\ne-2. In the SEE you are rarely asked to state this, but you must never put a value in that makes a denominator zero when you check an answer.

Equations with algebraic fractions

Solving a fractional equation

Multiply every term by the LCM of the denominators to clear the fractions, solve the resulting equation, and reject any answer that makes a denominator zero.

Solve

Problem. Solve \frac{1}{x-1}+\frac{1}{x+1}=\frac{4}{3}.

Solution. The left side is \frac{2x}{x^2-1} (done above). Then

\frac{2x}{x^2-1}=\frac{4}{3} \;\Rightarrow\; 6x=4x^2-4 \;\Rightarrow\; 2x^2-3x-2=0

Factorise: (2x+1)(x-2)=0, so x=-\frac{1}{2} or x=2. Neither makes a denominator zero, so both are valid.

Check x=2. \frac{1}{1}+\frac{1}{3}=\frac{4}{3}. ✓

Solve with a fraction on each side

Problem. Solve \frac{3}{x-1}-\frac{2}{x+1}=1.

Solution. Multiply by (x-1)(x+1):

3(x+1)-2(x-1)=(x-1)(x+1) \;\Rightarrow\; x+5=x^2-1

So x^2-x-6=0, (x-3)(x+2)=0 and x=3 or x=-2.

More patterns to practise

Trinomials on both sides

Problem. Simplify \frac{x^2+x-12}{x^2-16}.

Solution. x^2+x-12=(x+4)(x-3) and x^2-16=(x+4)(x-4). Cancel (x+4):

\frac{x^2+x-12}{x^2-16}=\frac{x-3}{x-4}

Check with x=5. Original: \frac{25+5-12}{25-16}=\frac{18}{9}=2. Answer: \frac{5-3}{5-4}=2. ✓

Splitting the middle term

Problem. Simplify \frac{2x^2+5x-3}{2x^2-5x+2}.

Solution. For the top, 2\times(-3)=-6 and 5=6-1: 2x^2+6x-x-3=2x(x+3)-1(x+3)=(2x-1)(x+3). For the bottom, 2\times2=4 and -5=-4-1: 2x^2-4x-x+2=2x(x-2)-1(x-2)=(2x-1)(x-2). Cancel (2x-1):

\frac{x+3}{x-2}

A perfect square over a difference of squares

Problem. Simplify \frac{a^2-4a+4}{a^2-4}\times\frac{a+2}{a-2}.

Solution. a^2-4a+4=(a-2)^2 and a^2-4=(a-2)(a+2). So

\frac{(a-2)^2}{(a-2)(a+2)}\times\frac{a+2}{a-2}=\frac{(a-2)^2(a+2)}{(a-2)^2(a+2)}=1

Two fractions, one hidden factor

Problem. Simplify \frac{x}{x-3}-\frac{x+3}{x^2-9}.

Solution. x^2-9=(x-3)(x+3), so the second fraction reduces first: \frac{x+3}{(x-3)(x+3)}=\frac{1}{x-3}. Now the denominators are equal:

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

Always look for a fraction that can be reduced before you find the LCM; it saves work.

Three terms with x^2-9

Problem. Simplify \frac{1}{x-3}+\frac{1}{x+3}+\frac{6}{x^2-9}.

Solution. LCM =(x-3)(x+3).

\frac{(x+3)+(x-3)+6}{(x-3)(x+3)}=\frac{2x+6}{(x-3)(x+3)}=\frac{2(x+3)}{(x-3)(x+3)}=\frac{2}{x-3}

A division with a common factor

Problem. Simplify \frac{x^2-9}{x^2-4x+3}\div\frac{x+3}{x-1}.

Solution. x^2-4x+3=(x-1)(x-3) and x^2-9=(x-3)(x+3). Invert the divisor:

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

Same denominator, difference of squares on top

Problem. Simplify \frac{x^2}{x-y}-\frac{y^2}{x-y}.

Solution. \frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=x+y.

Cubes

Problem. Simplify \frac{a^3-b^3}{a^2-b^2}\times\frac{a+b}{a^2+ab+b^2}.

Solution. a^3-b^3=(a-b)(a^2+ab+b^2) and a^2-b^2=(a-b)(a+b). Hence

\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}\times\frac{a+b}{a^2+ab+b^2}=1

Fractions inside

Problem. Simplify \frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}}.

Solution. The numerator is \frac{y-x}{xy}. The denominator is \frac{y^2-x^2}{x^2y^2}=\frac{(y-x)(y+x)}{x^2y^2}. Dividing,

\frac{y-x}{xy}\times\frac{x^2y^2}{(y-x)(y+x)}=\frac{xy}{x+y}

An equation that gives two roots

Problem. Solve \frac{x}{x-2}+\frac{x-2}{x}=\frac{5}{2}.

Solution. Multiply by 2x(x-2): 2x^2+2(x-2)^2=5x(x-2). Expand: 2x^2+2x^2-8x+8=5x^2-10x, so x^2-2x-8=0 and (x-4)(x+2)=0. Hence x=4 or x=-2.

Check x=4. \frac{4}{2}+\frac{2}{4}=2.5 ✓. Check x=-2. \frac{-2}{-4}+\frac{-4}{-2}=0.5+2=2.5 ✓.

Equation with shifted numerators

Problem. Solve \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}.

Solution. Multiply by 3(x-3)(x-5): 3(x-2)(x-5)+3(x-4)(x-3)=10(x-3)(x-5). Expanding, 3(x^2-7x+10)+3(x^2-7x+12)=10(x^2-8x+15), i.e. 6x^2-42x+66=10x^2-80x+150, which gives 4x^2-38x+84=0 or 2x^2-19x+42=0. Then (2x-7)(x-6)=0 and x=\frac{7}{2} or x=6.

11.2 Exercises of the SEE type

Method for any question

  1. Factorise each numerator and denominator completely.
  2. For \times,\div: invert the divisor, cancel common factors, multiply what remains.
  3. For +,-: LCM of denominators, bracket every numerator being subtracted.
  4. Check by substituting a small number (such as x=2 or x=3) in the question and the answer. They must agree.
  5. Present the answer in lowest terms.
SEE type: simplify

Problem. Simplify \frac{a}{a-b}-\frac{b}{a+b}-\frac{2ab}{a^2-b^2}.

Solution. LCM =(a-b)(a+b)=a^2-b^2.

\frac{a(a+b)-b(a-b)-2ab}{a^2-b^2}=\frac{a^2+ab-ab+b^2-2ab}{a^2-b^2}=\frac{a^2-2ab+b^2}{a^2-b^2}

The numerator is (a-b)^2, so the expression equals \frac{(a-b)^2}{(a-b)(a+b)}=\frac{a-b}{a+b}.

SEE type: combine three fractions

Problem. Simplify \frac{1}{x+1}+\frac{2}{x+2}-\frac{3}{x+3}.

Solution. The LCM is (x+1)(x+2)(x+3). The numerator is

(x+2)(x+3)+2(x+1)(x+3)-3(x+1)(x+2)

Expand each part: x^2+5x+6, 2x^2+8x+6 and 3x^2+9x+6. So the numerator is (x^2+2x^2-3x^2)+(5x+8x-9x)+(6+6-6), which equals 4x+6. Hence

\frac{4x+6}{(x+1)(x+2)(x+3)}=\frac{2(2x+3)}{(x+1)(x+2)(x+3)}

Check with x=1. Left: \frac{1}{2}+\frac{2}{3}-\frac{3}{4}=\frac{5}{12}. Right: \frac{2\times5}{2\times3\times4}=\frac{5}{12}. ✓

SEE type: show that

Problem. Show that \frac{x^2+3x+2}{x+1}+\frac{x^2-x-6}{x-3}=2(x+2).

Solution. x^2+3x+2=(x+1)(x+2), so the first fraction is x+2. x^2-x-6=(x-3)(x+2), so the second fraction is x+2. The sum is (x+2)+(x+2)=2(x+2). Proved.

SEE type: area of a rectangle

Problem. A rectangle has area x^2+7x+12 and length x+4. Find its breadth, and the ratio of the breadth to the length.

Solution. Breadth =\frac{\text{area}}{\text{length}}=\frac{x^2+7x+12}{x+4}=\frac{(x+3)(x+4)}{x+4}=x+3. The ratio is \frac{x+3}{x+4}.

Chapter summary

  • Factorisation formulas: a^2-b^2=(a+b)(a-b); (a\pm b)^2; a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2); trinomials by sum and product or by splitting the middle term.
  • Cancel only factors, never added terms.
  • \frac{p}{q}\times\frac{r}{s}=\frac{pr}{qs}; \frac{p}{q}\div\frac{r}{s}=\frac{p}{q}\times\frac{s}{r}.
  • To add or subtract: factorise the denominators, take the LCM, rewrite, combine numerators with brackets, simplify.
  • Complex fractions: simplify the inside first; x^2-\frac{1}{x^2}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right).
  • Fractional equations: multiply by the LCM; reject roots that zero a denominator.
  • Always check with a small number such as x=2.

Practice questions

Group A — Multiple choice (1 mark each)

  1. Factorise x^2-16. [1]a(x-4)^2b(x+4)(x-4)c(x+8)(x-2)dx(x-16)
  2. \frac{x^2-1}{x-1} in its simplest form is [1]ax+1bx-1cxd1
  3. The LCM of x-2 and x^2-4 is [1]ax-2bx+2cx^2-4d(x-2)(x^2-4)
  4. \frac{3}{x}+\frac{2}{x}= [1]a\frac{5}{2x}b\frac{5}{x}c\frac{6}{x}d\frac{5}{x^2}
  5. \frac{a}{b}\div\frac{c}{d}= [1]a\frac{ac}{bd}b\frac{ad}{bc}c\frac{bc}{ad}d\frac{ab}{cd}
Answers to Group A

1. (b) — difference of squares.

2. (a) — \frac{(x+1)(x-1)}{x-1}=x+1.

3. (c) — x^2-4=(x-2)(x+2) already contains x-2.

4. (b) — same denominator, add numerators.

5. (b) — invert the divisor.

Group B — Short answer (2 marks each)

  1. Factorise x^2+5x+6. [2]
  2. Simplify \frac{x^2-9}{x^2+6x+9}. [2]
  3. Simplify \frac{x}{x+1}+\frac{1}{x+1}. [2]
  4. Simplify \frac{x+2}{x-3}\times\frac{x-3}{x^2-4}. [2]
Answers to Group B

1. (x+2)(x+3).

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

3. \frac{x+1}{x+1}=1.

4. \frac{x+2}{x-3}\times\frac{x-3}{(x-2)(x+2)}=\frac{1}{x-2}.

Group C — Short answer (3 marks each)

  1. Simplify \frac{1}{x-1}+\frac{1}{x+1}. [3]
  2. Simplify \frac{x^2-5x+6}{x^2-4}\div\frac{x-3}{x+2}. [3]
  3. Simplify \frac{x}{x-2}-\frac{2}{x+2}. [3]
  4. Simplify \left(x+\frac{1}{x}\right)^2-\left(x-\frac{1}{x}\right)^2. [3]
Answers to Group C

1. \frac{2x}{x^2-1}.

2. \frac{(x-2)(x-3)}{(x-2)(x+2)}\times\frac{x+2}{x-3}=1.

3. \frac{x^2+4}{x^2-4}.

4. 4.

Group D — Long answer (4 marks each)

  1. Simplify \frac{a}{a-b}-\frac{b}{a+b}-\frac{2ab}{a^2-b^2}. [4]
  2. Simplify \frac{x^2+3x+2}{x^2+x-2}\times\frac{x^2-1}{x^2+2x}. [4]
  3. Simplify \frac{1}{1-\frac{1}{1+\frac{1}{x}}}. [4]
  4. Solve \frac{3}{x-1}-\frac{2}{x+1}=1. [4]
Answers to Group D

1. \frac{a-b}{a+b}.

2. \frac{(x+1)^2}{x(x+2)}.

3. x+1.

4. x=3 or x=-2.

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