Further algebraic methods - Pearson Education

402 Algebra

This chapter will show you how to✔ simplify, add, subtract, multiply and divide algebraic

fractions and use them to solve equations✔ change of subject of a formula where the variable

appears on both sides of the equation

Further algebraic methods

Chapter 22

22.1 Simplifying algebraic fractionsYou will need to know● how to factorise … common factor, quadratic, difference of two

perfect squares

Algebraic fractions have letters, or combinations of letters and numbers, in their numerator and denominator.

4a b

2 2pqr x 1 6

x 4x 2 5 18x3

3x 2x 2 610x are all algebraic

fractions.

Algebraic fractions can sometimes be simplifi ed. Look for a common factor of the numerator and the denominator.

x 1 6x is an algebraic fraction that cannot be simplifi ed.

18x3

3x can be simplifi ed. 18x3

3x 5 18 × x × x × x

3 × x 5 6x2

2x 2 610x can be simplifi ed. 2x 2 6

10x 5 2(x 2 3)

10x 5 x 2 3

5x

To simplify algebraic fractions

(1) factorise(2) divide by the HCF (or cancel common factors).

You cannot cancel the x terms … never cancel across a 1 or a 2 sign.

You can see that there are common factors of 3 and x.

Factorise then divide top and bottom by 2.

Remember to factorise first.

M22_CME_SB_IGCSE_6867_U22.indd 402 28/8/09 13:15:09

Algebra 403


Simplify the following expressions, where possible.

1 3x 1 69 2 4x 1 5

2 3 4x2x 2 10

4 4a2

6ab 5 3a 1 2ba 6 10 2 5f

2f

7 ab 1 2aa2 8 p 1 4

3p 1 6 9 5 2 p5 1 p

10 x 1 2x2 2 x 2 6 11 x2 1 7x 1 6

x 1 1 12 y2 2 y 2 20y2 1 y 2 12

13 x2 2 36x2 1 8x 1 12 14 m2 2 1

5m2 2 2m 2 3 15 2x2 2 5x 2 123x2 2 7x 2 20

22.2 Multiplying and dividing algebraic fractions

To multiply numerical fractions you multiply the numerators and multiply the denominators, simplifying if possible.

To divide numerical fractions you invert the second fraction and multiply, simplifying if possible.

EXAMPLE 1

Simplify the following, where possible.

(a) 2xx 1 3 (b) 3ab

6a2 (c) 4x 2 168

(d) x2 1 3x 2 4x 1 4 (e) m2 2 25

m2 2 3m 2 10

(a) 2xx 1 3 cannot be simplified

(b) 3ab6a2 5

3 × a × b6 × a × a 5

b2a

(c) 4x 2 168 5

4(x 2 4)8 5

x 2 42

(d) x2 1 3x 2 4x 1 4 5

(x 1 4)(x 2 1)(x 1 4) 5 x 2 1

(e) m2 2 25m2 2 3m 2 10 5

(m 1 5)(m 2 5)(m 2 5)(m 1 2) 5 m 1 5

m 1 2

Always look for a common factor first.Then look for other forms of factorising such as quadratics and the difference of two perfect squares.All of these are present in these examples.

Remember that you can only cancel whole brackets.

EXERCISE 22A


404 Algebra

Work out the following, simplifying your answers.

1 a4 ×

a3 2 b

2 4 a5 3 pq

r × r2

p 4 xy2 4

xyz

5 y3 ×

y 1 22 6 x 1 1

4 × 1

3x 1 3

7 10a3

3b 4 5a6b2 8 x2 1 5x

x2 1 2x× x 1 2

4

9 2x 2 314 4

4x 2 67 10 m2 2 9

4 × 8

m 1 3

11 a2 2 2aa2 2 1 4

3a 2 6a2 1 4a 1 3 12 y2 2 16

y2 2 y 4 y2 2 9y 1 20

2y 2 2

22.3 Addition and subtraction of algebraic fractions

When adding or subtracting algebraic fractions, the rules are the same as for numerical fractions. But before you find the LCM you must factorise the numerators and/or denominators if possible.

EXAMPLE 2

Work out the following, simplifying your answers.

(a) 2pqr

× pr2

3q2 (b) 4a

4 b2

(c) 2x 1 6x2 1 4x

4 x2 1 2x 2 3

x2 2 16

(a) 2pqr

× pr2

3q2 5 2 × p × q

r × p × r × r3 × q × q 5

2pqr

× 2p2r3q

(b) 4a

4 b2 5

4a

× 2b

5 8ab

(c) 2x 1 6x2 1 4x

4 x2 1 2x 2 3

x2 2 16 5 2x 1 6x2 1 4x

× x2 2 16

x2 1 2x 2 3

5 2(x 1 3)x(x 1 4) ×

(x 1 4)(x 2 4)(x 1 3)(x 2 1)

5 2(x 2 4)x(x 2 1)

In (c), invert the second fraction and change 4 to ×.

Then factorise and cancel any common factors.

EXERCISE 22B

The rules are exactly the same for algebraic fractions, but always remember to look for any factors before you simplify.


Algebra 405


EXAMPLE 3

Work out (a) 25 1

37 (b) 2

a 1

3b

(c) x 1 45 1

x 2 64

(a) 25 1

37 (b) 2

a 1

3b

5 1435 1

1535 5

2bab

1 3aab

5 2935 5

2b 1 3aab

(c) x 1 45 1

x 2 64

5 4(x 1 4)

20 1 5(x 2 6)

20

5 4(x 1 4) 1 5(x 2 6)

20

5 4x 1 16 1 5x 2 30

20 5

9x 2 1420

Look at a numerical example and an algebraic example side by side … you will see the similar approach.

Just as 5 × 7 5 35 is the LCM of 5 and 7 then a × b 5 ab is the LCM of a and b.

Notice the equivalent fractions 2a 5 2b

ab and 3b 5 3aab.

The LCM of 4 and 5 is 20. Notice the equivalent fractions on the first line of the solution.

EXAMPLE 4

Simplify (a) 52x

1 3

8y (b) 4

x 1 1 2 3

x 2 5

(a) 52x

1 3

8y

5 20y8xy

1 3x

8xy

5 20y 1 3x

8xy

(b) 4x 1 1 2

3x 2 5

5 4(x 2 5)

(x 1 1)(x 2 5) 2 3(x 1 5)

(x 1 1)(x 2 5)

5 4(x 2 5) 2 3(x 1 1)

(x 1 1)(x 2 5)

5 4x 2 20 2 3x 2 3

(x 1 1)(x 2 5)

5 x 2 23

(x 1 1)(x 2 5)

Notice that in (a) the LCM of 2x and 8y is 8xy.

You do not need to use 2x × 8y 5 16xy although it is not wrong to do so … you will need to simplify later if you do.

When the denominator has more than one term, put brackets round the expression. It helps you to spot the LCM and get the correct equivalent fractions.

Take care with the signs when you expand the brackets in the numerator.


406 Algebra

Simplify each of the following.

1 2x3 1

x5 2 3

a 1 4b

3 5x 2 3

2y 4 34x 2

23x2

5 65x 1

32xy

6 x 2 13 1

x 1 54

7 2x 2 15 1

x 1 24 8 x 1 6

2 2 x 1 1

3

9 y 2 35 2

y 2 23 10 2m 1 3

5 2 3m 2 4

2

11 2x 1 3 1

5x 2 1 12 3

x 1 42 1

x 1 3

22.4 Equations involving algebraic fractions

You can use algebraic fractions to solve quite complicated equations.

The equation in the next example can be solved in two ways.

EXERCISE 22C

EXAMPLE 5

Solve the equation 3x 1 12 2

x 1 43 5 5

Method A

3x 1 12 2

x 1 43 5 5

3(3x 1 1) 2 2(x 1 4)6 5 5

9x 1 3 2 2x 2 86 5 5

7x 2 56 5 5

7x 2 5 5 30

7x 5 35

x 5 5continued .


Algebra 407


Method B

3x 1 12 2

x 1 43 5 5

The LCM of the terms in the denominator is 6

Multiply all terms by 6

6(3x 1 1)2 2

6(x 1 4)3 5 5 × 6

3(3x 1 1) 2 2(x 1 4) 5 30

9x 1 3 2 2x 2 8 5 30

7x 2 5 5 30

7x 5 35

x 5 5

Mutiplying all terms by the LCM of the denominators (in this case 6) will give you an equation with no fractions in it.

EXAMPLE 6

Solve the equation 42x 2 12

1x 1 1 5 1

Method A

4

2x 2 12 1

x 1 1 5 1

4(x 1 1) 2 1(2x 2 1)

(2x 2 1)(x 1 1) 5 1

4x 1 4 2 2x 2 1(2x 2 1)(x 1 1) 5 1

2x 1 5

(2x 2 1)(x 1 1) 5 1

2x 1 5 5 (2x – 1)(x 1 1)

2x 1 5 5 2x2 1 x 2 1

0 5 2x2 2 x 2 6

0 5 (2x 1 3)(x 2 2)

Either 2x 1 3 5 0 giving x 5 21.5 or x 2 2 5 0 giving x 5 2

continued .

Notice the equivalent fractions. Take care with the signs when expanding the numerator.

Multiply both sides by (2x – 1)(x 1 1) then expand and rearrange to get a quadratic equation.

These questions often have two solutions.


408 Algebra

The methods of solution are quite similar to each other and you can use either ... just choose the one you prefer.

Solve each of these equations.

1 x 1 12 1

x 2 43 5 5 2 x 1 5

4 1 x 2 1

2 5 3

3 x 2 23 2

x 1 14 5 1 4 2x 2 3

3 1 3x 1 1

4 5 12

5 5x 1 14 2

x 2 46 5 2 6 2x 2 5

2 2 4x 2 1

3 5 0.5

7 3x 1 1 1

22x 2 3 5 1 8 9

x 2 2 1 5

x 1 2 5 2

9 7x 2 3 2

6x 2 1 5 2 10 9

2x 2 7 1 6

x 2 1 5 3

11 114x 2 5 2

8x 1 1 5 1 12 7

2x 2 1 1 11

x 1 1 5 6

22.5 Changing the subject of a formulaIn this section you will see how to change the subject of a formula when the subject appears more than once and on both sides of the formula.

You use the same methods as for solving equations. You will see that a pattern emerges for the order in which you do the operations.

Method B

4

2x 2 12 1

x 1 1 5 1

The LCM of the terms in the denominator is (2x 2 1)(x 1 1)

Multiply all terms by (2x 2 1)(x 1 1)

4(2x 2 1)(x 1 1)(2x 2 1) 2

(2x 2 1)(x 1 1)(x 1 1) 5 (2x 2 1)(x 1 1)

4(x 1 1) 2 (2x 2 1) 5 (2x 2 1)(x 1 1)

4x 1 4 2 2x 1 1 5 2x2 1 x 2 1

0 5 2x2 2 x 2 6

0 5 (2x 1 3)(x 2 2)

Either 2x 1 3 5 0 giving x 5 21.5 or x 2 2 5 0 giving x 5 2

Multiply throughout by the LCM of the terms in the denominator i.e. by (2x 2 1)(x 1 1).

Then simplify each term, expand and rearrange to get a quadratic equation.

This is a standard method with which you need to be familiar.

EXERCISE 22D

You learned how to change the subject of a formula in Chapter 8. In all of the questions the subject only appeared once.


Algebra 409


EXAMPLE 7

(a) Solve the equation 2(2m 1 1) 5 7(m 2 1).

(b) Make m the subject of the formula a(bm 1 c) 5 d(m 2 e).

(a) 2(2m 1 1) 5 7(m 2 1) (b) a(bm 1 c) 5 d(m 2 e)

4m 1 2 5 7m 2 7 abm 1 ac 5 dm 2 de

2 1 7 5 7m 2 4m ac 1 de 5 dm 2 abm

9 5 3m ac 1 de 5 m(d 2 ab)

93 5 m

ac 1 ded 2 ab

5 m

3 5 m

Expand.

Rearrange.

Factorise.

Divide through by (d 2 ab).

EXAMPLE 8

(a) Solve the equation 5 5 y 1 3y 2 4 .

(b) Make y the subject of the formula t 5 hy 1 3ky 2 4 .

(a) 5 5 y 1 3y 2 4 (b) t 5

hy 1 3ky 2 4

5(y 2 4) 5 y 1 3 t(ky 2 4) 5 hy 1 3

5y 2 20 5 y 1 3 tky 2 4t 5 hy 1 3

5y 2 y 5 3 1 20 tky 2 hy 5 3 1 4t

4y 5 23 y(tk 2 h) 5 3 1 4t

y 5 234 y 5

3 1 4ttk 2 h

y 5 5.75

Expand.

Rearrange.

Factorise.

Divide through by (tk 2 h).

Multiply through by (ky 2 4).


410 Algebra

In this section each example also includes a traditional equation so that you can see the similarity in the method.

EXAMPLE 9

(a) Solve the equation 2aa 2 1 5 3.

(b) Make a the subject of the formula 2aa 2 x 5 3x.

(a) 2aa 2 1 5 3 (b) 2a

a 2 x 5 3x

2aa 2 1 5 9 2a

a 2 x 5 9x2

2a 5 9(a 2 1) 2a 5 9x2(a 2 x)

2a 5 9a 2 9 2a 5 9x2a 2 9x3

9 5 9a 2 2a 9x3 5 9x2a 2 2a

9 5 7a 9x3 5 a(9x2 2 2)

97 5 a

9x3

(9x2 2 2) 5 a

Expand.

Rearrange.

Factorise.

Divide through by (9x2 2 2).

Multiply through by (a 2 x).

Square both sides.

EXERCISE 22E

You can see the pattern of the order in which you do the operations. The order is always the same.

Make y the subject of each of these formulae.

1 my 1 b 5 d 2 my 2 py 1 3 5 7 1 qy

3 ay 2 x 5 w 1 by 4 w(y 1 a) 5 k(y 1 b)

5 h(ay 2 c) 5 m(y 1 d) 6 h 2 yh 1 y 5 k

7 y 1 3y 2 2

3615 5

de 8 2y 1 a

y 5 m

9 3ym 2 y 5 2w 10 T 5 2kyg

11 ky2 1 2e 5 f 2 hy2 12 m(2y2 1 a) 5 h(w 2 3y2)


Algebra 411


EXAMINATION QUESTIONS

1 (a) Write 1

x 2 3 2 1x as a single fraction in its simplest form. [2]

(b) Use your answer to part(a) to make y the subject of

1y 5

1x 2 3 2

1x [2]

(CIE Paper 2, Jun 2000)

2 Write 2x 2 10x5 2 x as a single fraction. [2]

(CIE Paper 2, Nov 2002)

3 Work out as a single fraction

2

x 2 3 2 1

x 1 4. [3]


4 (a) Write 3x 2

2x 1 1 as a single fraction in its simplest form. [3]

(b) Solve the equation

3x 2

2x 1 1 5 0. [1]

(CIE Paper 2, Nov 2003)

5 Simplify x 1 2

x 2 x

x 1 2.

Write your answer as a fraction in its simplest form. [3]



Documents

Further algebraic methods - Pearson Education