Basic Mathematics
For Basic Mathematics consult
Foundation Maths
—Anthony .Croft and Robert Davison Prentice Hall 3th Edition 2003
510.P3;1 ISBN 0 130 45423-5
Read:
Chapter 1 Arithmetic of whole numbers
Chapter 2 Fractions
Chapter 5 Algebra
Chapter 6 Indices
Chapter Simplifying algebraic expressions
Chapter 10 Transposing formulae
Chapter 11 Solving equations.
A more substantial book on Basic Mathematics is
Foundation Mathematics -Dexter Booth Prentice Hall 1998 3 th
Edition
ISBN 978-0-201-34294-9.
See http://www.mathcentre.ac.uk----good refresher course free of charge
Also consult your leaving certificate course textbooks which I am sure occupy the premier position in your bookcase.
Discrete Mathematics for computing -Peter Grossman- Macmillan-
Essential Mathematics for Economics and Business--Teresa Bradley and Paul
Patton-Wiley
Statistics for Managers--Levine, Stephan, Krehbiel and Berenson--Prentice-Hall
International,Inc

History of Mathematics
In my opinion a great way to know what mathematics is all about is to know something about the history of mathematics. For example you cannot fail to know something about calculus if you know a bit about its history, of why it was developed by those great mathematicians Archimedes(287 BC--212 BC), Gottfried Wilhelm Leibniz(1646-1716) and Sir Isaac Newton(1642-1727). So to this end for any student who is really interested mathematics here are a few books on the subject of the history of mathematics which I am sure will make wonderful bedtime reading .
(1) The story of mathematics(from creating the pyramids to exploring infinity)--Anne
Rooney--Arcturus Publishing Limited 2010 ISBN:978-1-84193-940-7
(2)Men of Mathematics Vol 1 and 2 --E.T.Bell Penguin
(3) What is Mathematics? Richard Courant and Herbert Robbins revised by Ian Stewart
--Oxford University Press 1996
(4) Mathematics in Western Culture--Morris Kline Penguin
(5) A History of Mathematics--Carl B.Boyer,Uta C.Merzbach--John Wiley $ Sons
(6)Mathematical Thought from Ancient to Modern Times--Morris Kline Oxford
Universty Press 1972

Lecture Basic Mathematics
Arithmetic: is the branch of mathematics concerned with numerical
calculations such as addition, subtraction, multiplication, division and
the extraction of roots.
e.g.

3

24 , 5 32

2 .
General Rules for numbers
Addition of numbers with the same sign
When adding numbers with the same sign, the sign of the sum is the same as the sign on each of the numbers.
-7+(-5)=-12
+11+(+23)=+34
When adding numbers with the same sign, you can omit the brackets and the +sign for addition. When the first number is positive, it is usual to omit its +sign.
+11+(+23)=+34 can be written as 11+23 = 34
-7+(-5)=-12 can be written as -7-5 = -12.
Addition of numbers with different signs
To add numbers whose signs are different, subtract the numerically smaller from the larger. The sign of the result is the same as the sign of the numerically larger number.
-23+13=-10
41-56=-15.
When dealing with several numbers of different signs, separately add the positive and negative numbers together. The set of numbers then becomes two numbers, one positive and one negative, which you can add in the usual way.
-21+ 9-35+15+19= -56+43= -13
25-11+33-7+5=63-18=45.

Subtraction
To subtract numbers, change the sign of the number being subtracted and add the resulting number.
-4-(+11)=-4-11= -15
+6- (-23)= +6+23=+29.
Multiplication
The product of two numbers with similar signs is positive, while the product of two numbers with different signs is negative.
(-4)

(-6)= + 24. Negative
 negative=positive.
(+4)

(-6) = - 24 Positive

negative= negative.
If you multiply any number by zero, the answer is always zero.
123

0=0. n

0

0 where n is any number.
Division
When dividing, numbers with similar signs, give a positive answer and numbers with different signs give a negative answer.
-20

5


20
 
4 ,

20
 
5



20
5
 
4
5
If you divide a number by zero, you will always get an error, as this is impossible. n
If n is any number
0 is invalid .

Algebra: is the branch of elementary mathematics that generalizes
arithmetic by using variables to range over numbers ,for example in arithmetical identities such as x y y x .In particular algebra uses
symbols standing for unknown quantities in order to determine their
value by elementary operations of arithmetic and unknowns in equations such as
2 x 5 0 
2 x
 
5
 x
 
5
2 and .
What use is algebra ? Algebra enables us to derive and write formula that generalizes many phenomena and laws of nature. For example, V
 u
 at is a general expression connecting initial and final velocity ( u and V ) of a body, constant acceleration a and time t . The single formula allows you to calculate final velocity, V , of the body for any values of u , a and t ,not just for one specific set of data. Algebra enables us to generalize—something may seem to be true for many sets of specific data values, but is it true for all such sets?
Let us take an example Find the sum of three consecutive integers. Divide by 3.
What do you notice? Try a few examples
Example 1 5+6+7=18

18
3

6 .Example 2 22+23+24=69

101+102+103=306

306
3

102 .
69
3

23
Answer Sum always seems to be divisible by 3. Is this true in general?
Algebra enables us to answer this question because it allows us to generalize
as follows.

Let n be any integer. Then any three consecutive integers are n , n

1 and n

2 and their sum is s
 n
 n

1
 n

2

3 n

3

3

( n

1 ) . s
Hence
3

3

( n

1 )
3

( n

1 ) an integer.
Hence the sum of any three consecutive numbers is always divisible by 3.
Algebra also enables us to solve problems with more than one unknown.
Example : You have got a drawer full of odd socks: purple, pink and orange. You do not know how many of each colour: you pull out socks one at a time until you get the desired combination. If you want a pink pair, the most socks you need to pull out is 10.
If you want a orange pair, the most socks you need to pull out is 12.For a purple pair the most you need to pull out is 14.
How many socks are there of each colour? Irish Times 15/10/2008
Solution Let y x

 the number of
the number of purple pink
socks,
socks and z

the number of orange socks.
Now after you have picked out 10 socks you will have 2 pink socks.
Hence at this stage you will have 8 socks which must be either purple or orange .
Hence x x
 z
 y


8 .By similar reasoning we will get
10 and y
 z

12
To solve this system of equations subtract the first equation from the third y
 z

 y
 x x


4 z

12

8
. Add this equation the second so that x cancels to get x
 y

10 to give 2 y

14
 y

14

7
So there must be 7 pink socks. Substituting
2 y

7
. into y
7
 z

12
So we must have 5
 z
 orange
12

7

5 .
socks. Substituting y

7 x

7

10
 x

10

7

3 . into
 z x

12
 y we get

10 we get
So we must have 3 purple socks.
Example This problem is taken from the Egyptian Rhind papyrus. A quantity and its
1 are added together to give 21.What is the quantity?
5
Solution Let
 x

1
5 x

21
 x x required quantity, then the given problem in algebraic terms is
 x ( 1

7

2
5

1
5
)


35
2
21

 x (
17
1
2
5

5
1
)

21
 x

6
5

21
 x

21

5
6

Fractions
7 ( numerator )
A numerical fraction is of the form
8 ( deno min ator )
An algebraic fraction is of the form z
3
2 x
2


1 ( numerator ) y ( deno min ator )
.
Note: The letters x , y and z are used to represent numbers so the rules for the manipulation of algebraic fractions are the same as for ordinary numerical fractions.
Rules for multiplication : To multiply fractions you multiply their corresponding numerators and denominators.
Example a b
 c d
 a b

 c d
 ac bd
,
10
3

2
7

20
21
.
Rules for Division: To divide by a fraction you turn it upside down and multiply. a
Example b
 c d
 a b c d

Example Calculate (a)
2
3
 a b
 d c
5
4

, (b) ad bc
5
.
7

10
21

5
7

21
10

105
70

21
14

2

6
13
, (c)
6
7

4
21
, (d)
1
2

3
1
.
3
2
.
Solution (a)
6
(c)
7
2
3

5
4

4
21

10
12


6
7

5
6
. (b) 2

6
13
21

4
126
28

18
4


9
2
2
1

6
13

12
13
.
.
1
(d)
2

3
1

1
2

1
3

1
.
6
Rules for the addition and subtraction of fractions
To add(or subtract) two fractions you put them over a common denominator and add(or subtract) their numerators.
Example
Solution
Calculate (a)
(a)
(b)
(c)
1
5
1
4
7
12



2
5
2
3
5
8



1
3
1

5
2

7
5

1

2
5

3
2
5


4
8


3
5
12

, (b)
8

4

12
1
4
3



12
2
3
8
56
, (c)


12

11
12
60
8

7
12


5
8
4
12

8
.


1
.
24
Note: Provided that you can manipulate ordinary fractions there is no reason why
you should not be able to manipulate algebraic fractions just as easily
since the rules are the same.

Find expressions for each of the following simplifying your answer.
Example (a) x x

1
 x
 x
2

4
 (b) x
2

1
 x x

1
(c) x x
2

1

1
 x x
2

6

1
(d) x x

2
 x
1

1
.
Solution (a) x x

1
 x
 x
2

4


 x

1
2
  x x

4


 x

1
2
 x

4
 .
(b) x
2

1
 x x

1
(c) x x

1
2 
2

 x x

6
2 
2 x
2

1


 x
 x

1
 x
1
 
1 x
2



2 x
. x
2

6
 
1
(d) x x

2
 x
1

1
 x

 x x


1
2
 
 x x


1

2


 x
 x
2
2


 x

1
 x
2 x

2

6 x
2

1
 .

2 x

5 x
2 
2

Law of Indices
The Laws of indices
In mathematics we often need to find a shorthand way of representing information or data. Nowhere is the need more obvious than when we wish to represent something like the product of 2 multiplied by itself 2,6,10 or even 20 times.
Instead of writing a
 a
 a
 a
 a
 a , we write a
6 . This is read and said as ‘2 to the power 6’; 6 is the index of the power. In general this means a n  a
 a
 a
 a

  to n factors where n is called the index.
Both a and n can be positive or negative numbers : a
 n can be written as
1 a n
.
From this simple definition, we can now go on to look at power notation, or index notation as it is often called, in more detail. We will establish the five rules that are used in this notation
Law 1 a a m
3 
 a a
2 n

 a m
 n
( a
 a
.

 a )

2
3 
( a

2
4 a )
Thus a
3  a
2  a
( 3

2 )  a
5
.


2 a
5
3

4 
2
7 
128
This rule works for both positive and negative indices. a

4  a
6 
1 a
 a
 a
 a
 a
 a
 a
 a
 a
 a
 m a
6  a
3  a
Law 2
 a n a a
Thus a
6  a
3
 a
 a
 m
 n a

, a
 a
 a a
 a
( 6

3 )

 a a
3
.


3
4
3
2 a
 a

 a
3
4

2
 a
3

.
3
2 a
 a
 a
2

9 .
This rule works for both positive and negative indices, but we need to take care when
dividing by a negative power. a
3  a

7 
 a
( a

 a a
 a

 a )
1 a


( a a

 a a

 a a

 a a

 a a

 a a
 a )
Thus a
3  a

7  a
( 3

(

7 ))  a
10
.
 a
10
.
Rules 1 and 2 only work if the powers involve the same factor , for example you cannot simplify a 6  b 7 or a 10  b 6 .

Law 3 a
0 
1 .

This is because a
 a
 
0

1 , a
5

1
 a
5 
1 and a n  a n 
1
And using the division Rule 2 we get a n  a n  a
( n
 n )  a
0 
1
Note: we can now explain why a
 n  a
( 0
 n ) a
 n 
1 a n
.

5

2 and using Rule 2 this is the same as

1
5
2 a
0  a n

1
25
. which by Rule 3
is the same as 1
 a n 
1 a n
.
In the same way we can show that a n 
1 a
 n
;
1 a
 n
1 a
 n can be written as
= 1
 a
 n  a
0
1
 a
 n
 a
 n so using Rule 2 and Rule 3
 a
( 0

(
 n ))  a n .
Law 4 ( a m
) n  a mn
.

 
2
We know that
So ( a
3
)
2  a
3 a
2
 a
3

 a
 a a .
( 3

3 )  a
6 

2
3

2 a
( 3

2 )
.

2
6 
64 .
This rule works for both positive and negative indices.
For example ( a

5
)
3  a

15
.
Law 5 a n m
 n a m 
( a m
) n
1

( a n
1
) m 
8 3
4
 3 4
8

( 8
1
3 )
4 
2
4 
16
1
( a 2 )
2  a 2
1
 a 2
1
 a
(
1
2

1
2
)  a
1  a
1
( a 3 )
3
1
( a 4 )
4
 a 3
1
 a 3
1
 a 3
1
 a 4
1
 a 4
1
 a 4
1
 a
(
1
3

1
3

1
3
)
 a
1  a
 a
1
4
 a
(
1
4

1
4

1
4

1
4
)  a
1  a
1 1
Since a 2 squared is equal to a , a 2 is the square root of a and is sometimes written
as a .
1 1 a 3 cubed is equal to a , a 3 is the cube root of a and is sometimes written
as 3
1 a .
1 a 4 to the power of 4 is equal to a , which means that a 4 is the fourth root of a and is
sometimes written as 4 a .

1
Hence it follows that a n is the n th of a and can be written n

32 5
1

  1
5

2
5

1
5

2
1 
2 a .
If we use Rule 4 to write a m n as ( a m
) n
1 or
1
( a n ) m you should be able to see that a m n is the n th root of a m and can be written as n m m
.
Rules 1,2,3,4 and 5 all work for fractional indices.
Example 1 Simplify each of the following:
3

2
(a) 2 125 3 ,(c) 4 a
2 b
3 
5 a
5 b
4
, (d) 48 a
8 
6 a
4
, (e) 3 a
3 
12 a
9
,
(f) a
9  a
3 a
6
,(g)

 a
3 a
5 b
2


3 and (h)


16
9 a a
4
6


1
2
.
Solution
(a) 4 2
3


4 2
1
 3 which can be written
 
3 
2
3 
8 .
(b) 125

2
3

1
125 3
2
(c) 4 a
2 b
3 
5 a
5 b
4


1

1
5
2

1
.

125 3
1 

2
20 a
( 2

5 ) b
( 3

4 )
=
25 a
7
20 b
7 
20 ( ab )
7
.
(d) 48 a
8 
6 a
4 
48 a
8
6 a
4

8 a
4
.
(e) 3 a
3 
12 a
9
3 a
3

12 a
9
(f) a
9  a
3 a
6
(g)

 a
3 a
5 b
2


3

 a
( 9

3 ) a
3 a
6
3
5

3 a b
3
2

3


 a
12 a
6
1
4 a
( 9

3 )
 a
( 12

6 ) a
15 b
3
27 a
6


1
4 a
6
 a
6

1

6
4 a a
( 15

6 ) b
3
27
 a
9 b
27
3
.
(h)


16
9 a a
4
6


1
2
.

16 2
1 a
6

1
2
9
1
2 a
4

1
2

4 a
3 a
2
3

4 a
( 3

2 )
3

4 a
3

Example 2
Simplify (a)
2 n 
4
( n

2 )
8
( n

1 )
Solution
(a)
2 n 
4
( n

2 )
8
( n

1 )

, (b)
2 n 
( 2
2
)
( n

2 )
( 2 3 ) ( n

1 )
1
( 8 2


3 2
3 2
) 3 and (c)
2 n 
2
2 n

4
2
( 3 n

3 )

2
2
8
( n

2 )
2 n

6 ( 2
( 3 n

3 )
)

4
( n

2 )
( 3 n

4 )
( 3 n

3 )

2
( 3 n

4

( 3 n

3 ))
.

2
7 
128 .
(b)
1
( 8 2

3 2
3
) 3
2

8
1
2

2
3

3
3
2

2
3

1
8 3

3
1 
( 2
3
) 3
1

3

2
1 
3

6 .
(c)
8
( n

2 )
2 n


6 ( 2
( 3 n

3 )
4
( n

2 )
)

(( 2 )
3
)
( n

2 )
2 n

3

2 ( 2
( 3 n

3 )
)

( 2
2
)
( n

2 )

2
( 3 n

6 )
2 n


3 ( 2
3 n

3

1 )
2
( 2 n

4 )
)

2
( 3 n

6 ) 
3 ( 2
2
( n

2 n

4 )
3 n

4 )
)

2
( 3 n

4 )
( 2
2 
2
( 3 n

4 )
3

1 )

4

3

1 .
Note: Although you can do above examples differently you must always get the same answer.
Example 3
Simplify each of the following expressions, giving your answer with positive
indices only.
(a)
2
5
2
3
2

4
Solution
(b) QQ
1 .
5
P (c)
3

4
3
5
3
4
.
(a)
2 5
2
3
2

4
=
2 5
2
3

4
=
2 5
2

1
= 2
5 
(b) QQ
1 .
5
P = Q
1

1 .
5
P
2 .
5
= Q P .
2
1
=
6
2 =64.
(c)
3
5
3

4
3
4
=
  1
2
3

4
3
4
=
3
3
5

1
2
0
=
3
1
5
2
= 3 2
5
=15.588. http://www.mathcentre.ac.uk
and then click on Algebra
and then click on Powers or indices.

Changing the subject of an equation or Transposition of formulae
We may wish to rearrange an equation, to change its subject. Some equations may have more than one algebraic symbol that is an unknown. The equation may be expressed in terms of x , say x
 y

56 . We may wish to know what the equation would look like if it were expressed in terms of y .The important thing is to remember an equation is a balance, so that any calculation done on one side of a equation, the same calculation must be done on the other side of the equation to maintain the balance.
Example 1 Make y the subject of x
 y

56 .
Subtract 56 from both sides of the equation x

56
 y

56

56 x

56
 y y
 x

56 .
Example 2 Make y the subject of x
 y
5

2
Multiply across the equation by 5
5
 x

5
 y
5

5

2

5 x
 y

10
Add 10 to both sides of an equation we get
5 x

10
 y

10

10

5 x

10
 y
Rearrange we get y

5 x

10 .

Example 3 Transpose the following equations to express x in terms of y .
(a)
(a) ax
 bx
 ax cy

 d bx

 cy ax

 d bx
(b)
 bx y

 x
 bx x


1
2 cy
 d
 ax
 bx
 cy
 d

( a
 b ) x
 cy
 d
 x

( a c
 b ) y

( a d
 b )
.
(b) y
 x x

1

2

( x

2 )
 y

( x

2 )
 x x

1

2
 yx

2 y
 x

1
 yx
 x

2 y

1
 x ( y

1 )

2 y

1
 x

2 y

1
( y

1 )
.
Example 4 Consider T

2

L where T is the period of oscillation in seconds of a g pendulum, L is the length of the pendulum in metres, g is the gravitational constant
9.81 m / sec
2
.Transpose this formula so that L is the dependent variable and T is the independent variable.
T
2


L g

T
2

2



L g


2




 


L g


1
2




2

L g

T
4

2
2

L g
Hence
L g

T
4

2
2

L
 gT
4

2
2
.
For further material and examples click on http://www.mathcentre.ac.uk
and then click on
Transposition of formulae

Solving simple equations with one unknown
We will solve an equations with one unknown x to illustrate the method. Always remember that an equation is a balance, and what you do on one side of the equation you must do the exactly the same on the other side of the equation so as to maintain the balance.
Example 1
Consider the equation
5 x

30 .
We want to find the value of x which makes the statement 5 x

30 true.
Divide the equation on both sides by 5 to get
5
5 x

30
5
 x

6 .
Verification 5

6

30 .
Example 2
Solve x

11

20
Subtract 11 from both sides of the equation x


11 x
Verification


9
9
11

.

11

20
20

.
11 .
Example 3
Solve x

32

49
Add 32 to both sides of the equation x

 x
32
Verification


32

81
81

.
32
49


49
32
.
.
Example 4
8
Solve
9 x

64
Multiply across by 9 to get 9

8
9 x

9

64

8 x

576
8
Divide across by 8 to get
8
Verification
8
9

72

576
9 x

576
8

64 .
 x

72 .

Example 5
1
3 x

1
5 x

2 . Multiply across by 3

3

1
3 x

3

1
5 x

3

2 .
 x

3
5 x

6 . Multiply across by 5

5
 x

5

3
5
 x

5

6

5 x

3 x

30

8 x

30 . Divide across by 8

8
8 x
Verification
1
3

15
4

1
5

15
4

15
12

15
20


30
8
15

 x
20


30

15
8

12
12

20
15
4


3 .
300

180
12
.
75

20

480
12

20

40
20

2 .

Solving a Quadratic Equations
Remember important formula for solving a quadratic equation, that is that gives roots of a
general quadratic equation ax
2  bx
 c

0 . ax 2  bx
 c

0
 x

 b
 b
2 
4 ac
.
2 a
Example Find roots of x
2 
4 x

12

0 .
Solution By formula above x


4

16

4 ( 1 )(

12 )
2 ( 1 )


4

2
64


4

8
2
Check (

6 )
2 
4 (

6 )

12
 x

36



12 and
24
2

12

0 x

4
2 and

( 2 )
2 x

 
6
4 ( 2 ) and x

2 .

12

4

8

12
Example Solve x
2 
4 x

4

0 .
Solution by formula above is x

4

Here we get two equal roots x

2 .
Example Solve x 2 
4 x

5

0 .
16

4 ( 1 ) 4
2

4

0
2

2 .

0 .
Solution by formula above is x

2
 
1

2
 i where i
 x

4

1

16
2

4 ( 1 ) 5

4
 
4
2

4

2
2
.Here the two roots are complex.

1
.
Example A rectangular field has an area of 60 square units. If one side is 11 units longer than the other, how long is the shorter side.
Solution Let x
 the length of the shorter side. Then
Hence by definition of the area of a rectangle x

( x x

11

11 )

 the length of longer side.
60
 x 2 
11 x

60

0 .
Using the general quadratic formula x

 b
 b
2 
4 ac we get for our problem
2 a a

1 , b

11 and c
  x


11

Answer is x
121
2

4
 as
240 x

60 so that x


11

( 11 )
2 
2

1
4

1

(

60 )
 x


11

361
 x


11
2 2

15 has no physical realization.

19

4 or

15 .
Thus the length of the shorter side is 4 and the longer side is 15

Calculating Percentages
5
When we talk of %
100 e.g Find 5% of 255500.
 number .
5
Answer
100

255500

5

2555

12775
When we say a number increases by x % , then the increase in the number
 x
100
 number and the increased number is
 number
 the increase
 number
1

( number
1
 x
100
)
 number ( 1
 x
100
)
Example (a)Calculate (i) 23% of 1534. (ii) 100% of 1534
(b)A salary of €55240 is to be increased by 12%.Calculate(i) the increase and(ii)
the new salary.
(c) In 1999 an apartment was valued at €63400.This is 5% higher than the
price paid for the apartment in 1997. Calculate the price paid in 1997.
23
Solution (a) (i)
100

1534

352 .
82 . (ii)
100
100

1534

1534
12
(b)(i) increase=
100

55240

6628 .
8
(ii) new salary=55240+6628.8=€61868.8
(c) Let price of apartment in 1997=€ x .
Then price in 1999=€63400= x

5
100
 x
Hence x ( 1 .
05 )

63400
 x

63400
 x ( 1

0 .
05 ) .

60380 .
95
105

Simultaneous equations in two unknowns
Simultaneous equations in two unknowns are pairs of equations that are both true(i.e. they are simultaneously true).We solve these, by reducing the two equations in two unknowns to one equation in one unknown which we know how to solve. We do this by algebraic manipulation which is called Gaussian elimination. We will illustrate this by doing a few examples.
Example
Solve
3 x
5 x

4 y

7 y

11
 
9
.
Multiply across the first equation by 5 and the second equation by 3 so that the resulting equations have x coefficients with the same value.

5
3


3 x
5x

3
5


4
7y y


3
5


11
 

15
15 x x


20
21 y y


55

27
Subtract so that the unknown x is eliminated to give just one equation with one unknown y .

( 20

(

21 )) y

55

(

27 )

41 y

82 y

82

41
Substitute this value of y

2

3 x

4

Solution is
2 x


11
1
 and
3 y x


2
11
.
 into either equation
3

3 x

3
 x

1 .
2 .
3

1

Verification
5

1

4

2
7

2


3
5

8

14

11
 
9
.

Example y
Solve
3 y


7 x x


4
Rearrange

7


4
7


7 x x
 y

3 y

7 x x

 y
3 y



4

7
Multiply across the first equation by 3 and the second equation by 1 so that the resulting equations have y coefficients with the same value.

3
1

 x
7 x


3

(
 y )
1

(

3 y )

3


1

(

4 )
(

7 )

21 x x


3 y
3 y



7

12
Subtract so that the unknown y is eliminated to give just one equation with one unknown x .
( 21

1 ) x
 
12

(

7 )

20 x
 
5
 x
 
5
20
Solution is
Verification x
9
4

3



9
4
7
Example Solve
1
4


,
5 x
17 x y
(

27
4



1
4
)
9
 y
9
.
4
19


 y
4

1

4

14
7
16
4


7
7


9
4
4

4

498
.
.
1

 
1
4
Substituting y x


7

(

1
4
)

4

1
4 into the first equation we get

4

7
4

16

7
4

9
4

2 .
25 .
27
4
 
0 .
25
Multiply across the first equation by 19 and the second equation by 9 so that the resulting equations have y coefficients with the same value but opposite in sign
.
5 x

9 y
17 x


95
19 x y

498

171 y
 
266
153 x
 
14

171 y


19
9
4482

17

5 x
 x

19


( 95
(

9 ) y
9

19 y


153 )
 x

9


19

(

14 )
498

266

4482

248 x

4216
 x

4216

17 .
248
Substituting x

17 into the first equation we get 5

17

9 y
 
14

85

(

14 )

9 y
9 y

85

14

9 y

99
 y

99
9
Solution x

17 , y

11 .

11 .
Verification
5

17
17

17


9

19
11


11
85


99
289

 
14
209

498

Solution of Simultaneous Equations and their Graphs
Q(1) Draw the graphs of the following straight lines
  
2 y
 
0 ,
 
6 x

3 y

33
and hence find their solution graphically.
Verify your answer algebraically.
Q(1) Solution
(i) x

2 y

2

0

2 y
 x

2
 y

1
2 x

1 .
Slope=
(ii) 6 x

3 y
1
and y-intercept=1.

2
33

3 y
 
6 x

33
 y
 
2 x

11 .
Slope=-2 and y-intercept=11.

By Gaussian Elimination x

6 x

2 y
3 y



2

6
33

1

6 x
6 x

12 y

3 y
 
12

33
____________

15 y
 
45 y x


3
2 y

2

2 ( 3 )

2

4
Answer x=4,y=3. x
Verification
6 x

2 y

3 y


4

2 ( 3 )
6 ( 4 )
 
2

3 ( 3 )

24

9

33

Q(2)Solve the system of linear equations by Gaussian Elimination. x
1
2 x
1

 x
2 x
2

 x
3

4 x
3
3 x
1

6 x
2
 x
3
6

12

12 .
Q(2)Solution x
1
2 x
1

 x
2 x
2

 x
3

4 x
3
3 x
1

6 x
2
 x
3
6

12

12 .
Add the first and third equations x
1
 x
2
 x
3

6
2 x
1
 x
2

4 x
3

12
__________ ______
4 x
1

7 x
2

18 x
1
2 x
1

 x
2 x
2

 x
3

4 x
3
6

4


12

1

4 x
1
2 x
1


4 x
2 x
2


4 x
3

4 x
3

24
12
2 x
1
Solve
4 x
1
2 x
1

5 x
2

12

7 x
2


5 x
2

18

1

12

2

4 x
4 x
1
1

7 x
2

10 x
2

18

24
______________
Subtract

3 x
2
 
6
Hence x
2

2
Substituting
Substituting
4 x
1 x
1


7 x
2
 

 x
3
18


1

4 x
1
2

 x
3
18


14
6

4
 x
3
Solution is x
1

1 , x
2

2 and x
3

3 .
 x
1

1 .

6

3

3
For further material and examples click on http://www.mathcentre.ac.uk
and then click on
Simultaneous Equations.