algebra booklet

0

Algebra

1

Table of contents

Topic Page Number

Substitution

3

Brackets

5

Linear Equations

7

Indices

9

Changing the subject (1)

11

Changing the subject (2)

13

Brackets and quadratics

15

Fractorisation of quadratics

17

Algebraic fractions

19

Algebraic fractions and operations

21

Quadratic equations

23

Quadratic formula

25

Negative and fractional indices

27

Simultaneous equations

29

Simultaneous equations and the method of substitution

32

Conversion Factors and Rates

34

Prefixes

37

Extra exercises and answers

39

2

Self assessment Topic Notes/targets

Substitution

Brackets

Linear Equations

Indices

Changing the subject

Brackets and quadratics

Fractorisation of quadratics

Algebraic fractions

Quadratic equations

Negative and fractional indices

Simultaneous equations


Prefixes

3

Substitution Target: On completion of this worksheet you should be able to substitute values into algebraic expressions and formulae and hence evaluate unknowns.

When we replace a letter in an algebraic

expression by another term, that we know

has the same value, we are said to be

substituting this letter.

Often we replace a letter by a number.

E.g. If x = 2 find 4x.

4 2 = 8.

Remember 4x means 4 x.

We need to remember that we must do division and

multiplication before addition and subtraction for more

difficult questions.

Examples.

1) If a = 4 and b = 6 find 3a + 2b.

3 4 + 2 6 = 12 + 12 = 24.

2) If c = 3 and d =4 find 5(c + 2d).

5(3 + 2 4) = 5 11=55.

When we substitute negative numbers

for letters we have to be careful. Using

brackets often helps.

E.g. If x = -5, y = -9 and z = 4 find

3x - 2y + z.

3(-5) - 2(-9) + 4 = -15 – (-18) + 4

= -15 + 18 + 4

= 7.

See the number sheet on negative numbers if you have

difficulty with this.

Exercise.

If a=2, b=4, c=7, d=1, e=0 find the value of:

1. 3+a

2. b+4a

3. a+3c+d

4. 2a+4b – 2c 4e

5. 3(a+2b)

6. 2(a + b) 3(2c d)

(Answers: 5, 12, 24, 6, 30, -27)

Exercise.

If a=-4, b=6, c=-8, d=2 find the value of:

1. b + a

2. b + 4a

3. 3c – b + 2a

4. 2 + 2a – 3b + c

5. 6(2a - 3b)

6. 9(5c - d) - 4(a - 2b)

(Answers: 2, -10, -38, -32,

-156, -314)

A3

4

When letters are multiplied together or divided by

each other the same principles apply. To avoid

mistakes with minus signs we should again use

brackets.

E.g. If a = 2, b = -6, and c=5 find the value of

4ab + 7c 2 .

4 2 (-6) + 7 5 2 = -48 + 175

=127.

Remember 4ab means 4ab.

Substitution is frequently used with formulae. A

formula expresses one variable in terms of other

variables.

E.g. V = IR, is a formula.

We are often told the values of some of the

variables and asked to find the value of the others.

E.g. If V = IR what is the value of V when I = 15

when R = 16.

V = 1516 = 240. Exercise.

If a = 5, b = -6, c = 2, d = -10 find the value of

1. ac

2. ab - d 2

3. 3(c - 4a)

4. d

c

5. ad

ab

(Answers: 10, -130, -54, -5

1 , 2)

Exercise.

1. If S = UV and U=5 and V=12 find S.

2. If H = c

b and b=4 and c=8 find H.

3. If p=2q

rs and r = 4, s=12 and q=4 find p.

(Answers: 60,2

1 , 3)

5

Brackets Target: On completion of this worksheet you should be able to expand algebraic expressions involving brackets.

w s

There are two ways of finding the area of

this rectangle.

Find the areas of the two smaller

rectangles and add them together.

Multiply the height by the total width of

the rectangle.

We can use brackets to ensure that we do

the correct operation first. (Don’t forget

BIDMAS.)

The resultant algebraic expressions are:

hw + hs.

h(w + s).

These expressions must be equal. Therefore

.hshwswh

From this we can see that when we are

expanding brackets the term outside the

brackets is multiplied by everything inside

the brackets.

Examples.

1. .2apapzappapzpzap

2. .343323 zxyxyyxzyxxy

3. ptprrsrptrpspr )(

.2 ptrsrp

Exercise.

Expand the following:

1. d(f+g).

2. p(p+g)+g(h+p).

(Answers: df+dg, p2+2pg+gh.)

Often we have to expand brackets where some of

the terms are negative. We must be careful with

the signs. Remember that Mixed Means Minus.

Examples. Simplify 1) -x y, 2) –p -q and 3) –r (s - t).

1. –x y = -xy.

2. –p -q = pq.

3. –r ( s - t ) = -rs + rt.

If there is no term outside the brackets then we

can assume that it is one lot of the bracket and

insert the number1.

E.g.

x(x + y) – (yx - z) = x(x + y) - 1(yx - z)

= x2+ xy – yx + z

= x2

+ z.

See the number sheet on negative numbers if you have


Exercise.

Expand the following brackets.

1. )( qpx .

2. ).2( rpxzp

3. –3p(rp-xy).

4. -z(z+xy)-(z2-4xy).

5. 3c(d-2s)+x(c-s).

(Answers: xp-xq, zpx-zp2+2zpr, -3rp

2+3pxy,

-2z2-zxy+4xy, 3cd-6cs+xc-xs).

h

A4

6

,

If we have a bracket inside a bracket it

often helps to expand the inside bracket first

and then simplify before expanding the

outside bracket.

E.g. x(yz - z( y + x)) = x(yz – zy - zx)

=x(-zx)

=-x2z.

Exercise.

Expand the following.

1. r(pq - 2p(q + r)).

2. 2s(st + 3t(r - s) - (rt + s)).

(Answers: -rpq - 2pr2, 4rts - 4s

2t -2s

2.)

a

b

c d

There are several ways of finding the area of

this rectangle. Here are two of them.

Find the areas of the four smaller

rectangles and add them together.

Find the total height and the total width

of the rectangle and multiply them

together.

The resultant algebraic expressions must be

equal. Therefore

.bdbcadacdcba

From this we can see that every term in the

first bracket must be multiplied by every

term in the second bracket.

Examples.

1) (x + 2)(x + y) = x2

+ xy + 2x + 2y.

2) (pz - r)(z + q) = pzz + pzq – rz - rq

= pz2+ pzq – rz - rq.

3) (3x - 2y)(x - 3y) = 3x2- 9xy - 2xy + 6y

2

= 3x2

- 11xy + 6y2.

Exercise.

Expand the brackets in the following expressions.

1. (2a+b)(3c+2d).

2. (a+b)(a+b).

3. (a+b)(a-b).

4. (3x+y)(x+2y).

5. (4x-y)(x-2y).

6. (2y+x-z)(x-2y).

7. (2x-y+z)(3x-y+2z).

8. (x+2y)(x+3y(x-2y)).

(Answers: 6ac+4ad+3bc+2bd, a2+2ab+b

2,

a2-b

2, 3x

2+7xy+2y

2, 4x

2-9yx+2y

2,

-4y2+x

2-zx+2zy, 6x

2-5xy+7xz+y

2-3zy+2z

2,

x2+3yx

2+2yx-12y

3.)

When we have brackets and indices we must write

the expression out in full before expanding to

avoid mistakes.

E.g. (x+2y)2

= (x+2y)(x+2y)

= x2+4xy+4y

2.

Exercise.


1. (x - y)2.

2. (2x+3y)2.

3. (r - 2s)2

(Answers: x

2-2xy+y

2, 4x

2+12xy+9y

2, r

2-4rs+4s

2).

When we have more than two brackets

multiplied together we first expand two brackets

and then, using brackets, multiply the result by the

third bracket etc.

Examples.

1. 3(x+2)( y-x) = 3(xy - x2+ 2y - 2x)

= 3xy-3x2+6y-6x.

2. (r+s)(2r+t)(3s-2t)=(2r2+rt+2rs+st)(3s-2t)

= 6r2s-4r

2t+3rts-2rt

2+6rs

2-4rts+3s

2t-2st

2

= 6r2s-4r

2t-rts-2rt

2-6rs

2+3s

2t-2st

2.

To avoid mistakes we should always simplify as

soon as possible.

Exercise.


1. 4(s-2t)(3s-t).

2. (x+y)(x+2y)(x-y).

3. (c-2t)2(t-3c).

4. (x+y)3.

(Answers: 12s2-28st+8t

2, x

3+2x

2y-xy

2-2y

3,

13c2t-3c

3-16ct

2+4t

3, x

3+3x

2y+3xy

2+y

3.)

7

Linear equations.

Target: On completion of this worksheet you should be able to solve linear equations, including those with fractions.

A linear equation is an equation where the

highest power of the variable (usually x) is one.

Examples

3x+2=6 is a linear equation.

2x2+3x+1=8 is not a linear equation as the

highest power of x is two.

The solution to an equation is the value(s) of

the variable that make the equation hold.

E.g. 2 is the solution of the equation 2x + 1 = 5,

since 2 2 + 1 = 5.

Notice that not every value of x makes this

true. E.g. 2 3 + 1 5.

x x 1 5

When we solve a linear equation it helps to

consider the equation as a set of balanced

scales. The two sides are equal. To retain the

balance we can

Add the same amount to both sides.

Subtract the same amount from both sides.

Multiply both sides by the same amount.

Divide both sides by the same amount.

Clearly to find what x is we want to end up

with just x on one side of the scales.

Example. Solve the following.

.2

242

1512

x

x

x

Here the square brackets contain the operation

that we are doing to both sides of the equation.

More examples.

1.

.3

339

5354

3534

x

x

x

xx

2.

.16

282

x

x

3.

.2

336

22

33

52

358

2

35

2

38

x

x

x

x

xx

Exercise. Solve the following:

1. 3a = 12.

2. x + 3 = 7.

3. b – 2 = 5

4. .4

3

b

5. 2a + 5 =9.

6. 5a – 3 = 22.

7. .7

411

x

(Answers: 4, 4, 7, 12, 2, 5, 16.)

A5

8

When the equation contains more than one

term involving the variable, e.g. 3x+2=10-x, we

should first try to get all the terms involving x

on the same side of the equation.

The same rules apply; we do the same to both

sides.

Examples.

1.

.2

336

4432

34623

x

x

x

xxx

2.

.1

121212

75712

33579

x

x

x

xxx

Exercise. Solve the following:

1. 5b – 25 = 3b - 11.

2. 15 + 3x = 10 - 2x.

3. 7 - 3x = 2 - x.

4. .1

24 x

x

5. .42

2

310 x

x

(Answers: 7, -1, 2.5, 2, 4).

If there are brackets in the linear equation then

we should expand the brackets and simplify

before beginning to solve the equation.

Example.

.2

5

4

10

4104

61664

216262

2175562

)3(7)1(5)3(2

y

y

y

yyy

yyy

yyy

Exercise. Solve the following equations.

1. 4(g + 1) = 8.

2. 3(b – 1) –2(3b – 2)=4.

3. 4(x + 3) = 2(x – 3) + 10.

4. 4(a - 5) = 7 – 5(3 – 2a).

5. .9)11(3

256

x

x

6. .1

542

5

23

xx

(Answers: 1, -1, -4, -2, 1, -5)

Fractions in linear equations can cause

confusion. It is often helpful to multiply both

sides of the equation by the denominator of

the fraction (to get rid of the fractions), and

then use brackets to avoid mistakes.

Examples.

1.

.4

3123

2212

62

262

x

x

xxx

xx

xx

2.

.23

15815

484155

)42(2)3(5

55

)42(23

25

42

2

3

x

x

xxx

xx

xx

xx

Exercise. Solve the following equations.

1. .210

5

7x

x

2. .3

3

2

5

3

xx

3. .

3

105

4

53

xx

4. .

4

1

3

4

4

3

pp

5. .

3

11

4

43

5

32u

uu

(Answers: 50/17, 45, -5, -4, 36).

9

Indices

Target: On completion of this worksheet you should be able to evaluate terms with indices and simplify algebraic expressions using indices.

Indices are a mathematical shorthand used when

a quantity is multiplied by itself a number of

times.

a a a can be written as a 3 .

The number 3 is said to be the index of a. In

general

a m = a a a,

where a is written m times.

E.g.

z 4 y 2 = z z z z y y.

E.g.

p p p q q q q q = p 3 q 5 .

We read am as a to the power m.

Remember that p 1 =p.

Exercise.

Write the following in index form.

1. c c c c.

2. d d d.

3. a a b b b b.

4. w w z.

5. r r s s s s t t.

Write the following in full, (without indices).

1. d 3 .

2. k 4 z 3 .

3. p 2 q 3 r.

4. w 2 y 3 w y 4 .

5. yx

yx3

26

.

(Answers: c 4 , d3

, a 42 b , w z2 , r 242 ts ,

d d d, k k k k z z z, p p q q

q r, ,yyyywyyyww

yxxx

yyxxxxxx

.)

wwyyywyyyy,

We can evaluate terms involving indices by

using the xy button on our calculators.

E.g. Evaluate 35.

Type 3 xy

5, to obtain 243.

Exercise.

Evaluate the following.

1. 24.

2. 324

3.

3. 5622

3.

(Answers: 16, 576, 1440).

To simplify expressions with indices we

should write them out in full and then

remember that multiplication and division are

commutative (that is a b = b a).

E.g.

z3y

2 z

4y = z z z y y z z z z y

= z z z z z z z y y y

=z7y

3.

Exercise.

Simplify the following.

1. p5

q6 p

2q

4.

2. w s4 t2 s t

7.

3. l3

h 2t l

6 t.

(Answers: p7q

10, ws

5t9, l

9h

2t2

).

A6

10

When we are working with fractions we should

remember that we simplify fractions by dividing or

multiplying the whole of the numerator and the

whole of the denominator by the same amount.

E. g. 22

b

a

ab , by dividing top and bottom by a.

When we are simplifying we should first write the

expression out in full.

Example.

.3

25

c

baa

cbaaa

bbaaaaa

bca

ba

This is achieved by dividing the numerator and the

denominator by a a a.

The result can then be written as .2

c

ba

Example.

rrstttt

sssttrssttt

srt

strst

24

3223

rrtt

ssrst

.22

2

rt

stsr

Notice that we are dividing both terms in the

numerator by t t s.

Exercise. Simplify the following.

1. .4

26

br

br

2. .24

243

abt

abt

3. .44

29

rbs

tbs

4. .23

s

sts

5. .35

63423

tsb

stbsb

6. .432

234

xzzx

xzyxz

(Answers: ,1

,,.

,,2

4322

2

522

tsb

sbtts

rb

ts

t

abbr

.1

2

22

zxz

yxz

)

There are some rules of indices that enable us

to simplify without writing the expressions out

in full.

.nmnm aaa

nmnm aaa or .nm

n

m

aa

a

.mnnm aa

.10 a

Example.

.523 aaa

When more than one letter is involved we

consider each letter separately.

Examples.

Simplify .324 abba

a4

b2 a b

3 = 2314 ba a

5b

5.

In the same way, when numbers are involved

we consider them separately.

Example.

Simplify .24 2432 yxyx 4

4 x2

y3 2 x

4 y

2 =2x

-2y.

Exercise.

Simplify the following.

1. .34 bb

2. .322 baab

3. .2524 yzxzyx

4. .48 23 xx

5. .423a

6. .523b

7. .520 32 xzpzyx

(Answers: ,,, 1153 yzxbab 2x,

,25,4 66 ba .)4 13 pxy

11

Changing the subject 1.

Target: On completion of this worksheet you should be able to change the subject of formulae including those with fractions, indices and roots.

An algebraic formula is an expression with

an equals sign linking several variables.

E.g. V=IR.

The subject of a formula is the variable

preceding the equal sign. In the previous

example V was the subject.

It is often useful to change the subject of a

formula. We should follow the same rule, do

the same to both sides, as we follow with an

equation. That is we can:

Add the same amount to both sides.

Subtract the same amount from both

sides.

Multiply both sides by the same amount.

Divide both sides by the same amount.

Square the whole of both sides.

Square root the whole of both sides.

Examples.

1) Make d the subject of the formula .dC

.dC

dC

Remark: “” is notation for implies.

2) Make h the subject of the formula

hrV 2 .

.2

22

hr

V

rhrV

3) Make b the subject of the formula .c

ba

.bac

cc

ba

Exercise. Make the letter in brackets the subject.

1. S = t + a (t)

2. PV = T (V)

3. 2A = PQ (Q)

4. V2

= 4gh (h)

5. q

pa (p)

6. v = u + at (u)

(Answers:

.),,4

,2

,,2

atvuaqpg

Vh

P

AQ

P

TVaSt

.,,4

,2,2

atvuaqpg

VhP

AQP

TV )

Frequently more then one operation is required.

Examples.

Make t the subject of v = u + at.

.ta

uv

aatuv

uatuv


1) dlSH (l)

2) q

pa (q)

3) 500

VATS (T)

4) Ax

FlY (x)

5) rR

pEI

(r)

(Answers: ,500

,,VA

ST

a

pq

d

HSl

.),I

IRpEr

YA

Flx

A8

12

When the formula contains brackets we deal

with it in exactly the same way. Sometimes it

is useful to expand the brackets first.

Example.

Make x the subject in the formula S = h (x-t).

.xh

htS

hhxhtS

hththxS

Notice that we could have done this without

expanding the brackets.

.

)(

xth

S

ttxh

S

htxhS

Exercise. Make the letter in brackets the

subject.

1) R2=R1(1-at) (t)

2) )(2 qzpz (q)

3) Z

PST

)(2 (S)

4) C

CxT

)(

(x)

5) )(2 hrrA (h)

6) dnan

S )1(22

(d)

(Answers

,2

,2

,1

21 PTZ

Sp

zpzq

aR

RRt

.))1(

22,

2,

nn

anSdr

r

AhC

TCx

When a formula contains powers or roots we

must be careful about the order of the operations

we use to transform the formulae.

In general rooting or taking powers should occur

as late as possible to avoid algebraic mistakes.

Examples.

Make r the subject of the formulae

.2),3

4) 3

r

ltbrVa

.4

3

4

3

443

33

4)

3

33

3

3

rV

rV

rV

rVa

.2

2

2

22)

2

2

2

rt

l

ll

rt

l

rt

l

rtb


1. S = t2 (t)

2. )(yyx

3. )(2 tuts

4. )(uasup

5. ghV (h)

6. dkv (d)

7. y

xht 3 (x)

8. VPQR 2 (Q)

9. 12

2

2

2

q

y

p

x (p)

10. 2

2

1atVS (t)

(Answers: ,,, 2 ustxySt

,3

,,,)(

2222

h

tyx

k

vd

g

Vhaspu

.))(2

,,22 a

VSt

yq

xqp

PV

RQ

)

13

Changing the subject 2.

Target: On completion of this worksheet you should be able to change the subject of formulae where the desired subject occurs more than once in the formulae.

Sometimes when we are changing the subject of

a formula the variable that we want to make the

subject occurs more than once. In this case it is

necessary to transform the formula so that the

variable is only written once. We can do this by

Getting all the terms with the variable on one

side of the formula.

Taking the variable out as a common factor.

Example.

Make x the subject of the formula

a) cx + dx = m and b) xy = x – bx + c

.

)()()

dc

mx

dcmdcxa

.1

)1()1(

)

by

cx

bycbyx

cxbxxy

xcxbxxy

bxcbxxxyb

Exercise.

Make the letter in brackets the subject.

1. heyby (y)

2. mststct 3 (t)

3. pxmxmnx 423 (x)

4. gsftysrt (t)

5. 543 sqtqppq (q)

(Answers:

,23

4,

3,

mmn

px

sc

mst

eb

hy

.)3

45,

stp

pq

fr

ysgst

It often simplifies a problem involving fractions

if we first transform the formula to one without

fractions by multiplying the entire formula by the

denominator of the fraction.

Example.

Make x the subject of the formula .rx

yxI

.

)()(

)(

)(

xIy

Ir

IyIyxIr

IxyxIr

IxyxIrIx

yxrxI

rxrx

yxI

A9

14

Exercise.


1. d

sdg

(d)

2. pf

pf

d

D

(p)

3. 2

222

y

ryx

(y)

4. gmM

lmMT

)2(3

)3(4

(m)

(Answers:

,(

,1 22

)22

2 Dd

dDfp

g

sd

,1 2x

ry

.)

)8(6

)163(22

22

gTl

lgTMm

Exercise.


1. plxmx (x)

2.rcRL

uLm

(L)

3.S

nemeP

22 (e)

4.y

yxa

2 (y)

5.Kc

bK

x 53

241

(K)

(Answers: ,,mu

mrcRL

lm

px

,1

)(2,

xa

xay

nm

Pse

.)

54

23

x

bxcK

When powers and roots are involved we should

deal with them as late as possible.

Example. Make p the subject of the formula

.4

ugp

qptg

.

4

44

4

4

44

4)(

)(4

4

4

2

2

222

22

222

22

2

2

2

t

gg

t

guq

p

t

gg

t

guq

t

ggp

t

guqpp

t

gg

t

guq

t

gupp

t

gg

pqpt

gup

t

gg

qpt

gugp

ugpugp

qp

t

g

ugp

qp

t

g

tugp

qptg

15

Brackets and Quadratics

Target: On completion of this worksheet you should understand how expanding brackets can lead to quadratic expressions.

A polynomial expression is an algebraic

expression involving positive whole number

powers of a letter.

Example. 5113734 23456 xxxxxx

is a polynomial.

The coefficient of xn is the constant in the term

involving xn.

Example.

The coefficients of x6, x

3 and x in the example

above are 4, -3, and 1 respectively.

Note that the sign stays with the coefficient.

Exercise.

State the coefficients of y4, y

2 and y in the

following polynomials.

1. .13724 23457 yyyyyy

2. .42

43 25 y

yy

(Answers: -2, -1, 3; 0, -4, .2

1)

A quadratic expression is a polynomial that

has 2 as the highest power of its variable.

Example.

1. 3x + 4 is not a quadratic expression.

2. 4x3

+ 3x2

+ 2 is not a quadratic expression.

3. 9x2

+ 3x + 4, -2x2

+ 8, 8x2

- 3x, 6x2 are all

quadratic expressions

Sometimes we will not immediately be able to

tell that an algebraic expression is quadratic

because it contains brackets. However when

we expand the brackets it becomes clear.

E.g. .44 2 xxxx

Exercise.

Expand the brackets and hence write these

quadratic expressions in the form

,2 cbxax where a, b and c are just

constants.

1. ).23( xx

2. ).84(3 xx

3. ).26( xx

4. ).8)(4( xx

5. ).6)(32( xx

6. .)8( 2x

(Answers:

.)6416,18152,324

,62,2412,23

222

222

xxxxxx

xxxxxx

It is sometimes useful to be able to do the

reverse process, which is put a quadratic

expression into the form of two linear

expressions multiplied together.

We will make some observations that will

enable us to do this with greater ease.

Exercise.

Expand the brackets and write the quadratic

expressions in the form .2 cbxax

1. .)2( 2x

2. .)3( 2x

3. .)4( 2x

4. .)2( 2x

5. .)8( 2x

(Answers: ,96,44 22 xxxx

.)6416,44,168 222 xxxxxx

A10

16

Exercise.

In the previous exercise what did you notice

about the coefficient of x2 and x and the units?

(Answers: The coefficient of x2 was always 1.

The coefficient of x was twice the units in the

bracket.

The units were the square of the units in the

bracket.)

The quadratic expressions involved in the last

two exercises are called perfect squares. This

is because they can be written as the square of

a linear expression.

Exercise.



1. ).2)(2( xx

2. ).3)(3( xx

3. ).4)(4( xx

4. )2)(2( xx

5. ).83)(83( xx

6. ).12)(12( xx

(Answers: ,9,4 22 xx

.)14,649,4,16 2222 xxxx

Exercise.



(Answers: The coefficient of x2 was always

the square of the coefficient of x in the

brackets.

The coefficient of x was always 0.

The units were the square of the units in the

bracket.)

The quadratic expressions involved in the last

two exercises are called the difference of two

squares. This is because they can be written

as the square of one algebraic term subtracted

from the square of another algebraic term.

Exercise.



1. ).3)(2( xx

2. ).5)(3( xx

3. ).6)(4( xx

4. ).4)(2( xx

5. ).7)(8( xx

(Answers: ,158,65 22 xxxx

.)5615,82,2410 222 xxxxxx

Exercise.



(Answers: The coefficient of x2 was always 1.

The coefficient of x was the units in the

brackets added together.

The units were the units in the brackets

multiplied together.)

Exercise.



1. ).33)(22( xx

2. ).55)(3( xx

3. ).65)(42( xx

4. ).45)(23( xx

5. ).74)(87( xx

(Answers: ,15205,6126 22 xxxx

.)568128

,8215,243210

2

22

xx

xxxx

Exercise.


about the coefficient of x2 and the units?

(Answers: The coefficient of x2 was always

the coefficients of x in the brackets multiplied

together.

The units were the units in the brackets

multiplied together.)

17

Factorisation of quadratics

Target: On completion of this worksheet you should be able to factorise any quadratic expression.

A factor is a term that divides an expression

leaving no remainder.

Examples.

1. 3 is a factor of 6.

2. 1, x, 3, x2, 3x, 3x

2 are factors of 3x

2.

When we factorise an expression we write it as a

product some of its factors.

E.g. 6 = 3 2.

There are three main methods for factorising

quadratic expressions.

By finding a common factor.

By identifying the expression as the difference

of two squares.

By considering guessing the factors and

checking by expanding the brackets.

To find a common factor, in an algebraic

expression, we must first find the factors of its’

constituent terms. Then we find the (largest) factor

that its terms have in common.

E.g. Find the largest common factor of 3x+6x2.

3x has factors 1, 3, x and 3x,

6x2 has factors 1, 2, 3, 6, x, 2x, 3x, 6x, x

2, 2x

2, 3x

2,

6x2.

The largest factor they have in common is 3x.

To factorise an expression with a common factor

we need to find what the common factor should be

multiplied by in order to get the original expression.

Examples.

1. Factorise 3x + 6x2.

3x + 6x2

= 3x(1 + 2x).

2. Factorise xy 2- yz.

xy 2- yz = y(xy - z).

See the algebra sheet on common factors if you have difficulty

with this.

Exercise.

Factorise the following expressions by finding a

common factor. In each case check your answer

by expanding the brackets.

1. 4x2 + 2x.

2. 3x2 - 6x.

3. 2x2

- 4x.

4. 6r2 + 4r.

.

5. 4sw2 - sw.

6. 4x2 - 5x.

7. r(x + y)2 + p(x + y).

(Answers: 2x(2x+1), 3x(x-2), 2x(x-2), 2r(3r+2),

sw(4w-1), x(4x-5), (x+y)(r(x+y)+p).)

Sometimes an algebraic expression does not

have any common factors. We must then try one

of the other methods.

A perfect square is a term that has a square

root.

E.g. 9, x2

and 4y2

are all perfect squares.

See the algebra sheet on indices if you have difficulty with

this.

If an algebraic expression consists of two perfect

squares one subtracted from the other then we

say it is the difference of two squares. We can

factorise these expressions simply by first

finding the square root of its’ terms.

Example. One plus and one minus sign.

x2

- y2 = ( x – y ) ( x + y ) .

Square root of 1st term. Square root of 2nd term.

Example.

4z2 - 9 = (2z-3)(2z+3).

A11

18

Exercise.

Factorise the following. Check your answers by

expanding the brackets.

1. y2

- s2.

2. z2

- 16.

3. 9 - p2.

4. 4r2

- 25s2.

5. x2y

2 - 36r

2

6. (x + y)2

- y2.

(Answers: (y - s)(y + s), (z - 4)(z + 4),

(3 - p)(3 + p), (2r - 5s)(2r + 5s),

(xy - 6r)(xy + 6r), x(x + 2y).)

When a quadratic algebraic expression has no

common factors and is not the difference of two

squares we must guess the factors. We know that it

must be written as the product of two brackets. We

should then check our guess by expanding the

brackets.

Clearly this could take a long time. To make a

sensible guess we should consider the following.

Multiply a and b together to get ab.

(x+a)(x+b)=x2+(a+b)x+ab.

Add a and b together to get the coefficient of x.

Example

Factorise 1. x2 + 9x + 20 and 2. y

2 2y – 8.

1. The possible values of ab satisfying ab=20 are

120, 210, and 54.

Of these only 5 and 4 add together to give 9,

therefore

x2 + 9x + 20 = (x + 4)(x + 5).

2. The possible values of ab satisfying ab=-8 are

1-8, 2-4, 4-2 and 8-1.

Of these only 2 and –4 add together to give –2

therefore

y2 - 2y – 8 = (y - 4)(y + 2).

Notice that it helps to list the possibilities of ab

first.) Exercise. Factorise the following:

1. y2

+ 7y + 12.

2. x2

+ 6x + 9.

3. r2

+ 15r + 36.

4. x2 – 8x + 16.

5. y2

- 4y - 32.

6. p2 + p –12.

7. z2

+ 30z - 64

(Answers: (y+3)(y+4), (x+3)(x+3), (r+12)(r+3),

(x-4)(x-4), (y-8)(y+4), (p+4)(p-3), (z + 32)(z-2).)

When the coefficient of x2 is not one we have to

guess the factors more carefully.

We should consider the following.

Multiply p and q together to get the coefficient of x2.

(px+a)(qx+b)=pqx2+(pb+qa)x+ab.

Multiply a and b together to get ab.

The following process is helpful.

List the possibilities for p q.

List the possibilities for a b.

Try each possible pair of brackets and check

them by expanding the brackets.

Example. Factorise 2x2

+ 11x + 12.

Possible values of p q are 21.

Possible values of a b are 112, 26, 34,

43, 62, 121. (Notice that the ordering

matters)

Try (2x+1)(1x+12). Expanding gives

2x2+25x+12, so this is wrong.


2x2+14x+12, so this is wrong.


2x2+11x+12 so this is correct.

Therefore,

2x2

+ 11x + 12 = (2x + 3)(x + 4).

Example. Factorise 3x2

+ 25x - 18.

31.

1-18, 2-9, 3-6, 6-3, 9-2, 18-1,

-181, -92, -63, -36, -29, -118.

Trying the possibilities (3x+1)(1x-18),

(3x+2)(x-9), etc gives us

3x2

+ 25x – 18 = (3x - 2)(1x + 9).

Exercise. Factorise the following:

1. 3x2 + 11x + 6.

2. 5x2

+ 36x + 7.

3. 7x2

+ 26x - 8.

4. 3x2

- 13x + 12.

5. 2x2

+ 2x - 12.

(Answers: (3x+2)(x+3), (5x+1)(x+7),

(7x-2)(x+4),(3x-4)(x-3),(2x+6)(x-2).)

19

Algebraic Fractions

Target: On completion of this worksheet you should be able to simplify fractions involving algebraic terms.

An algebraic fraction is a fraction involving

algebraic terms.

Example.

)3(4

32,

3

2,

1 2

x

xxy

x are all algebraic fractions.

We can obtain equivalent algebraic fractions

by dividing or multiplying the whole of the

numerator and the whole of the denominator by

the same thing.

Examples.

1. .2

63

xx [ Multiplying top and bottom by 2]

2. .33

2 xx

x [Dividing top and bottom by x.]

3. .2

3

6 2

x

y

yx

y [Dividing top and bottom by 3y.]

4. .146

73

146

73

yp

y

xyxp

xyx

[Dividing the top and bottom by x. Notice

that all the terms are divided.]

If you are having difficulties with this see the

number sheet on equivalent fractions.

We can use equivalent fractions to simplify

algebraic fractions. We should:

Find the highest common factor of all the

terms involved.

Divide the whole of the top and the whole of

the bottom of the fraction by this factor.

Example.

.49

2

1227

6323

2

p

p

pp

pp

Here we divided top and bottom by the common

factor of ,27,6,3 32 ppp and 12p namely 3p.]

Exercise.

Simplify the following:

1. .3

6

x

2. .10

5 2

y

y

3. .6

3 2

zp

zz

4. .24

633

2

pppa

ppq

5. .23

5

x

yzx

(Answers: ,24

63,

6

3,

2,

22

pa

pq

p

zy

xcannot be

simplified.)

A12

20

If a term involves two things multiplied together then

when we divide it we just divide one of the two things.

E.g. .22 xx

This works equally well when one of the things

involves brackets.

Examples.

1. ).3(2)3(2 xx

2. ).4()4( xxxx

3. ).3()4()4)(3( xxxx

Exercise.

1. 2)3(4 x

2. xyxx )3(

3. )92()83)(92( yyy

(Answers: 2(x + 3), (3x + y), (3y – 8).)

When we simplify fractions with brackets we again divide the

whole of the numerator and the whole of the denominator by the

same thing.

Example.

1. .3

)2(

9

)2(322 x

x

x

x

[Dividing top and bottom

by 3.]

2. .3

)4(

)62(3

)62)(4(

x

x

xx[Dividing top and

bottom by (2x-6).]

Exercise. Simplify the following algebraic fractions.

1. .)3(3

)3(2

x

x

2. .)5(3

)5)(2(

x

xx

3. .)32)(4(

)32( 2

xx

x

4. .)13)(24(

)24(2

xxx

xx

(Answers: .)13(

,)4(

)32(,

3

)2(,

3

2

x

x

x

xx)

It may at first appear that some algebraic fractions

cannot be simplified but on further inspection they

can be.

E.g. .)74(

)4(5)2(3

x

xx

It helps to follow the following procedure:

Expand the brackets.

Simplify the numerator.

Simplify the denominator.

Factorise the numerator and the denominator.

Simplify the fraction.

Example.

Simplify )74(

)4(5)2(3

x

xx.

.274

)74(2

74

148

74

20563

)74(

)4(5)2(3

x

x

x

x

x

xx

x

xx

Exercise.


1. .63

2

x

x

2. .5

102

x

x

3. .155

32

2

xy

xy

4. .123

8232

2

pp

pp

5. .12

12

xx

x

6. .98

92

xx

x

7. .12

32

xx

x

8. .107

562

2

xx

xx

9. .)4(3

)303()10(3 2

x

xx

(Answers:

).5(,2

1,

)4(

1,

)1(

1,

)1(

1,

3

2,

5

1,2,

3

1

x

x

x

xxxp)

21

Algebraic fractions and operations

Target: On completion of this worksheet you should be able to add, subtract, multiply and divide algebraic fractions.

When we multiply fractions we multiply the

numerators together and the denominators

together.

Examples.

.3

10

6

20

1

4

6

54

6

5

.21

12

7

6

3

2

We do exactly the same with algebraic

fractions. We should always simplify our

answers.

Examples.

1. .33

sd

ab

d

b

s

a

2. .3

2

18

12

18

1222332 baba

ab

a

b

b

a

3. )4)(3(6

)3(2

)4)(3(6

)3(2

xxx

xxy

xx

xy

x

x

)4(3

x

y

Exercise.


1. .9

6

124

3

2

2

x

y

y

x

2. .34

2

3 ad

b

b

as

3. .2

4cp

hs

4. .)2(5

)3(4

3

2

px

zy

y

x

(Answers: .5

4,

2

4,

12,

182 p

z

cp

hs

bd

s

x

y

)

When we divide fractions we invert the

second fraction and then multiply the

numerators together and the denominators

together.

Examples.

.24

5

4

1

6

5

1

4

6

54

6

5

.7

8

21

24

3

4

7

6

4

3

7

6

We do exactly the same with algebraic

fractions. We should always simplify our

answers.

Examples.

1. .33

3

sb

ad

b

d

s

a

d

b

s

a

2. .43

12

3

123123

323

232a

ab

ba

ba

ab

b

a

ab

ba

b

a

3. x

xx

x

x

3

)3)(5(

6

)3(2

)3)(5(

3

6

)3(2

xx

x

x

x

)5)(3(6

)3(6

xxx

xx

.)5(

1

x

A13

22

Exercise.


1. .9

5

122

3

2

2

y

x

y

x

2. .10

6

55

3

2

4

e

d

e

cd

3. .8

24

48

162

5

2

3

x

yz

y

xyz

4. .9

6

184

3

2

2

q

r

p

r

5. .40

15

5

88

5

2

4

q

pr

q

pr

6. .22

3 2

pq

rp

7. .)(4

6

)(

)(3 2

ssr

q

sr

qqp

8. .)(3

)(2

2 22

22

ssr

qp

srsr

qp

(Answers:

.))(2

)(3,)(2

,4

3,

15

64,

3,

9,

12,

15

4 6

2

4

22

33

sr

qpsqsqp

q

rp

r

q

rp

q

zy

xcde

x

When we add or subtract fractions we find a

common denominator, express each of the

fractions as equivalent fractions with this

common denominator and then add the

numerators.

Example.

.21

23

21

14

21

9

3

2

7

3

If you have difficulty with this refer to the number

sheet on fractions.

We do the same with algebraic fractions.

First we need to be able to find a common

denominator. The easiest way to do this is to

multiply the denominators together.

Example.

A common denominator of 3a

s and

)(

)(

srqp

sp

is (a+3)(qp(r+s)).

Notice that putting all the denominators in

brackets avoids mistakes.

Exercise. Find the common denominator of

)(

)(,

sr

qp

d

a

and .

ba

sr

(Answer: (d)(r+s)(a+b) .)

We then need to be able to express each fraction

as an equivalent fraction whose denominator is

the common denominator.

Remember that equivalent fractions are obtained

by multiplying the top and the bottom of the

fraction by the same thing.

Example. Multiply by (r+s)(a+b)

.))((

))((

basrd

basra

d

a

Multiply by (r+s)(a+b)

To add or subtract algebraic fractions we must:

Find a common denominator.

Express each fraction as an equivalent

fraction with this denominator.

Add or subtract the numerators.

Simplify

Example.

.)(

2)(

)(

)(2

)(

)(2

srb

cbsra

srb

bc

srb

sra

sr

c

b

a

Exercise. Simplify the following:

1. .32

aa

2. .5

2

4

baab

3. .c

b

d

s

4. .2

fa

c

c

a

5. .2

p

q

qp

q

6. .2

yx

yx

yx

yx

(Answers:

.)))((

4

,)(

)(,

2,,

20

3,

6

5

2

22

yxyx

y

pqp

qpq

ca

fcaca

dc

dbscaba

23

Quadratic equations

Target: On completion of this worksheet you should be able to recognise and solve quadratic equations.

A quadratic equation is an equation where the highest

power of the variable (usually x) is two.

Examples.

2x2 + 3x + 1 = 8 is a quadratic equation.

3x + 2 = 6 and x3 + 2x – 4 = x are not quadratic

equations.

A solution of a quadratic equation is a value of the

variable that makes the equation hold.

E.g. 5 is a solution of the equation ,01522 xx since 52-25-

15=0.

-3 is also a solution of the equation. (Check this for

yourself.)

Often there are two solutions to a quadratic equation, but

sometimes the solutions are identical or don’t exist in

our number system. There will never be more than two

solutions.

In order to solve a quadratic equation we must use an

important property of numbers:

If the product of two values is zero then one of the two

values itself must be zero.

That is if

=0

either = 0 or = 0.

This means that if we can write a quadratic expression

that is equal to zero as the product of two linear

expressions then the value of one of the linear

expressions must be zero. Since we can solve linear

equations we would then be able to solve the quadratic.

Therefore to solve a quadratic we

Make one side of the equation zero.

Factorise the quadratic expression.

Solve the resulting linear equations.

Examples. Solve

a) x2 2x – 15 = 0, b) x

2 - 2x + 1 = 0,

c) 2x2 3x = 0, d) 2x

2 + 7x = -3.

0)3)(5(

0152) 2

xx

xxa.

.0)3()5( xx

Either x – 5 = 0 or x + 3 = 0.

We now solve these linear equations.

.5

]5[05

x

x

.3

]3[03

x

x

.0)1()1(

.0)1)(1(

012) 2

xx

xx

xxb

Either x 1 = 0, or x 1 = 0. Hence x = 1 or

x = 1.

The two solutions are identical.

.032

0)32(

032) 2

xx

xx

xxc

Either x = 0 or 2x 3 = 0.

Solving these linear equations gives

x = 0 or x = .2

3

.0)3()12(

)3)(12(

0372

3372)

2

2

xx

xx

xx

xxd

Either 2x + 1 = 0 or x + 3 = 0.

Solving these linear equations gives

2

1x and x = -3.

A14

24

Exercise.

Solve the following quadratic equations.

1. .0652 xx

2. .020122 xx

3. .021102 xx

4. .0452 xx

5. .0822 xx

6. .1892 xx

7. .06132 2 xx

8. .012102 2 xx

9. .0164 2 x

10. .0259 2 y

11. .046 2 xx

12. .083 2 xx

13. 492 2 xx .

14. .11122 2 xx

15. .65)32( xxx

16. .24)2(212)1( xxxx

(Answers: -3, -2; -10, -2; -7, -3; -4, -1; 4, -2; 6, 3;

-1/2, -6; 2, 3; -2, 2; -5/3, 5/3; 0, 2/3; 0, 3/8; -4, -1/2;

4, -3/2; -5/2, 1; -4, 3.)

25

Quadratic Formula

Target: On completion of this worksheet you should be able to solve quadratic equations by using the quadratic formula.

Sometimes it is impossible to factorise the

quadratic expression. Fortunately there is a

quadratic formula that enables us to find the

solutions, in such cases.

If ax2+bx+c=0, where a, b and c are constants then

.2

42

a

acbbx

To use this formula we should:

Identify a, b and c.

Substitute them into the formula.

Example.

Solve the equation 4x2

+ 5x – 6 = 0 using the

formula.

Here a = 4, b = 5 and c = -6.

The formula gives the solutions.

.8

115

8

96255

42

)6(4455 2

x

Taking the “+” and “” in turn gives two solutions.

4

3

8

6

8

115

x

and

.28

16

8

115

x

Exercise.

Solve the following quadratic equations giving your

answers to three decimal places.

1. 0132 xx

2. 012 xx

3. 0323 2 xx

4. 0252 xx

5. 05136 2 xx

6. xx 1136 2

7. xx 215 2

(Answers: -0.282, -2.618; 0.618, -1.618;

0.721, -1.387; -0.438, -4.562; -0.5, -1.667; 1.5,

0.333; 0.290, -0.690.)

Often we will need to use a calculator to evaluate x.

Example

Solve x2

+ 5x + 2 = 0 giving your answer to 3

decimal places.

Here a = 1, b = 5 and c = 2.

438.02

175

12

21455 2

x 3dp

and

562.42

175

12

21455 2

x 3dp

Remember that we should use brackets when putting

this into calculators to ensure that the whole of the

numerator is divided by the whole of the

denominator.

See the number sheet on the use of a calculator if you are having


A15

26

Some equations with algebraic fractions may be

rearranged to form quadratic equations. To

rearrange these equations we should multiply the

whole equation by the denominator of a fraction

and use brackets to avoid mistakes.

Examples.

2

1

0)12)(12(

014

01

4)

2

x

xx

x

xx

xa

or .2

1x

2

0)5)(2(

0107

010)7(

707

10)

2

x

xx

xx

xx

xx

xb

or .5x

194940

4863241129

)243162(22718567

)8)(32(2)32(9)8(7

8)32(28

)32(97

3228

9

32

7)

2

2

2

xx

xxxx

xxxxx

xxxx

xxx

x

xxx

c

401.0 x (3 d.p.) or x = -11.849 (3 d.p.).

See the algebra sheet on algebraic fractions if you have


Exercise.

Solve the following.

1. .02

1

xx

2. .35

12

1

5

xx

3. .81

3

5

2

xx

x

4. .32

4

5

1

x

x

5. .12

23

34

13

t

t

t

t

(Answers: 0.707, -0.707; -7/5, 4; 3.936, -0.636;

12.633, -0.633; 3.423,0.243.)

Some quadratic equations do not have any real

solutions.

Example.

Solve .042 xx

Here a = 1, b = -1and c = 4. Using the formula

gives:

1 1 16 1 15.

2 2x

As 15 doesn’t exist in the real numbers there

are no real solutions to this equation.

We can tell how many solutions there are to a

quadratic equation by looking at the value of

.42 acb

If 042 acb it has two solutions.

If 042 acb it has one solution.

If 042 acb it has no real solutions.

27

Negative and Fractional Indices

Target: On completion of this worksheet you should understand and be able to use properties of negative and fractional indices.

Sometimes in algebraic expressions we see a term

with a negative or fractional or zero index. It is

important to understand what these indices mean.

There are two rules that help us understand their

meaning:

.nmnm aaa

.nmnm aaa

We can write a-2

a different way, by choosing two

numbers, say 4 and 6, that differ by two.

.1

26

4642

aa

aaaa

That is

.1

2

2

aa

We could repeat the same process for a-3

, a-1

, etc. In general

.1m

m

aa

Examples.

Express the following as fractions without

negative indices .)(,)(, 2216 xyxyxx

1. .1

6x Note the size of the index remains the

same but the term becomes the denominator of

a fraction.

2. .1

xy

3. .11

)(

1222

2

22

2

2

2

yyx

x

yxx

xyx Note that

to avoid mistakes each term is dealt with

separately and then multiplication of fractions

is used.

Exercise.

Express without negative indices and simplify the

following:

1. .7b

2. .)( 2cd

3. .)( 1pq

4. .12 px

5. .4yz

6. .)( 23 lml

7. .))(( 122 spsp

(Answers: .))(

1,,,,

1,

)(

1,

124

2

27 spm

l

z

y

p

x

pqcdb

We can write 0a a different way, by choosing two

numbers, say 3 and 3 that differ by zero.

.13

3330

a

aaaa

Thus

.10 a

Examples.

Simplify .2 323 yxyx

.222 330323 yyxyxyx

Exercise.


1. .24 226 rssr

2. .623 3222 qpqqp

(Answers: 2r5, 1)

A16

28

Before we can express fractional indices a different

way, we need to remember what the square root of a

term is. It is a term which when multiplied by itself

gives the original term.

E.g. 2 is the square root of 4 since .422

x is the square root of x2 since .2xxx

For any term t, it is clear that

.ttt

In the same way the cube root of a term is the term

that can be multiplied by itself three times to obtain

the original term.

.333 tttt

We can express 21

x a different way.

.121

21

21

21

xxxxx

So 21

x when multiplied by itself gives x. In other words

.21

xx

We can express 31

x a different way.

.131

31

31

31

31

31

xxxxxx

So 31

x when multiplied by itself and itself again

gives x.

In other words

.331

xx

In general

.1

mm xx

Examples.

Express without fractional indices and simplify

.,,)(, 31

23

21

41

pecdb

1. .4 b

2. .cd

3. .)( 321

323

eee (Notice that we are using the

rule mnnm aa )( here.)

4. .11

331

31

ppp

Exercise.

Simplify and express the following without

fractional indices.

1. .81

c

2. 51

d .

3. .)( 61

cd

4. .32

f

5. .45

g

6. .)( 32

pq

7. .21

xx

8. .32

yy

9. .21

p

10. .32

s

11. .35

232

pqqpp

12. .824 2231

prrpp

(Answers:

.)

1,

1,

1,

1

,,,)(,,,,,

3 43 2

333 24 53 2658

prqsp

yxpqgfcddc

To avoid confusion, when we are asked to

simplify expressions with fractional indices we

should

First simplify using the rules of indices.

Then write the simplified form without

fractional indices.

29

Simultaneous Equations

Target: On completion of this worksheet you should be able to solve two linear simultaneous equations by elimination.

An equation with two unknowns has many

solutions.

Example.

a + 2b =10 has solutions:

a = 1, b = 4.5;

a = 2, b = 4;

a = 4, b = 3;

and many more!

If we have two equations with two unknowns then

there is often one solution that satisfies both

equations.

Example.

a + 2b = 10 and b + a = 6 both have a solution

a = 2 and b = 4.

Similarly if there are three equations with three

unknowns there is often one solution that satisfies

all three equations etc.

Finding the solutions that satisfy all the equations

given is called solving the equations

simultaneously.

In order to solve simultaneous equations we need

to remember two algebraic tools that we have met

previously.

We can multiply or divide the whole of an

equation by an amount.

We can solve equations with one unknown.

If you have difficulty with this then refer to the algebra sheet

on linear equations.

In addition we need to learn a new algebraic tool:

We can add or subtract a pair of equations to obtain a new

equation.

We can see that this makes sense by considering a

pence to be the price of an apple and b pence to be

the price of a banana.

If someone buys one apple and two bananas for 64p

and his friend buys one apple and one banana for

42p then two apples and three bananas must cost

106p. That is

1a + 2b = 64

1a + b = 42

Adding equations and gives:

.4264)11()21( baba

On simplifying we obtain:

2a + 3b = 106.

Exercise.

If x + y = 5 and 3x + 4y = 18 solve the following:

1 4x + 5y =

2 2x + 3y =

3 2x + 2y =

4 x + 2y =

5 3x + 3y =

6 y =

7 6x + 8y =

(Answers: 23, 13, 10, 8, 15, 3, 36.)

A17

30

When we have a pair of equations with two

unknowns we try to add or subtract the equations

to get rid of one of the unknowns. This is called

the process of elimination. This leaves us with an

equation with one unknown that we can then

solve.

Examples.

1. If 2x + y = 5 and x + y = 3, find x.

52 yx

3 yx

35)()2( yxyx

.2 x

2. If x – y = 7 and 2x + y = 23, find x.

7 yx

22 yx

+ 30)2()( yxyx

.10

]3[303

x

x

Notice that the numbers circled at the side name

the equations and then explain what we are doing

with them in each successive line. This avoids

confusion in more complex examples.

If you have difficulty with this then refer to the algebra sheet on

brackets.

To find the solution to the equations we must find

both unknowns. Once we have found one

unknown we can substitute the value of it into one

of the original equations to find the other

unknown.

Example.

In example 1) above we substitute x = 2 into

2x + y = 5 to obtain

4 + y = 5.

We solve this equation to obtain

y = 1.

Therefore the full solution is x = 2, y = 1.

Exercise.

Solve the following simultaneous equations.

1. 3w + s = 10, 2w + s = 7.

2. 4t + r = 12, 2t + r = 8.

3. 5p + 2y = 11, 2y + p =3.

4. 6h + l = 19, 5h - l = 14.

5. 3j – 2k = 3, 5j + 2k = 37.

6. 3c-4d = 32, 4d + c = 0.

(Answers: w=3, s=1; t=2, r=4; p=2, y=1/2;

h=3, l=1; j=5, k=6; c=8, d=-2)

So far the equations we have been given have had

the same number of one of the unknowns. In general

this will not be the case. We therefore need to

multiply one or both of the equations by an amount

so that we obtain two new equations with the same

amount of one unknown.

We should:

Obtain two equations with the same number of

one unknown.

Eliminate one of the unknowns and solve.

Substitute the value of this unknown into one of

the original equations and solve.

Examples.

1. Solve 2x + 3y = 18 and x + 2y = 11.

1832 yx

112 yx

2 2242 yx

2218)42()32( yxyx

.4

4

y

y

Substituting y = 4 into gives

2x + 12 = 18.

Solving gives x = 3.

2. Solve 2x + 3y = 20 and 3x - 2y = 17.

2032 yx

1723 yx

3 6096 yx

2 3446 yx

]13[2613 y

.2 y

Substituting y = 2 into gives 2x + 6 =20.

Solving gives x = 7.

31

Exercise.

Solve the following pairs of simultaneous

equations.

1. 2x – y = 4, x + 2y = -2.

2. 3x – y = 6, 2x + 3y = 4.

3. 2x – 3y = 9, 4x – y = 8.

4. 2x + 3y = 11, 4x + y = 12.

5. 3x + 4y = 25 4x – 3y = 0.

6. 2x + 5y = 16, 3x – 2y = 5.

7. a + 3b = 7, a + b = 3.

8. 3x – 2y = 10, x + 2y = 6.

9. 5a – 3b = 16, 4a + 2b = 4.

10. 2a + 5b = 11, 7a + 3b = -5.

(Answers: x=1.2, y=-1.6; x=2, y=0; x=1.5, y=-2;

x=2.5, y=2; x=3, y=4; x=3,y=2; a=1, b=2;

x=4, y=1; a=2, b=-2; a=-2, b=3.)

32

Simultaneous equations and the method of substitution

Target: On completion of this worksheet you should be able to solve quadratic and linear simultaneous equations by substitution.

It is impossible to solve some pairs of equations by

elimination.

Example.

3x + y = 11 and x2 + 2y

2 = 59.

We therefore have an alternative method- the

method of substitution.

As with elimination the method of substitution

reduces the two equations to one equation with one

unknown.

To apply the method of substitution we should:

Rearrange one equation to make one of the

unknowns the subject.

Substitute the expression for this unknown into

the other equation.

Example.

Solve the simultaneous equations:

3r + 2s = 8 and 2r – 4s = 16.

First name the equations to avoid confusion.

3r + 2s = 8

2r - 4s =16.

Rearrange to make r the subject.

.3

28

]3[283

]2[823

sr

sr

ssr

Substitute into . Use brackets to avoid mistakes.

.1643

282

s

s

.1643

282

s

s

Then solve, expanding the brackets carefully!

.2

]16[1632

]48[164816

]4[1248416

]3[4163

416

]4[1643

416

s

s

s

sss

ss

sss

Then we substitute this value of s into one of the

original equations to obtain the full solution,

3r – 4 = 8.

Hence r = 4.

Exercise.


equations using the method of substitution.

1. a + b = 7, 2a + 3b =18.

2. s + 2t = 14, 3s – t = 0.

3. 2a 3b = 0, 3a – 2b = 5.

4. 4w –3t = 34, 2t +3w = 17.

5. 7p + 6 = 2r, 3r – 2p = 26.

6. z+2l-13= 7, 3z –2l = -4.

(Answers: a=3, b=4; s= 2, t=6; a=3, b=2;

w=7, t=-2; p=2, r=10; z=4, l=8.)

This method may appear to be more complicated

than elimination but it enables us to solve non-

linear simultaneous equations.

A18

33

Whenever there are indices involved in

simultaneous equations we should solve them by

substitution. We must be careful when we expand

our brackets. We should always write an

expression out in full in order to avoid confusion.

When we solve a linear equation and a quadratic

equation by substitution we should always first

rearrange the linear equation.

Example.

Solve the following equations simultaneously.

.12

313 22

yx

yxyx

First name the equations.

313 22 yxyx

.12 yx

Rearrange to make y the subject.

.12 xy

Substitute into .

.)12()12(3 22 xxxx

Solve.

.2,11

15

0)2)(1511(

030711

]31[311711

3114437

31)12)(12(36

2

2

22

22

xx

xx

xx

xx

xxxx

xxxxx

For each of these solutions we now need to find

the value of y.

Substituting 11

15x into gives

.11

41

111

30

111

152

y

y

y

Similarly substituting x = 2 into gives y = 3.

(Check this for yourself.)

Exercise.


equations.

1. .423

12

22

yxyx

yx

2. .113

823

22

yx

yx

3. .562

5

2

yxy

yx

4. .3

832

22

yxyx

yx

(Answers: x=7, y=3; x=2, y=1 and x=-18, y=31;

x=9, y=4 and x=-9, y= -14;

x=1, y=2 and x=1.947, y=1.368.)

34


All measurements consist of a numerical value AND a unit.

Conversion factors allow us to change a measurement from one kind of unit to another. For example, 5 minutes can be changed to seconds by using the conversion

60 seconds=1 min

Which can also be written as a conversion factor

60 seconds per min or 60 s×min-1

A conversion factor always has two different units of the same type (for example minutes and seconds).

Use the conversion factor to convert 5 minutes into seconds. Answer

.300

605

min60min55 1

s

s

sinutesm

A rate is like a conversion factor but involves units of different types. For instance, if a jogger runs 1 kilometre in 8 minutes we say that she runs at a rate of

.min8

1min

8

1

8

1 1 kmperkminutesm

kilometre

We also call this her speed. The reverse (inverse, reciprocal) is also a rate

.min8min81

8 1 kmkmperkilometre

inutesm

In the picture below identify any conversion factors or rates!

Use one of these rates to find out how long the jogger takes to run 2 kilometres. Answer

.min16min82 1 kmkm

35

The first thing to do is to identify and write down any rates and conversion factors in the problem we are given (when in doubt write as a conversion). Rates and conversion factors can be multiplied together to give new rates and conversion factors. We just have to remember to cancel the units properly when we multiply.

Example A recipe uses 0.225 kg of sugar. How many cups are needed if there are 2 cups of sugar per 0.5 kg of sugar? We have the conversion/conversion factor

145.02 kgcupskgcups

So

.900.0

4225.0225.0 1

cups

gkcupsgkkg

Example A typical line of English text has 12 words, with an average of four letters per word. How many letters would there be on a page that has 35 lines? From the question we have the rates 4 letters×word-1 12 words×line-1 35 lines×page-1 Multiplying these and cancelling 4 letters×word-1×12 words×line-1×35 lines×page-1 = 4×12×35 letters×page-1 = 1680 letters×page-1. Example Calculate the cost of gasoline for a 420 mile trip if your car averages 20 miles per gallon of gas and the gas costs $0.95 per gallon. We have the rates 20 miles×gal-1 and $0.95 × gal-1 but in order to get the answer we need to take the inverse of the first rate

.95.19$42095.0$

95.0$420

201

11

201

lagelimlagelim

Exercise In each of the following identify which is a conversion factor and which is a rate (some are neither). Write any conversion as a conversion factor. Simplify any fractions. a) 12 kilometres

b) 12 inches = 1 foot

c) 3 books per week

d) BHD 300

e) 500 bytes per second

f) 1000 kilobytes = 1Megabyte

g) 300 fils per 100 grams

h) 12 eggs/1 dozen

i) 60 kilometres/hour

j) 1000 m per kilometre

Answers a) Neither

b) Conversion Factor: 12 inches × foot-1

c) Rate: 3 books × week-1

d) Neither

e) Rate: 500 bytes × s-1

f) Conversion factor: 1000 kilobytes per Megabyte

g) Rate: 3 fils × gm-1

h) Rate: 12 eggs × dozen-1

i) Rate: 60 kilometres × hr-1

j) Conversion factor: 1000 m × km-1

36

Exercise 1. A computer can process a data record in 0.1

seconds. How many seconds will it take to process a file of 5000 data records?

2. A book has 300 pages. Each page has 40 lines, with approximately 14 words per line and an average of four letters per word. Each letter is stored in computer memory using one byte. How many bytes would be required to store the book on your memory stick?

3. Jamil works at a fast food restaurant wrapping hamburgers. Every 3 hours he wraps 350 hamburgers. He works 8 hours per day. He works 5 days a week. He gets paid every 2 weeks with a salary of BHD 400. Approximately how many hamburgers will he have to wrap to make his first one million dinars?

Answers 1. 500 seconds. 2. 672’000 bytes. 3. 23’333’333 hamburgers.

37

Prefixes

Instead of writing numbers out in full we sometimes use prefixes to simplify. For example: 1 kilobyte = 1’000 bytes 1 megabyte = 1’000 kilobytes = 1’000’000 bytes 1kilometre = 1’000 metres 1 kilogram = 1’000 grams The prefixes ‘mega’, ‘kilo’, ‘milli’ are usually abbreviated:

Prefix Symbol Factor Factor

giga G 1’000’000’000 109

mega M 1’000’000 106

kilo k 1’000 103

milli m 0.001 10-3

From the table: 1 Gbyte = 1’000’000’000 byte 1 kbyte = 1’000 byte so we have the conversion factor

1’000 byte per kbyte or 1’000 byte×kbyte-1 To convert from Gbytes to kbytes we should divide by this conversion factor 1 Gbyte = 1’000’000’000 byte×0.001 kbyte×byte-1 = 1’000’000 kbytes .

Exercise 1. Using the rates from question 2 above: how

many books could you store on a typical, single-sided, single-layer DVD? (a typical DVD has a capacity of 4.7 Gbytes)

2. A computer can process a data record in 100 milliseconds. How many seconds will it take to process a file of 5000 data records?

3. A file on a computer hard drive consists of 5000 data records each 128 bytes long. The computer can process 1 kilobyte of data in 100 milliseconds. Find the number of seconds to process the entire file.

4. An older modem transfers data over a telephone line at a rate of 14,400 bits per second. How many seconds will it take to transfer a megabyte of data? Use the fact that a byte is composed of eight bits.

5. A graphic image is being sent over the Internet. The image, consisting of 5 megabytes of information, is transferred along a data line at the rate of 1.5 megabits per second. How many seconds will the transfer take?

6. What is the total storage capacity in kilobytes of an old single-sided, single-density floppy disk used with original IBM PC? It had the following characteristics: Heads: 1 Tracks per head: 40 Sectors per track: 8 bytes per sector: 512

38

Exercise (continued) 7. What is the total capacity in gigabytes,

rounded to the nearest tenth of a gigabyte, for a computer hard drive with the following specifications: Total heads: 16 Tracks per head: 6400 Sectors per track: 128 Bytes per sector: 512

8. The CMOS setup of a computer auto detects an installed hard drive on bootup and outputs the following information: Heads: 16 Cylinders: 2484 Sectors: 63 The cylinder count represents the number of tracks per head, and the sector count represents the number of sectors per track. Use the definitions given in problem 7 for bytes per sector to calculate the total capacity of the hard drive in gigabytes, rounded to the nearest tenth of a gigabyte.

9. Pictures are stored on computers as a series of dots (also called pixels). Each black and white pixel can be represented as a 16-bit number. How much memory will a black and white picture of size 1024x800 pixels take up. Give your answer in megabytes (1 d.p.).

10. How many pictures, of size 2048x1024, in black and white format, could you store on a 2Gb camera memory card?

Exercise (continued)

11. A video monitor has square pixels, the side of each pixel is 0.28 mm. If the monitor displays 1024 pixels by 768 pixels, and this represents the number of pixels along the horizontal and vertical sides of the screen, what are the dimensions of the display area of the screen in inches, to the nearest tenth of an inch? Use the fact that 1 millimeter is approximately

inch. 12. A computer CPU can process 1 million

machine language instructions per second. If a program contains 1000 lines of code and each line represents an average of 10 instructions, calculate the approximate amount of time in milliseconds for the CPU to process the entire program.

13. Once the desired track of data is located on a disk drive, data can be transferred to memory at a rate of 1.5 megabytes per second. How many milliseconds are needed to transfer 200 kilobytes of data?

Answers

1. 6994 books. 8. 1.3 Gbytes.

2. 500 seconds. 9. 4.2 Mbytes

3. 64 seconds. 10. 476 pictures.

4. 556 seconds. 11. 11.5 in × 8.6 in.

5. 27 seconds. 12. 0.01 seconds

6. 163.84 kbytes. 13. 133 ms.

7. 6.7 Gbytes.

39

Extra Exercises

52

Answers for Extra Exercises

59

References This resource was produced using materials from: [1] Coventry University Mathematics Support Centre,

https://cuportal.coventry.ac.uk/C13/MSC/Document%20Library/worksheets.aspx [2] The Centre for Innovation in Mathematics Teaching,

http://www.cimt.plymouth.ac.uk/ [3] Justin’s Chemistry Corner,

http://www.chemistrycorner.org/Site/Chem_I_files/Dimensional%20Analysis%20.pdf [4] Questions supplied by Bahrain Poly IT department Compiled by Ihsan Eltom and Mark Harmer with some ‘Rates/Conversions’ materials created by Mark Harmer.