HANDOUT_Outcome 1 No 2 - Algebra 1

8/8/2019 HANDOUT_Outcome 1 No 2 - Algebra 1

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Mathematics for TechniciansAlgebra 1

An equation is a statement that two quantities are equal, for instance,1000 mm = 1 m. More often an equation contains an unknown quantitywhich is represented by a symbol. It is the value of this unknown quantitywhich we desire to find. Consider the following:

In the equation 3x 4 = 23,xis the unknown quantity.

There is only one value ofxsuch that the left hand side of theequation is equal to the right hand side (this value isx= 9).

When we have calculated this value ofxwe have solved the equationand the value ofxso obtained is called the solution (ie the solution isx= 9).

In the process of solving an equation the appearance of the equation maybe considerably altered but the values on both sides must remain thesame. We must maintain this equality and hence whatever we do to oneside of the equation we must do exactly the same to the other side.

There are many types of equations which occur in mathematics and theseare classified according to the highest powerof the unknown quantity.

3x 4 = 23 contains only the first power ofx.

5x2 3x+ 5 = 0 containsx2 as the highest power ofx, that is thesecond power ofx.


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Mathematics for Technicians -Algebra 1

2

Simple Equations

Simple equations contain only the first power of the unknown quantity.

7457 += tt

or

2

52

3

5 +=

xx

After an equation is solved, the solution should be checked by substitutingthe result in each side of the equation separately. If each side of theequation then has the same value the solution is correct.

In the detail which follows, LHS means left hand side and RHS means righthand side.


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3

Solving Simple Equations

Equations Requiring Multiplication and Division

Example 1

a) Solve the equation

Multiplying each side by 6, we get:

Check: whenx= 18, LHS = 36

18= RHS = 3

b) Solve the equation 5x= 10

Dividing each side by 5, we get:

Check: whenx= 2, LHS = 5 2 = 10 RHS = 10

36=

x

6366

=x

510

55

=x

18=x

2=x


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4

Exercise 1

Solve the following equations:

a) 24 =x b) 2

9 =x c) 4

1.5 =x

d) 53=

re) 3

2=

yf) 4

3=

m

g) 3x= 9 h) 2x= 10 i) 7x= 28

j) 4t= 12 k) 7V= 42 l) 1.5J= 6

m) 0.3d= 1.8


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5

Equations Requiring Addition and Subtraction

Example 2

a) Solvex 4 = 8

If we add 4 to each side, we get:

The operation of adding 4 to each side is the same as transferring -4to the RHS but in so doing the sign is changed from a minus to a plus.Thus,

Check: whenx= 12, LHS = 12 4 = 8, RHS = 8

b) Solvex+ 5 = 20

If we subtract 5 from each side, we get:

Alternatively moving + 5 to the RHS

Check: whenx= 15, LHS = 15 + 5 = 20, RHS = 20

4844 +=+x

12=x

84=x

48+=x

12=x

52055 =+x

15=x

520 =x

15=x


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6

Exercise 2

Solve the following:

a) x+ 2 = 7 b) t 4 = 3 c) x 8 = 12

d) q + 5 = 2 e) 7 +x= 21 f) 4 +x= -2

g) t+ 42 = 80 h) z 4 = -8 i) y+ 4 = -10

j) p 7 = -4 k) -4 +x= 10 l) -x+ 7 = 3

m) 7 t= 24 n) x 7 = -3 o) -x 8 = -4


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8

Exercise 3


a) 4x+ 13 = 3x+ 15 b) 6x 4 = 3x+ 8

c) 5x 10 = 3x+ 2 d) 3x 22 = 8x+18

e) .6m + 11 = 25 m f) 2d 1 3d= 4d 21

g) 2a 4 = 3a 2 5a h) 3p 18 = 8p + 22

i) 3u 2 = 7u 6 j) 0 = 2 + 4b

k) 3x+ 1 4x= 0 l) 2r+ 5 3r 2 = 0

m) 1.2x 0.8 = 0.8x+ 1.2 n) 2.6x 0.4 = 0.9x+ 1.3


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9

Equations Containing Brackets

When an equation contains brackets remove these first and then solve asshown previously.

Example 4

a) Solve 2(3x+ 7) = 16

Removing the bracket,

Check whenx=

3

1,

LHS = 2

+7

31

3x = 2 (1 + 7) = 2 8 = 16

RHS = 16

16146 =+x

14166 =x

26 =x

62

=x

31

=x


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10

b) Solve 3(x+ 4) 5 (x 1) = 19

Removing the brackets:

Check: whenx= -1,

LHS = 3 (- 1 + 4) 5 (- 1 1) = 3 3 5 (- 2) = 9 + 10 = 19

RHS = 19

1955123 =++ xx19172 =+ x

17192 = x

22 = x

22

=x

1=x


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11

Exercise 4


a) 2(x+ 1) = 8 b) 5(m 2) = 15

c) 4(x+ 2) = 14 d) 5(x+ 2) = 14

e) 3(x 1) 4 (2x+ 3) = 14 f) 5(x+ 2) 3 (x 5) = 29

g) 3x= 5 (9 x) h) 4(x 5) = 7 5 (3 2x)

i) 3(x 1) = 6 j) 4(3x+ 2) = 14

k) 3(e 1) 4(2e + 3) = 15 l) 5(g+ 2) = 29 + 3(g 5)

m) 5(h 3) = 2h n) 4(3 2x) 5 = -3 (x 2)

o) 3(2y 1) 2 = 17 p) 4(3x 4) (x+ 1) = 10

q) 0 = 2 (3a 5) 5 r) 0 = 2(b 2) 3 (3b + 1)

s) 5(2x+ 3) = 6(x+ 2) 2 (x 4)

t) 20 + 8(3c 2) = (c 1) + 3(2c 4)

u) 16 (p + 2) = 3(p 7) + 10

v) 8 + 5(d 1) 6(d 3) = 3(4 d)


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Equations Containing Fractions

When an equation contains fractions, multiply each term of the equation bythe Lowest Common Multiple (LCM) of the denominators.

Example 5

a) Solve 22

353

4=+

xx

The LCM of the denominators 2, 4 and 5 is 20. Multiplying each term by20 gives:

Therefore:

202202

3205320

4=+ xx

2552

=x

4030125 =+ xx

1240305 = xx

5225 = x

08.22552

==x


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13

b) Solve the equation 42

123

4=

xx

In solving equations of this type remember that the line separating the

numerator and denominator acts as a bracket. The LCM of thedenominators 3 and 2 is 6. Multiplying each term of the equation by 6,

6462

126

34

=

xx

429

=x

( ) ( ) 2412342 = xx

243682 =+ xx

2454 = x

5244 += x

294 = x

25.74

29==x


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14

c) Solve the equation2

452

5+

=+ xx

The LCM of the denominators is (2x+ 5) (x+ 2). Multiplying each

term of the equation by this gives:

Therefore:

2)(5)(22

42)(5)(2

525

+++

=+++

xxx

xxx

5)4(22)5( +=+ xx

208105 +=+ xx

10-208-5 =xx

103x- =

310

=x

33.33

10==x


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Exercise 5


a) 235 =xx b)

65

543 =++xxx

c)6

2332

mmm+=++ d)

32

243

3x

x +=+

e) 33

=m

f) 25=

x

g)324 =

th)

357 =

x

i) 25

3

7

4= yy j)

20

7

4

1

3

1=+

xx

k) 25

34

3=

+ xxl)

23

203

126

152

=

xxx

m)354

432 mm =

n)

343 yy

=

o)6

535

=

xx p) 3

32=

x

x

q)4

42

3+

= xx

r)425

213

314 ppp

=

s)5

21

3

= xx

t) 0329

453

=

mm

u))2(3

572

3

=+ xx

v) 02

523

=

xx

w)6

2

5

73

3

=

xxxx) x

xx=

6

12

2

54


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Exercise 5a


a) 5231 =+a [9]

b)21

31

243

+=+ xx [-353

]

c)21

4)12(31

=+a [-443

]

d)43)12(

313 =+ y [

872 ]

e) 0)4(31

)32(21

=+ ff [81

2 ]

f) 0 = )5(43

)12(32

+ mm [74

7 ]

g) 0)72(51

)15(41

)23(31

=++ aaa [1594

]

h) )5(31

151

)14(51

)43(91

+=+ xxx [-2 ]61

31

i) 253=

tt[15]

j)430

4735

VVV=+ [2]

k)3

16

33

mmm=+ [-4]

l) 105=

x

[

2

1]


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17

m)103

54

21

2 =+ tt [-32

]

n) 5

93=

w [1 3

2

]

o)15

1

5

2

9

5=

xx[73

]

p)203

41

31

=+aa

[98

3 ]

q) 5

324

3 +=

+ cc[13]

r)152

12)6(

203

23 yyy

+

=+ [-10]

s)207

545

+= dd

[2]

t)5

624 += aa [2]

u) 432=

f

f[3

31

]

v)4

42

3+

= yy

[20]

w)1

35

2

= aa

[13]

x))3(5

723

4

=+ bb

[-74]

y)21

121

43

=

a

[221

]


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z) 035

123

1=

++

xx[-

81

]

aa) 23

13

32

2

+

= b

b

b

b

[ 5

1

]

bb) 0)3(5

2)2(3

1=

+

yy[27]

cc)34

123233

=+

+

x

x[7]

dd) (c 3)(2c+ 6) = 2c(c 18) [21 ]

ee) 91

1=

+

a

a[-

54

]

ff)43

25=

x

x[14]

gg)5

62

34

+=

+

aaa[-6]

hh) )3(22

)4(36

=+

xxx

[0]

ii)5

14

322

312 fff

=+

[-2121

]


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19

Equations Involving Squares and Square Roots

Example 6

a) Solve the following forx

205 =x

In order to take out the square root, we have to square both sides ofthe equation.

b) Solve the following forx

This time, we have to take care of thex2 term. In order to do this, wesquare root both sides.

(7x2) = 63

7 x2 = 63

x= 637

x= 3

40025 = x

22

2

205 =x

1625

400==x

6372 = x


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Exercise 6

Solve the following equations that involve squares and square roots.

a. 124 =a

b.3

52 =x

c. 12

6

=

x

d.

+= 234x

e. 41

2=

b

b

f.

= 1

248

a

g.2

624

a=

h.

2

2

2

3

4

+=

x

x

i. 916

2e

=

j.2528 x

x=

k.2

24

2

9+=


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Making Expressions

It is important to be able to translate information into symbols thus makingup algebraic expressions. The following examples will illustrate how this is

done.

Example 7

a) Find an expression which will give the total mass of a box containingxarticles if the box has a mass of 7 kg and each article has a mass of1.5 kg.

The total mass ofxarticle is 1.5x.

Therefore total mass of the box of articles is 1.5x+ 7.

b) If the price of an article is reduced fromxpence to ypence make anexpression giving the number of extra articles that can be bought for80 pence.

Atxpence each, the number of articles that can be bought for 80

pence isx

80

At ypence each, the number of articles that can be bought for 80

pence isy

80

The extra articles that can be bought isxy

8080

c) Ifxnails can be bought for 6 pence, write down the cost ofynails.

Ifxnails cost 6 pence

Then 1 nail costsx

6pence

Hence, ynails costx

yy

x

66= pence


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Exercise 7

a) A boy isxyears old now. How old was he 5 years ago?

b) Find the total cost of 3 pencils at a pence each and 8 pens at b penceeach.

c) A man worksxhours per weekday except Saturday when he works yhours. If he works zhours on Sunday how many hours does he workper week?

d) What is the perimeter of a rectangle lmm long and b mm wide?

e) A man A has a and a man B has b. If A gives B xhow munch willeach have?

f) A factory employs Mmen, Nboys and Pwomen. If a man earns xper week, a boy yper week and a woman zper week what is thetotal wage bill per week?

g) A man earns uper week when he is working and he is paid vperweek when he is on holiday. If he is on holiday for 3 weeks per yearfind his total annual salary.

h) The price ofm articles was Mbut the price of each article isincreased by n pence. How many articles can be bought for N?

i) A man starts a job at a salary of uper week. His salary is increasedby ypence per week at the end of each years service. What will behis salary afterxyears?

j) A numberm is divided into two parts. Ifa is one part what is the

product of the two parts?


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Construction of Simple Equations

It often happens that we are confronted with mathematical problems thatare difficult or impossible to solve by arithmetical methods. We then

represent the quantity that has to be found by a symbol. Then byconstructing an equation which conforms to the data of the problem wecan solve it to give us the value of the unknown quantity. It is stressed thatboth sides of the equation must be in the same units.

Example 8

a) The perimeter of a rectangle is 56 cm. If one of the two adjacent sidesis 4 cm longer than the other, find the dimensions of the rectangle.

Let xcm = length of the shorter side

Then (x+ 4) cm = length of the longer side

Total perimeter =x+x+ (x+ 4) + (x+ 4) = (4x+ 8) cm

But the total perimeter = 56 cm

Hence 4x+ 8 = 56

4x= 56 8

4x= 48

x= 12

Hence the shorter side is 12 cm long and the longer side is 12 + 4 =16 cm long.

b) A certain type of lathe costs five times as much as a certain make ofdrilling machine. If two such lathes and five such drilling machinescost 7500, find the cost of each machine.

Let the cost of a drilling machine be x

Then the cost of a lathe = 5x

Cost of 2 lathes and 5 drilling machines =


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(2 5x+ 5 x) = (10x+ 5x) = 15x

Since the cost of 2 lathes and 5 drilling machines is 7500

15x= 7500

x= 500

Hence the cost of a drilling machine is 500 and the cost of a lathe is5 500 = 2500.

Exercise 8

a) A foreman and 5 men together earn 366 per week. If the foremanearns 6 per week more than each of the men, how much does eachearn?

b) One side of a triangle is 4 cm shorter, and another 3 cm shorter thanthe longest side. If the perimeter of the triangle is 25 cm, find thelengths of the three sides.

c) The three angles of a triangle arexo, (x+ 30)o and (x 6)o. The sumof the three angles is 180o; find each angle.

d) The perimeter of a rectangle is 56 cm. If one of the two adjacent sidesis 4 cm longer than the other, find the dimensions of the rectangle.

e) A certain type of lathe costs seven times as much as a certain makeof drilling machine. If two such lathes and three such drillingmachines cost 8500, find the cost of each machine.

f) The perimeter of a triangle ABC is 26 cm. BC is two-thirds of AB and

it is also 2 cm longer than AC. Find the lengths of the three sides.

g) A transformer compound of rectangular plan is to be constructed using100 m of chain link fencing for one long and two short sides and anexisting wall for the other side. If the length of the compound has tobe 10 m longer than twice the breadth, find the dimensions of thecompound.

h) Three plugs and five electric light fittings together cost 3.30. If a plug

costs twice as much as a light fitting find the cost of a plug and a lightfitting.


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i) Two taps are used to fill a cooling tank, which has a capacity of 1200

litres. If it takes 16 minutes to fill the tank and one tap delivers waterat twice the rate of the other, find how many litres per minute each tap

delivers.

j) A house is fitted with 3 electric radiators and 5 convector heaters at atotal cost of 370. If a convector heater costs 10 more than aradiator, find the cost of each.


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Summary

a) To solve an equation the same operation must be performed on bothsides. Thus the same amount can be added or subtracted from each

side or both sides can be multiplied or divided by the same amount.

b) After an equation has been solved the solution may be checked bysubstituting the result into the original equation. If each side of theequation has the same value the solution is correct.

c) To construct a simple equation, the quantity to be found isrepresented by a symbol. Then using the given data the equation isformed. Note that both sides of the equation must be in the sameunits.

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HANDOUT_Outcome 1 No 2 - Algebra 1