Download Basic Mathematics For Basic Mathematics consult Foundation Maths

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- MacmillanEssential 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  2  24, 5 32  2 .
3
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
 20
 4 ,  20  5 
 4
5
5
If you divide a number by zero, you will always get an error, as this is impossible.
n
If n is any number is invalid.
0
-20  5 
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
5
2
2 x  5  0  2x  5  x   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
18
69
 6 .Example 2 22+23+24=69 
 23
3
3
306
 102 .
101+102+103=306 
3
5+6+7=18 
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  3n  3  3  (n  1) .
Hence
s 3  (n  1)

 (n  1) an integer.
3
3
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 x  the number of purple socks,
y  the number of pink 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  z  8 .By similar reasoning we will get
x  y  10 and
y  z  12
To solve this system of equations subtract the first equation from the third
y  z  x  z  12  8
 y  x  4 . Add this equation the second so that x cancels to get
14
x  y  10 to give 2 y  14  y 
 7.
2
So there must be 7 pink socks. Substituting y  7 into y  z  12 we get
7  z  12  z  12  7  5 .
So we must have 5 orange socks. Substituting y  7 into x  y  10 we get
x  7  10  x  10  7  3 .
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  required quantity, then the given problem in algebraic terms is
1
1
5 1
6
5
x  x  21  x(1  )  21  x(
)  21  x   21  x  21
5
5
5
5
6
75
35
1
x

 17
2
2
2
Fractions
7(numerator )
A numerical fraction is of the form
8(deno min ator )
2 x 2  1(numerator)
An algebraic fraction is of the form 3
.
z  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.
a c a  c ac 10 2 20
Example  
,
.

 
b d b  d bd
3 7 21
Rules for Division: To divide by a fraction you turn it upside down and multiply.
a
a c
a d ad 5 10 5 21 105 21 3
  

 .
Example   b   
. 
b d c
b c bc 7 21 7 10 70 14 2
d
Example Calculate (a)
2 5
6
6 4
1 3
 , (b) 2  , (c)  , (d)  .
3 4
7 21
2 1
13
2 5 10 5
6
2 6 12
 
 . (b) 2 
   .
3 4 12 6
13 1 13 13
6 4 6 21 126 18 9
  
  .
(c) 
7 21 7 4
28
4 2
1 3 1 1 1
(d)     .
2 1 2 3 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.
1 2
1 2
7 5
 .
Example Calculate (a)  , (b)  , (c)
5 5
4 3
12 8
1 2 1 2 3

Solution (a)  
5 5
5
5
1 2 3  1  2  4 3  8 11


(b)  
4 3
43
12
12
7 5 7  8  5  12 56  60
4
1
 



(c)
.
12 8
12  8
12  8
12  8 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.
Solution (a)
Find expressions for each of the following simplifying your answer.
Example (a)
(d)
Solution (a)
x
2
2
x
x 1 x  6
(b)
(c) 2



x 1 x 1
x  1 x2  1
x  1 x x  4 
x
1
.

x  2 x 1
2x
2
x
2
.



x  1 xx  4 x  1xx  4 x  1x  4
(b)
2
x
2
x 1 2



 .
x 1 x 1 x 1
x
x
(c)
x 1
x  6  x  1  1   x  6  1 x  1  x  6 2 x  5
 2


 2
2
x 2 x 2
x2  2
x2  2
x 2
(d)
x2  2
xx  1  x  2
x
1
.



x  2 x 1
x  2x  1 x  2x  1
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.
1
.
an
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
Both a and n can be positive or negative numbers : a  n can be written as
Law 1
a m  a n  a m  n .  2 3  2 4  2 3 4  2 7  128
a 3  a 2  (a  a  a)  (a  a)  a 5
Thus a 3  a 2  a (3 2)  a 5 .
This rule works for both positive and negative indices.
1
a 4  a6 
 a  a  a  a  a  a  a  a  a2
aaaa
34
am
mn
Law 2
 2  3 4 2  3 2  9 .
a ,
n
3
a
a

a

a

a

a

a
 a  a  a  a3 .
a6  a3 
aaa
6
3
a

a

a ( 6 3)  a 3 .
Thus
This rule works for both positive and negative indices, but we need to take care when
dividing by a negative power.
1
a3  a 7  a  a  a 
aaaaaaa
 (a  a  a)  (a  a  a  a  a  a  a)  a10 .
Thus a 3  a  7  a (3( 7 ))  a10 .
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 .  11  1
This is because a  a  1, a 5  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
1
1
1
Note: we can now explain why a  n  n .  5  2  2 
.
25
a
5
a  n  a ( 0 n ) and using Rule 2 this is the same as a 0  a n which by Rule 3
1
is the same as 1  a n  n .
a
1
In the same way we can show that a n   n ;
a
1
can be written as 1  a  n so using Rule 2 and Rule 3
n
a
1
= 1  a  n  a 0  a  n  a ( 0(  n ))  a n .
n
a
0
 
2
Law 4 (a m ) n  a mn .
 2 3  2 32  2 6  64 .
We know that a 2  a  a .
So (a 3 ) 2  a 3  a 3  a (33)  a 6  a (32) .
This rule works for both positive and negative indices.
For example (a 5 ) 3  a  15 .
m
1
2
1
a n n
Law 5
2
1
2
1
2
1
3 3
1
3
1
3
(a )  a  a  a
1 1
(  )
2 2
1
3
(a )  a  a  a  a
1
1
1
4
1
a m  (a m ) n  (a n ) m  8 3  3 8 4  (8 3 ) 4  2 4  16
1
1
 a1  a
1 1 1
(   )
3 3 3
1
(a 4 ) 4  a 4  a 4  a 4  a 4  a
1
 a1  a
1 1 1 1
(    )
4 4 4 4
 a1  a
1
Since a 2 squared is equal to a , a 2 is the square root of a and is sometimes written
as a .
1
3
1
3
a cubed is equal to a , a is the cube root of a and is sometimes written
as 3 a .
1
4
1
4
a to the power of 4 is equal to a , which means that a is the fourth root of a and is
sometimes written as 4 a .
1
n
Hence it follows that a is the n th of a and can be written
 
1
5
 32  2
1
5 5
2
1
5
5
n
a.
 21  2
1
m
1
If we use Rule 4 to write a n as (a m ) n or (a n ) m you should be able to see that
m
n
a 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:
2
3
3
2
(a) 4 , (b) 125 ,(c) 4a 2 b 3  5a 5 b 4 , (d) 48a 8  6a 4 , (e) 3a 3  12a 9 ,
1
3
16a 6  2
 a 5b 
a9  a3
(f)
,(g)
and
(h)
 2
 4  .
a6
 3a 
 9a 
Solution
3
 12 
(a) 4   4  which can be written
 
3
2
2
1
(b) 125 3 
2
3
1

2

 4
3
 23  8 .
1
1

.
2
25
5
1


125 3 




2 3
5 4
( 2  5 ) ( 3 4 )
b
(c) 4a b  5a b  20a
= 20a 7 b 7  20(ab) 7 .
125
(d) 48a 8  6a 4 
48a 8
 8a 4 .
6a 4
3a 3
1
1
1
(e) 3a  12a 
 ( 9 3)  6  a  6
9
4
4a
12a
4a
3
9
a 9  a 3 a (93) a12
 6  6  a (126)  a 6
6
a
a
a
3
53 3
5
a b
a 15b 3 a (156) b 3 a 9 b 3
a b
(g)  2   3 23 
.


27
27
27a 6
3 a
 3a 
(f)
1
1
6
1
16a 6  2 16 2 a 2 4a 3 4a (3 2) 4a

(h)  4  .  1 1  2 
4
3
3
3
a
 9a 
92 a 2
Example 2
2 n  4 ( n 2)
8 ( n 2)  6(2 (3n 3) )
2
2 3
,
(b)
and
(c)
.
(
8

3
)
8 ( n 1)
2 n  4 ( n 2)
1
Simplify (a)
3
2
Solution
2 n  4 ( n  2 ) 2 n  (2 2 ) ( n  2) 2 n  2 2 n  4 2 ( 3n  4 )
(a)


 (3n 3)  2 (3n 4(3n3))  2 7  128 .
(2 3 ) ( n 1)
8 ( n 1)
2 ( 3 n 3)
2
1
2
3
2
2
3
(b) (8  3 )  8
(c)
1 2

2 3
3
3 2

2 3
1
3
1
3 3
 8  3  (2 )  3  21  3  6 .
1
8 ( n 2)  6(2 (3n 3) ) (( 2) 3 ) ( n  2)  3  2(2 (3n 3) ) 2 (3n 6)  3(2 3n 31) )


2 n  (2 2 ) ( n  2 )
2 n  4 ( n 2)
2 n  2 ( 2 n 4)
2 (3n 6)  3(2 3n  4) ) 2 (3n  4) (2 2  3  1)


 4  3  1.
2 ( n 2 n 4)
2 ( 3n  4)
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)
35
25
1.5
(b)
(c)
.
QQ
P
2 3 2 4
34 34
Solution
25
25
25
(a) 3  4 = 34 = 1 = 2 5  21 = 2 6 =64.
2 2
2
2
(b) QQ 1.5 P = Q11.5 P = Q 2.5 P .
3
3  3 3
(c)  4 4 =  4 4 = 0 = = 3 2 =15.588.
3 3
1
3
3 3
5
1
5 2
5
1
2
5
2
5
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 .
y
Example 2 Make y the subject of x   2
5
Multiply across the equation by 5
y
5  x  5   5  2  5 x  y  10
5
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 .
x 1
(a) ax  bx  cy  d (b) y 
x2
(a) ax  bx  cy  d  ax  bx  bx  bx  cy  d  ax  bx  cy  d
c
d
.
y
 (a  b) x  cy  d  x 
( a  b)
( a  b)
(b) y 
x 1
x 1
 ( x  2)  y  ( x  2) 
x2
x2
 yx  2 y  x  1  yx  x  2 y  1  x( y  1)  2 y  1
x
2y 1
.
( y  1)
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.
Example 4 Consider T  2
2
T

2
 L
L
 T 

    

g
g
 2 


2
2
1


2


T2
L
L
L

     

2
 g  
g
g
4


Hence
L T2
gT 2
.
L



4 2
g 4 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
5x  30 .
We want to find the value of x which makes the statement 5x  30 true.
Divide the equation on both sides by 5 to get
5
30
x
 x  6.
5
5
Verification 5  6  30 .
Example 2
Solve x  11  20
Subtract 11 from both sides of the equation
x  11  11  20  11.
 x  9.
Verification 9  11  20 .
Example 3
Solve x  32  49
Add 32 to both sides of the equation
x  32  32  49  32 .
 x  81 .
Verification 81  32  49 .
Example 4
8
Solve x  64
9
8
Multiply across by 9 to get 9  x  9  64  8x  576
9
8
576
Divide across by 8 to get x 
 x  72 .
8
8
8
576
 64 .
Verification  72 
9
9
Example 5
1
1
1
1
x  x  2 . Multiply across by 3  3  x  3  x  3  2 .
3
5
3
5
3
3
x  6 . Multiply across by 5  5  x  5   x  5  6  5x  3x  30
5
5
8
30
30 15
x

 3.75 .
 8x  30 . Divide across by 8  x 
8
8
8
4
1 15 1 15 15 15 15  20  15  12 300  180
480
Verification    




3 4 5 4 12 20
12  20
12  20
12  20
 x

40
 2.
20
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 .
 b  b 2  4ac
ax  bx  c  0  x 
.
2a
2
Example Find roots of x 2  4 x  12  0 .
Solution By formula above x 
 4  16  4(1)( 12)  4  64  4  8


2
2
2(1)
 12
4
and x   x  6 and x  2 .
2
2
2
Check (6)  4(6)  12  36  24  12  0 and (2) 2  4(2)  12  4  8  12  0 .
Example Solve x 2  4 x  4  0 .
4  16  4(1)4 4  0

 2.
Solution by formula above is x 
2
2
Here we get two equal roots x  2 .
Example Solve x 2  4 x  5  0 .
4  16  4(1)5 4   4 4  2  1


Solution by formula above is x 
.
2
2
2
x  2   1  2  i where i   1 .Here the two roots are complex.
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 x 11  the length of longer side.
Hence by definition of the area of a rectangle x  ( x  11)  60  x 2  11x  60  0 .
x
 b  b 2  4ac
we get for our problem
2a
 11  (11) 2  4  1  (60)
a  1 , b  11and c  60 so that x 
2 1
 11  121  240
 11  361
 11  19
x
 4 or  15 .
x
x
2
2
2
Answer is x  4 as x  15 has no physical realization.
Thus the length of the shorter side is 4 and the longer side is 15
Using the general quadratic formula x 
Calculating Percentages
5
When we talk of 5% of a number we mean
 number .
100
e.g Find 5% of 255500.
5
Answer
 255500  5  2555  12775
100
When we say a number increases by x% , then the increase in the number
x

 number and the increased number is
100
number
number
x
x
(

)  number (1 
)
 number  the increase 
1
1
100
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
100
Solution (a) (i)
 1534  352.82 . (ii)
 1534  1534
100
100
12
 55240  6628.8 (ii) new salary=55240+6628.8=€61868.8
100
(c) Let price of apartment in 1997=€ x .
5
 x  x(1  0.05) .
Then price in 1999=€63400= x 
100
63400
 60380.95
Hence x(1.05)  63400  x 
105
(b)(i) increase=
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
3 x  4 y  11
Solve
.
5 x  7 y  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  3x  5  4 y  5  11
15 x  20 y  55

3  5x - 3  7y  3  - 9
15 x  21y  27
Subtract so that the unknown x is eliminated to give just one equation with one
unknown y .
82
 (20  (21)) y  55  (27)  41y  82 y 
 2.
41
Substitute this value of y  2 into either equation
 3x  4  2  11  3x  11  3  3x  3  x  1.
Solution is x  1and y  2 .
Verification
3  1  4  2  3  8  11
5  1  7  2  5  14  9
.
Example
Solve
y  7x  4
Rearrange 
 4  7x  y

7 x  y  4
 7  x  3y
3y  x  7
x  3 y  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.
21x  3 y  12
3  7 x  3  ( y )  3  (4)


x  3 y  7
1  x  1  (3 y )  1  (7)
Subtract so that the unknown y is eliminated to give just one equation with one
unknown x .
5
1
   0.25
(21  1) x  12  (7)  20x  5  x  
20
4
1
Substituting x   into the first equation we get
4
1
7 16  7 9
y  7  ( )  4  4  
  2.25 .
4
4
4
4
1
9
Solution is x   , y  .
4
4
9
1
16  7 9
 .
Verification  7  ( )  4 
4
4
4
4
9 27
1
7  4  1 27
3 
  7 

4
4
4
4
4
5 x  9 y  14
Example Solve
.
17 x  19 y  498
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  14
19  5 x  19  (9) y  19  (14)

17 x  19 y  498
9  17 x  9  19 y  9  498
 95 x  171y  266
 (95  153)  x  266  4482  248x  4216
153x  171y  4482
4216
x
 17 .
248
Substituting x  17 into the first equation we get 5  17  9 y  14  85  (14)  9 y
99
9 y  85  14  9 y  99  y 
 11 .
9
Solution x  17 , y  11 .
Verification
5 17  9 11  85  99  14
17 17  19 11  289  209  498
Solution of Simultaneous Equations and their Graphs
Q(1) Draw the graphs of the following straight lines
i  x  2 y  2  0,
ii 6x  3y  33
and hence find their solution graphically.
Verify your answer algebraically.
Q(1)Solution
1
(i) x  2 y  2  0  2 y  x  2  y  x  1 .
2
1
Slope= and y-intercept=1.
2
(ii) 6 x  3 y  33  3 y  6 x  33  y  2 x  11 .
Slope=-2 and y-intercept=11.
By Gaussian Elimination
x  2 y  2
6
6 x  3 y  33  1

6 x  12 y  12
6 x  3 y  33
____________
 15 y  45
y3
x  2 y  2  2(3)  2  4
Answer x=4,y=3.
Verification
x  2 y  4  2(3)  2
6 x  3 y  6(4)  3(3)  24  9  33
Q(2)Solve the system of linear equations by Gaussian Elimination.
x1  x 2  x3  6
2 x1  x 2  4 x3  12
3x1  6 x 2  x3  12.
Q(2)Solution
x1  x 2  x3  6
2 x1  x 2  4 x3  12
3x1  6 x 2  x3  12.
Add the first and third equations
x1  x 2  x3  6
2 x1  x 2  4 x3  12
________________
4 x1  7 x 2
x1  x 2  x3  6
4
2 x1  x 2  4 x3  12  1
 18
 4 x1  4 x 2  4 x3  24
 2 x1  x 2  4 x3
 12
__________
_______
2 x1  5x2
 12
Solve
4 x1  7 x2  18  1
 4 x1  7 x 2  18
2 x1  5 x2  12  2  4 x1  10 x2  24
______________
Subtract
 3x2  6
Hence
x2  2
Substituting 4x1  72  18  4x1  18  14  4  x1  1 .
Substituting x1  x2  x3  1  2  x3  6  x3  6  3  3
Solution is x1  1, x2  2 and x3  3 .
For further material and examples click on
http://www.mathcentre.ac.uk
and then click on
Simultaneous Equations.