FEngO

Foundation Degree

Mathematics

Pre-Course Study Package

Name: …………………………… Service No: …………………

2

MATHEMATICS PRE-COURSE STUDY PACKAGE

This package has been designed so that all students on the FEngO Foundation Degree course have the opportunity to practice basic algebraic manipulation before arriving at DCAE Cosford. This will enable all students to achieve a common entry level for the start of the mathematics module. The nine exercises are to be completed and handed in on the first day of the course. On the first mathematics lesson a diagnostic test on these topics will be set. The outcome of this test will be used to assess areas of potential strengths and weaknesses of each individual student. These can then be addressed during the 20 period Maths lead-in course.

TRANSPOSITION OF FORMULA

The process of rearranging a formula so that one of the other symbols becomes the subject is called transposing the formula. Example 1 Transpose the formula to make a the subject maF =

Divide both sides by m mma

mF =

amF = or

mFa =

Example 2 Transpose the formula to make x the subject cmxy += Subtract c from both sides ccmxcy −==− mxcy =−

Divide both sides by m mmx

mcy =−

xm

cy =− or m

cyx −=

3

Example 3

Transpose the formula tT

)hH(wQ−−= to make T the subject

Multiply both sides by tT − )tT(tT

)hH(w)tT(Q −×−−=−

)hH(w)tT(Q −=−

Divide both sides by Q Q

)hH(wQ

)tT(Q −=−

Q

)hH(wtT −=−

Add t to both sides tQ

)hH(wttT +−=+−

tQ

)hH(wT +−=

Subtract Q

)hH(w − from both sides

Q

)hH(wtQ

)hH(wQ

)hH(wT −−+−=−−

tQ

)hH(wT =−−

Q

)hH(wTt −−=

4

Example 4

Transpose the formula ba

aby−

= to make a the subject

Multiply both sides by ba − )ba(ba

ab)ba(y −−

=−

ab)ba(y =−

Multiple the brackets on the LHS abbyay =− Group all the terms containing an a on the LHS and all other terms on byabay =− the RHS. Factorise the LHS by)by(a =−

Divide both sides by )by( − by

by)by()by(a

−=

−−

by

bya−

=

Example 5 Transpose the formula to make n the subject

t)n( =− 21

Square root both sides t)n( =−1 Add 1 to both sides 111 +=+− t)n( 1+= tn

5

Example 6

Transpose the formula c

)bx(bd −= to make x the

subject

Square both sides c

)bx(bd −=2

Multiply both sides by c cc

)bx(bcd ×−=2

)bx(bcd −=2

Divide both sides by b b

)bx(bb

cd −=2

bxb

cd −=2

Add b to both sides bbxbb

cd +−=+2

xbb

cd =+2

Or bb

cdx +=2

6

EXERCISE 1

Transpose the following formulae 1) dC π= for d 2) for d dnS π= 3) for V 4) for r cPV = rlA π= 5) for h 6) I=PRT for R gh2v2 =

7) yax = for y 8)

REI = for R

9) aux = for u 10)

VRTP = for T

11) c3

b2

a1 += for b 12)

321 R1

R1

R1

R1 ++= for R2

13) 33000PLANH = for L 14)

4hdV

2π= for h

15) for P 16) 7.14Pp −= atuv += for t 17) crpn += for r 18) baxy += for x

19) 175xy += for x 20) for q qLSH +=

21) cxba −= for x 22) d281BD ⋅−= for d

23) rR

R2V−

= for r 24) rR

EC+

= for E

25) )hr(rS +π= for h 26) )tT(wSH −= for T

7

8

27) p2

nNC −= for N 28) L

)dD(12T −= for d

29) rR

R2V−

= for R 30) C

)FC(SP −= for C

31) gh2V = for h 32) dkw = for d

33) gL2t π= for L 34)

gfW2t π= for f

35) r

mvmgP2

=− for m 36) yx

xZ+

= for x

37) 1n2n3k

++= for n 38)

5t43a+

= for t

39) ⎟⎠⎞

⎜⎝⎛ −=

a1

x1k2v2 for x 40)

)1n(n)anS(2d

−−= for a

41) 2hhr22c −= for r 42) dD

dhx−

= for d

43) pfpf

dD

−+= for f 44)

gt)uv(WF −= for t

45) 2

2

t1t1y

+−= for t 46)

gHR2T −π= for R

47) 1n2n3k

++= for n 48)

baaby−

= for b

49) cb

ba+

= for b 50) ay

bax22 −= for y

LINEAR EQUATIONS

A linear equation is an equation that contains only one unknown. For example; 1) 17x53x7 +=+ 2) 16)7x3(2 =+ 3) 191x54x3 =−−+ )()(

4) 22x3

53

4x −=+

5) 42

1x23

4x =−−−

6) 2x

45x2

5+

=+

The method used to solve linear equations is exactly the same as for transposition of formula. Example 1 Solve 17x53x7 +=+ Rearrange by subtracting 5x and 3 from both sides so that the LHS contains all the x terms and the RHS all the non x terms. 3x517x53x53x7 −−+=−−+ 2x =14

2

14x = x = 7

Example 2 Solve 16)7x3(2 =+ Removing the bracket 1614x6 =+Subtract 14 from both sides 14161414x6 −=−+ 2x6 =

Dive both sides by 6 62

6x6 =

Therefore 31x =

9

10

Example 3 Solve 191x54x3 =−−+ )()( Removing the brackets 195x512x3 =+−+ 1917x2 =+− Re-arranging 19-17= - 2x Therefore 2= -2x x = -1

Example 4 Solve 22x3

53

4x −=+

Multiply by the LCD 20 202202x320

5320

4x ×−×=×+×

40x3012x5 −=+ x5x304012 −=+ x2552 =

2552x =

2522x =

Example 5 Solve 42

1x23

4x =−−−

Multiply by the LCD 6 6462

1x263

4x ×=×⎟⎠⎞

⎜⎝⎛ −−×⎟

⎠⎞

⎜⎝⎛ −

241x234x2 =−−− )()( 243x68x2 =+−− x2x62438 −=−+−

x429 =− 417x −=

Example 6 Solve 2x

45x2

5+

=+

Multiply both sides by ))(( 2x5x2 ++

24)2)(52(

)52(5)2)(52(

+×++=

+×++

xxx

xxx

)()( 5x242x5 +=+ 20x810x5 +=+ x5x82010 −=− x310 =−

310x −=

313x −=

Exercise 2 Solve the following linear equations. 1) 2) 177x6 =− 8x314 =− 3) 4) x2511x6 −=+ 81x30 ⋅=⋅ 5) 6) 21x8080x21 ⋅+⋅=⋅−⋅ 81x2 =+ )( 7) 8) 143x241x3 =+−− )()( 295x32x5 =−−+ )()( 9) 10) )( x95x3 −= )()( x23575x4 −−=−

11

11) 23x

5x =− 12)

65

5x

4x

3x =++

13) 6x23

3x

2x +=++ 14)

3x22

43x3 +=+

15) 3x3 = 16) 2y

53y

74 =−

17) 207

x41

x31 =+ 18) 2

53x

43x =−−+

19) 23

20x3

126x

15x2 =−−− 20)

3x54

43x2 −=−

21) 3y

4y3 =− 22)

65x35x −=−

23) 33x2x =

−− 24)

5x2

1x3

−=

−11

25) 4x

42x

3+

=−

26) )( 2x3

57x2

3−

=+

27) 6

2x5

7x33x −=−− 28)

4x25

21x3

31x4 −=−−−

29) 03

x294

5x3 =−−− 30) 02

5x23x =−−

31) x6

1x22

5x4 =−−− 32) 321

81x3

123x =−+− )()(

33) 324

35x2

9x42 =+−− )()( 34)

328

63x55

73x23 =+−+ )()(

12

35) 1541

3x43

51x3 =−++ )()( 36)

416

45x7

32x5 =−++

37) 321

37x4

42x3 =−+− )()( 38)

3211

51x42

31x25 =+−+ )()(

39) 6120

3x54

4x37 =+−− )()( 40) 22

5x232

24x35 =−−− )()(

13

SIMUTANEOUS EQUATIONS

Consider the following two pairs of equations 1) and 11y4x3 =+ 15y7x =+ 2) and 26y4x3 =+ 18y3x4 =− Example 1 Solve the equations: 3x + 4y = 11 ……. (1)

and x + 7y = 15 ……. (2) The first objective is to make the coefficients of either x or y to be the same. To achieve this we will multiply equation (2) by 3 3x + 21y = 45 ……. (3) We can now eliminate x by subtracting equation (1) from equation (3) 3x +21y = 45 3x + 4y = 11 17y = 34 y = 2 Substitute y = 2 into equation (1) 3x + 4y = 11 3x + 8 = 11 3x = 11 – 8 3x = 3 x = 1 Hence the solutions are : x = 1 and y = 2

14

Example 2 Solve the equations

3x + 4y =26 ……. (1) and ……. (2) 18y3x4 =−

To obtain the same coefficient of x, equation (1) is multiplied by 4 and equation (2) by 3 12x + 16y = 104 ……. (3) 54 ……. (4) −x12 =y9 25y = 50 y = 2 Substitute y = 2 into equation (1) 3x + 8 = 26 826x3 −= 3x = 18 x = 6 Hence the solution is: x = 6 and y = 2 Exercise 3 Solve the following simultaneous equations 1. 2. 29y7x5 =+ 17yx5 =− 7y2x =+ 6y3x4 =+ 3. 4. 22yx5 =+ 30y4x3 =− 25y5x2 =− 9y5x4 =+ 5. 6. 11y5x6 =− 24y3x10 −=+ 1y3x4 =+ 23y4x5 −=− 7. 8. 10y2x =+ 8y5x4 −−= 3y5x3 −= 26y2x6 += 9. 10. 2y3x2 =+ 1y3x12 =− 1y9x8 =− 5y6x4 =+

15

MULTIPYING BRACKETS

If we wish to multiply (a+b) by (c+d) then it can be written as (a+b)(c+d) If we now multiple each individual term of the first bracket by the second bracket then a(c+d) + b(c+d) ac+ad +bc +bd Similarly (2x+3y)(4x+5y) = 2x(4x+5y) + 3y(4x+5y) 8x2 + 10xy +12xy + 15y2 8x2 + 22xy + 15y2 And (x + y)2 = (x+y)(x+y) x(x+y) + y(x+y) x2 +2xy + y2

16

)

) )

) )

) )

)

) )

Expanding will become the first term squared, plus twice the product of the first and second terms, plus the second term squared.

( 2y4x3 −

So = 2y4x3 )( − 22 y16xy24x9 +− Exercise 4 Expand and simplify the following brackets 1. 2. ( ) ( 1x254x3 +++ ( ) ( y4x35y3x23 +−− 3. 4. ( ) ( x2573y7x210 −−+ ( ) ( 2222 b6a33b4a5 −−+ 5. ( ) 6. ( ) ( 3x2x ++ ( 5x33x2 −+ 7. ( 8. ( ) 2y3x2 + 2y3x2 − 9. ( ) 10. ( ) ( y3x2y3x2 −+ ( ) ( )( 4x33x53x25x3 +−+−+

Factorising Quadratic Functions

From the previous exercise ( )( ) ( ) mnxnmxnxmx 2 +++=++ And the general quadratic equation is cbxax 2 ++ Then if 1a = ( ) cbxxmnnmx 22 ++=+++ So nmb += and mnc = Example 1 Factorise 6x5x 2 ++ As and 632 =× 532 =+ = 6x5x2 ++ ( )( 3x2x ++ )

)

)

)

Example 2 Factorise 150x2x 2 −+ As and 1553 −=+×− ( ) 253 +=++− ( )( 5x3x15x2x 2 +−=−+ Example 3 Factorise 6x7x2 2 ++ As and and 623 =× x42x2 =× x33x =× Giving 4x+3x=7x ( )( 2x3x26x7x2 2 ++=++ Example 4 Factorise 15x4x3 2 −− As and and 1535 −=−× x55x =× x93x3 −=−× Giving x4x9x5 −=−+ )( ( )( 3x5x315x4x3 2 −+=−−

17

Exercise 5 Factorise the following quadratic functions 1) 2) 12x7x 2 ++ 15x2x 2 −+ 3) 4) 28x3x 2 −− 28x11x 2 +− 5) 6) 56x15x 2 +− 2xx1130 +− 7) 8) 2mm412 −+ 2aa20 −− 9) 10) 2x5x2 2 ++ 14x13x3 2 ++ 11) ` 12) 10x13x3 2 −− 4x11x6 2 ++

18

Solution of Quadratic Equations

Method one – Factorisation Example 1 Solve 014x5x 2 =−+

If then 014x5x 2 =−+ ( )( ) 07x2x =+−

Therefore and 02x =− 07x =+

So and 2x = 7x −= Example 2 010x23x5 2 =−− ( )( ) 05x2x5 =−+ and 02x5 =+ 05x =−

50or52x ⋅−−= and 5x =

Exercise 6 Solve the following quadratic equations by factorisation 1) 2) 0x4x 2 =− 04x 2 =− 3) 4) 016x25 2 =− 0x5x 2 =+ 5) 6) 010x7x 2 =+− 012xx 2 =−− 7) 8) 028x11x 2 =+− 030x11x 2 =+− 9) 10) 035x12x 2 =++ x263x 2 =− 11) 12) 05x11x2 2 =+− 010x23x5 2 =−−

19

Solution of Quadratic Equations by Formula Given that the general form of a quadratic equation is

0cbxax2 =++

Then a2

ac4bbx2 −±−=

Example 1 Solve using the quadratic formula

08x2x3 2 =−+

Comparing with 08x2x3 2 =−+ 0cbxax 2 =++ Therefore 3a = , 2b = and 8c −=

So 32

83422x2

×−××−±−=

6

9642x +±−=

6

102x ±−= 34x = and 2x =

Example 2 Solve 07x9x2 2 =−−

2a = , 9b −= and 7c −=

So ( ) ( ) ( )22

72499x2

×−××−−±−−

=

4

56819x +±= 41379x ±=

and 185x ⋅= 680x ⋅−=

20

Exercise 7 Solve the following quadratic equations by formula 1) 2) 02x3x4 2 =−− 01xx 2 =−− 3) 4) 05x7x3 2 =−+ 02x8x7 2 =−+ 5) 6) 01x4x5 2 =−− 3x7x2 2 =−

21

)7) 8) ( ) ( 53xx24xx =+++ ( ) ( ) 201x2x21xx5 =−−+

9) 51x

32x

2 =+

++

10) 42x

53

2x =+

−+

11) x

2x4

5x3 2 −=− 12) ( ) 665xx =+

12) 14) ( ) 133x2 2 =− 2x1

2x12 =−+

22

Indices

Laws of indices

nmnm xxx +=×

nmnm xxx +=÷ ( ) mnnm xx =

1x0 =

xx1 =

mm

x1x =−

m1

m xx =

nm

m n xx = Examples Simplify the following

1. 2

3 2

xxx × = 22

132

xx −×× 652

21

32 −=−+

65

x−

=

2. 43

3

21

52

x

xx

−

⎟⎠

⎞⎜⎝

⎛× = 4

381

52

xxx ×× 4051

43

81

52 =++

4051

x=

23

Exercise 8 Without a calculator find the numerical value of the following

1) 23

21

2 555−

×× 2) 21

44 ÷

3) 31

8 4) 61

64

5) 32

8 6) 23

25

7) 3

41

16 ⎟⎠

⎞⎜⎝

⎛ 8) 23

9

1−

9) 21

41 −

⎟⎠⎞

⎜⎝⎛ 10) 5016 ⋅

11) 5036 ⋅− 12) ( )21

34−

13) 25

41⎟⎠⎞

⎜⎝⎛ 14)

3

21

16

1−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

15) ( ) 231

− 16) 21

49 −

⎟⎠⎞

⎜⎝⎛

24

Exercise 9 Simplify the following expressions, expressing each as

a single power. 1) x 2) 5 4x

3) x

1 4) 3 4x

1

5) 3x − 6) 4 30x

1⋅−

7) ( )23 x− 8) 3

2

x

9) ( )32

x 10) 43

3 4x1

−

⎟⎠

⎞⎜⎝

⎛

11) xx

x2

3

× 12)

32

3

x

x −

13) 23

3

x

x−

14) 21

23

25

x

xx−

×

15) 25

43

x

x−

−

16) 22

7

23

25

xx

xx

×

×−

−

17) ( )

3

21

3

21

x

x ⎟⎠

⎞⎜⎝

⎛

18) 3xx

19) xx4 3

20) ( )xx

34

21) 7 2

4 2

xx

− 22)

25

33

x

xx ×