# JSS 3 FIRST TERM LESSON NOTE MATHEMATICS

SCHEME OF WORK FOR MATHEMATICS JSS

SESSION

WEEK TOPIC

1 Revision of JSS 2 work

2 The Binary number system

3 Binary number system continued

4 Algebraic Processes

5 Word problems

6 Change of subject of formulae

7 Revision of first half terms work and periodic test

8 Statistics

9 Statistics Continued

10 Simple equations involving fraction and simultaneous equations

11 Revision of 2nd half term’s lesson and periodic test

12-13 First term examination

REFERENCE MATERIALS
ESSENTIAL MATHEMATICS for junior secondary school, book 3 by A. J. S Oluwasanmi
EFFECTIVE MATHEMATICS for junior secondary school book 3 by M.K.Akinsola, M.C.Ejike and A.Tella

JSS 1 SECOND MATHEMATICS LESSON NOTE

WEEK 1
REVISION OF JS S 2 WORK

WEEK TWO
BINARY NUMBERS
Numbers in base two are called binary numbers at is made up two digit is 0 and 1
Converting base 10 numbers to base two number
We do this by dividing the base ten number repeatedly by 2, writing down the remainder until we get to zero and reading the remainder upwards.
Example: (a) Write 810 to a number in base two
b) Express 85 in a binary number
c) Convert 10710 to a number in the base two
d) Convert 152ten to a number in base two
e) Convert 3/8ten to a binary fraction (bicimal)
f) Express 15.12510 in binary notation

SOLUTION
(a) 2 8
2 4 R 0
2 2 R 0
0 R 1
810 = 10002

(b) 2 85
2 42 R 1
2 21 R 0
2 10 R 1
2 5 R 0
2 2 R 1
1 R 0
0 R 1
8510 = 1010101two

(c) 2 107
2 53 R 1
2 26 R 1
2 13 R 0
2 6 R 1
2 3 R 0
2 1 R 1
0 R 1
10710 = 110100112
(d) 2 152
2 76 R 0
2 38 R 0
2 19 R 0
2 9 R 1
2 4 R 1
2 2 R 0
2 1 R 0
0 R 1
152ten = 100110002
(e) 2 3
2 1 R 1
0 R 1
310 = 112

2 8
2 4 R 0
2 2 R 0
2 1 R 0
2 0 R 1
810 = 1000two

First express 3 and 8 in binary, 3810 = 112/10002= 0.0112
(f) 15.125 = 151251000 = 1518= 121810

2 121

2 60 R 1
2 30 R 0
2 15 R 0
2 7 R 1
2 3 R 1
1 R 1
0 R 1
12110 = 11110012
2 8 R
2 4 0
2 2 0
2 1 0
2 0 1
810 = 10002

(1218)10 = 1111001/10002 = 1111.0012

Exercise: Convert the following binary numbers.
(a) 72 (b) 34 (c) 0.875 (d) 32
Converting Base Two Numbers to Base 10 Numbers
We express the given binary numbers as a sum of multiples of powers of two 20, 21, 22, 23 etc.
Example: Convert (i) 101two (ii) 10.10012 (iii) 1112
SOLUTION
1012 = 1×22 + 0x21 + 1×20
= 4 + 0 + 1
= 510
1112 = 1×22 + 1×21 + 1×20
= 4 + 2 + 1
= 910
10.10012 = 1×21 + 0x20 + 1 x 2 -1 + 0 x 2-2 + 0 x 2-3 + 1 x 2-4
= 2 + 0 + 12 + 0 + 0 + 116
= 4116
= 291610
101012 = 1 x 24 + 0 x 23 + 1 x 22 + 0 x 21 + 1 x 20
= 16 + 0 + 4 + 0 + 1
= 2710
Exercise: If 1102 = P10. Find the value of P

Assignment
Write 1-10 in binary numbers
Convert to base 10 (a) 111012 (b) 11.01012 (c) 10110012
Convert to binary number (a) 43ten (b) 1280ten (c) 17610
ADDITION, SUBTRACTION, MULTIPLICATION AND DIVISION IN BINARY NUMBERS
Examples
Subtract 100112 from 1011112
11012 x 1112
100100012 1012

Solution
1. 1011112 2. 1011112 3. 11012
100112 +100112 x1112
111002 10000102 1101
1101
1101
1011011
4.Convert to base 10 to have 14510 510 = 2910 = 111012 or
111
1012 10010012
101
1000
101
101
101
0
Exercise: (1) add 1001102, 1010102 and 1110112
(2) Multiply 100112 x 112
(3) 10110102 – 1001112
(4) 10101112 1112

WEEK THREE
APLLICATION OF BINARY NUMBERS
Bunch Cards: it is used in business farms, examining from boards and other organization as a solution for sorting information.
Example: suppose that a failure in a subject is represented by the digit zero. Then for any 5 subjects English, Mathematics, French, Agric and C.R.K. The number 10110 represent a pass in English, a failure in mathematics, a pass in French, a pass in agric and a failure in C.R.K.
Bunch Tape
Letters Decimal No Binary Code
A 1 1
B 2 10
C 3 11
D 4 100
E 5 101
F 6 110
G 7 111
H 8 1000
I 9 1001
J 10 1010
K 11 1011
L 12 1100
M 13 1101
N 14 1110
O 15 1111
P 16 1000
Q 17 10001
R 18 10010
S 19 10011
T 20 10100
U 21 10101
V 22 10110
W 23 10111
X 24 11000
Y 25 11001
Z 26 11010
French tapes are used to enter information and instruction into some computers.

Exercise:
Using a letter per line code and with the words represented by the following
(a) 1 (b) 10000 (c) 100
100 101 101
100 10100 1100
10010 101 1001
101 10010 10110
10011 101
10011 10010

Write the following in a letter per line code.
(a) S (b) D (c) A (d) S (e) E
U I N C L
N V G H I
D I E O A
A N L O S
Y E L

WEEK 4
ALGEBRAIC PROCESSES
Objective: Use letter to generalize statements
Interpret mathematical symbols
Solve simple and problems
Generating statement
The following symbols are generally used in mathematics
“=” means is equal to”
Example: 13 – 6 = 7 means thirteen minus six is equal to seven
“>” means is greater than”
Example: 18 > 10 means eighteen is greater than ten”
“<” means is less than”
Example: -10 < 2 mean negative ten is less than two
“ ≅” means is approximately equal to”
Example: 227 is approximately equal to 3.142 that is 227≅ 3.14
“” means not equal to”
Example: 2 3 means two is not equal to three.
Exercise
Write the following using symbols
M is equal to 8 (ii) y lies between 8 and 12
(iii) Twice y equals fourteen, therefore y equals seven.

OPERATION IN ALGEBRA
EXAMPLE
Write down the meaning of the following and find the values if x = 8, y = 3, z = 9
(a) 2x (b) 2 + x (c) x2 (d) x – 5 (e) x2 (f) x + yz
Solution
(a) Multiply x by 2 (b) Add two to x (c) Divide x by 2
I.e. 2 x 8 = 16 i.e. 8 + 2 = 10 x2 = 82 = 4
(d) Subtract five from x (e) Square x (f) multiply y by z
x– 5 = 8 – 5 = 3 x 2 = 8 x 8 = 64 x + yz = 8 + 3(9) = 8 + 27 = 35
Exercise: If a = 5, b = -3 and c = 2. Find the value of (a) 2a + b – c (b) 5a – 8 (c) 8ac – 2b (d) a+b+c4
LIKE AND UNLIKE TERMS (WORD PROBLEMS)
Examples 1: A fence is made up of 3 different bundles of length 3m each, 5 different hurdles of lengths 5m each and 2 different hurdles of length 2m each. What is the total length?
SOLUTION
Let the 3 hurdles bex, y and z
Then total length force = 3x + 5y – 2z
2) A tourist walked xkm an hour for 3 hours on the first day and xkm an hour for y hours on the second day. How far did he walk in the two days?
Ans. (3x + xy) km

Exercise
A family eats a loaves of bread a day. How many loaves are need for (a) M-days (b) Kweeks (c) How long will p loaves last? (a) 4m loaves (b) 28k Loaves (c) pd days
Use of brackets
Examples 1: The angles of a triangle measured in decrees’ are 2 x, 3 x+ 5, 4 (x+ 10). What are the values of the angles?
SOLUTION
2 x + (3 x + 5) + 4 (x + 10) = 180
2 x + 3 x + 5 + 4 x+ 40 = 180
9x + 45 + 180 – 45
9×9 = 1359
x= 150
The angles are 300, 500 and 1000
2. The results of adding three consecutive numbers is one hundred and twenty. Find the number.

SOLUTION
Let the numbers bex, x + 1, x+ 2
x+ x+ 1 + x+ 2 = 120
3 x + 3 = 120
3 (x+1) = 120
x + 1 + 40
x + 40 -1
x= 39o
Exercise: The greatest of 5 consecutive even numbers is 2a. What is the sum of all the five numbers?

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WEEK 5
WORD PROBLEMS
Sum and Difference
When 30 is added to a number the result is -18. Find the number.
SOLUTION
The number be x
x + 30 = -18
x = – 48
The difference between a number and 5 is 20 what are the two positive numbers
SOLUTION
Let the number be x
x – (-5) = 20
x + 5 = 20
x – 15
If xis less than -5
-5 – x = 20
x = -5 – 20
x = -25
The sum of three consecutive integers is 138. Find the numbers.
SOLUTION
Let the numbers be n, n + r, and n + 2
n + (n + 1) + (n + 2) = 138
3n + 3 + 138
3n = 138
n = 45
n + 1 = 46
n + 2 = 47
The numbers are 45, 46, and 47
Exercise: Ex 2.1 No 6, 7, 8

Product
Examples: (1) Find the product of 134, -0.8 and – 212 (2) what number must be multiplied by 25 to make 34 ? (3) The product of 3 numbers is 3600. If two of the numbers are equal and the third number are 25. Find the two equal numbers.
SOLUTION
1. 74x (-810) x -52 (2) Let the number be x (3) 25 x x x x = 3600
= 74 25 ×x = 34 25×2 = 3600
= 312 100x = 3 x2 = 360025
x = 3100 x2 = 144
x = 0.03 x = 12
Solving combine products with sum and difference
Examples: (1) Add the sum of the 20 and 30.5 to the position difference between 25 and 45. (2) From the sum of 8 ad 7 subtract the negative difference between 15 and 24 (3) find the product of 334 and 56
SOLUTION
1. (20 + 30.5) + (45 – 25) (2) -8 + 7 – 115 – 25 (3) Sum = 154 + 56
= 50.5 + 20 = -8 + 7 + 9 45+1012
= 70.5 = 8 = 5512
Product = 25 x 5512
= 116 = 156
Exercise: Ex. 2.3 No 1, 2, 3, page 17.
Word problems with fractions
Examples:
Find the three-fifth of the sum of 45 and -60. 35 (45 – 60) = 35 (-15) = -9
Divide the difference between 25 and 10 by the product of 6 and 5. 25-106 x 5 = 1530 = 0.5
Find one-quarter of the sum of the product of 2 18 and 3 15 and the product of 112 and 49

Solution: 14178 x 165+ 32 x 49 = 2810 = 11315
Exercise: Ex 2.4 No 1, 2, 3, page 18
Examples
When 35 of a number is added to 30. The result is 20 added to the number find the number.
When the sum of 28 and a certain number is divided by 5. The result is equal to treble the original number. What is the number?
SOLUTION
(1) Let the number be x (2) Let the number be x
Hence 35x + 30 = x + 30 28+x5 = 3x
Multiply each term by 5 15x = 28 + x
3x + 150 = 5x + 100 14×14 = 2814
5x – 3x = 150 -100 x = 2
2x = 50
x = 25
ASSIGNMENT
EXERCISE 2.6; NO 1, 2, AND 3 PAGE 20.

WEEK 6

CHANGE OF SUBJECT OF A FORMULA
Subject of formula: A formula is always written in terms of the subject e.g. V = 13πr2h is a formula for volume of a cone, r is the radius, height (h) =227.
Examples 1: Make r the subject of the formula in A = πr2 (2) Make T the subject of the formula in PTV = K (3) Make x the subject of the formula in y =km/x
SOLUTION
a) A = πr2 b) PTV = K c) y = kmx
r2 =A PT = KV xy = km
r2 = √ A T = KVp x = kmy
Exercise
Make b the subject of the formula in A = 12 (a + b) x
Make x the subject of the formula in a = b ( 1 – x)
Make w the subject of the formula in L = wha(w+p)
Substitution in formula
Examples: The total surface area of a closed cylinder of base radius rcm and height hcm is given by A = 2 πr(r+h)
Write in terms of A and r
Find the height of the cylinder of base radius 7cm and the total surface area of 396cm2. = 227.
SOLUTION
A = 2 πr(r+h) b) h = 396227 x 7 x 2 – 7
r + h = A2 πr h = 2m
h = A2 πr – r
Exercise: The mass of water in a rectangular tank 1m long. 6m wide and him deep is Mkg where M = 1000 lbh (a) What is the mass of water in a tank 5m long, 4m wide and 5m deep? (b) What is the depth of the water in a tank 5m long and 3m wide of its mass is 24000kg?
Assignment
if T = 2 π√mk (a) Express m in terms of T an K (b) Find m when T = 20 and K = 50, = 227
make h the subject of the formula V = 13πr2h
hence find the value of h when V = 256, = 227, r = 21

WEEK 8
EVERYDAY STATISTICS
Data Presentation
Frequency Table
Example: The raw data below show the number of mobile phone calls made by a group of students in a certain day.
5 6 7 8 4 4 5 7 8 10
7 6 5 8 5 7 8 7 10 7
6 5 6 7 7 5 4 5 7 8
Use a tally mark to prepare a frequency table for this data
What calls occur most often?
What percentage of students made 8 calls?
SOLUTION
No of calls made Tally Frequency
4 3
5 7
6 4
7 9
8 5
9 0
10 2
7 calls
530 x 100
= 1623%

Pictogram
Example: The following table shows the colour of cars in a car park one morning. Draw pictogram to illustrate this data.
Colours of car frequency
Black 20
White 17
Red 8
Yellow 5
Green 10

SOLUTION
Colour of cars
Black
White
Red
Yellow
Green
Key: = 2 = 1

Bar Chart
Bar charts consist of series of bars with equal width.
Example: Draw bar chart to illustrate the data of the example above

Frequency colour of cars of the park

Compound bar chart
It is used to compare two or more different sets of information.
Example:
The following table shows the number of candidates who gained admission into higher institutions at a certain town over a period of years.
Year Boys Girls
1997 65 46
1998 50 55
1999 80 73
2000 70 92
2001 45 64
Illustrate this information on dual bar chart
Illustrate what year did girls leave the highest admission?
Illustrate what year did boys have the least admission?
How many more candidates gained admission in 2000 than 1998?
SOLUTION
No of candidates

ASSIGNMENT
EXERCISE 22.1; NO 1, 2, AND 3 PAGE 226.

Pie charts
A pie chart is a circle divided into sectors whose angle are used to display data
Example:
In a certain year, the expenditure of a university is shown in the table below.

Items Expenditure in Million Naira
Equipment 20
Salaries and wages 25
Building projects 70
Maintenance 25
Miscellaneous 10
Draw a pie chart to illustrate the information
What percentage of total expenditure goes on project
SOLUTION

Items Expenditure in Million Naira Angles
Equipment 20 20150 x 360°1 = 48o
Salaries and wages 25 25150 x 360°1 = 60o
Building Project 70 70150 x 360°1 = 168o
Maintenance 25 25150 x 360°1 = 60o
Miscellaneous 10 10150 x 360°1 = 24o
Total 150 360o

b) 70150×100°1 = 46.7o

Example 2

History 40% , Geography 30% , Further Mathematicss10% and Physics 20%
The pie chart shows the percentage of students taking Further Mathematics, Physics, History and Geography
What angle represented subject?
What fraction of students are taking history
If the total number of students is 500, how many students are taking physics?
SOLUTION
Subject Percentage Angles
Further Maths 10% 10100 x 360°1 = 36o
Physics 20% 20100 x 360°1 = 72o
History 40% 40100 x 360°1 = 144o
Geography 30% 30100 x 360°1 = 108o
Total 100% 360o
(b) Fraction for history = 40100 = 25
(c) No of Physics Students = 20100 x 500 = 100 Students
Exercise: Ex. 22.2 No 1 and 5 page 228 and 229

WEEK 9
MEASURE OF CENTRAL TENDENCY (MEAN, MEDIAN, MODE)
Example
In a test 10 pupils obtained the following marks 5, 7, 4, 8, 5, 7, 10, 9, 3. Find (a) The mean mark (b) Median mark (c) Modal mark.
SOLUTION
Mean = sum of mark No of marks = 5+7+4+8+5+7+10+9+6+3 10 = 64 10
Mean = 6.4
Arrange the marks in ascending order of magnitude 3, 4, 5, 5, 6, 7, 8, 9, 10
Median = 6+7 2 = 132 = 6.5
The mode is the value that occurs most the mode are 5 and 7 this is bimodal.
Calculating average from frequency tables
Examples: In a science test. The following score shown in the table below were obtained out of 10 by some students.
Marks No of Students (Frequency)
0 2
1 1
2 2
3 4
5 1
6 7
7 3
8 4
9 1
Find (i) the mode (ii) the median (iii) the mean of the frequency distribution
How many students scored at least 5 marks

SOLUTION
(i) Mode = 8
(ii) 0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 7, 8, 8, 8, 8, 9
Median = 5+6 2 = 5.5
(iii) Sum of values = 0 + 0 + 1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + 7 + 7 + 7 + 8 + 8 + 8 + 8 + 9 = 99
Mean = sum of values No of values = 9920 = 4.95

ALTERNATIVELY

Scores (x) frequency (f) frequency x score (fx)
0 2 2 x 0 = 0
1 1 1 x 1 = 1
2 2 2 x 2 = 4
3 1 1x 3 = 9
4 3 3 x 4 = 12
5 1 1 x 5= 5
6 2 2 x 6 = 12
7 3 3 x 7 = 21
8 4 4 x 8 = 32
9 1 1 x 9 = 99
f = 20 fx = 99
Mean = fxf = 9920 = 4.95
Exercise: Ex 22.3 No 1 and 2 page 231
Range: It gives a measure of how spread and the values are. Range = Highest value – Lowest value
Examples
Find the range of these numbers 9, 4, 7, 6, 12, 8, 15, 10
Solution: Range = 15 – 4 = 11

A student obtained the following marks each out of 100 in different geography test 42, 44, 50 40, 54, 48, 10 88. Find (a) The mean (b) the range (c) make a comment why the range in this case is not good to measure the spread.
SOLUTION
Mean = sum of values No of values = 42+44+50+40+54+48+10+88 8
Mean = 48
Range = 88 – 10
= 78
The two extreme values i.e. 10 and 88 affects the range, so at it not a good measure of spread in this particular case.
Exercise
Find the range of the following
35cm, 50cm, 45cm, 90cm, 30cm
67km, 50km, 20km, 48km, 55km
5.2, 4.7, 8.2, 9.3, 6.4, 5.5
ASSIGNMENT
EXERCISE 22.4; NO 8, 9 AND 10. PAGE 240.
WEEK 10
SIMPLE EQUATIONS INVOLVING FRACTIONS
Examples: Solve the following equations
2×5 – 8- x10
P = 35-3p2
a-52 = 5 + a3
SOLUTION
The L.C.M of 5 and 10 is 10 multiply both sides by 10
10 x2x5 = 10 (3- x10) OR we cross multiply
2 x 2x = 8 – x 2×5 = 8-x10
5×5 = 85 10 x 2x = 5 x (8 –x)
x = 135 20x = 40 – 5x
25×25 = 4025
x = 135
(b) P = 35-3p2 (c) Multiply both sides by 6
2p = 35 – 3p 6 x (a-5)2 = 6 x 5 + 6 x a3
5p5 = 355 3 (a-5) = 30 + 2a
P = 7 3a – 15 = 30 + 2a
a = 45
Exercise: Solve the following equations
(a) x12 = 5 (b) x9 = 43 (c) y-54 = 8 (d) x-72 + x-215 = 0
Fraction with binomial denominator
Examples
a)25-x – 4 = 0 (b) 42y-3 – 35 = 0 (c) 3y + 2 = 52y-1
SOLUTION
25-x – 4 = 0 (b) 42y-3 – 35 = 0 (c) 3y + 2 = 52y-1
25-x – 4 42y-3 = 35 3 (2y – 1) = 5 (y + 2)
4 (5 – x) = 2 3 (2y – 3) = 20 6y – 3 = 5y + 10
20 – 4x = 2 6y – 9 = 20 Y = 13
4x = 18 6y6 = 296
x = 45 y = 456
EXERCISE
a) 1 x+ 13 + 8 = -3 (b) 112x – 5 = 4 (c) 128 + 3y = 1 (d) 81 + 5x = 37 + x
Simultaneous linear Equations
These are equations such as 4x – y = 8 and 3x – y = 6
Graphical Method
To solve simultaneous equations graphically
Make a table of values for both equations
Draw the graphs of both equations on the same axes
Find the coordinate where both graph interest. This values (x-y) are the solutions of both axes
Examples: Solve the following simultaneous equations graphically (a) x-2y=4 and 2x-y=5 (b) 2x-y=10 and 4x-2y=4r
SOLUTION
Y = -2 + 0.5 x y=2x- 5
x 0 2 4 x -1 0 5 x = 2
y -2 -1 0 y -7 -5 5 x = -1

y=10-2x y=-2+2x
x -1 0 2 x -2 0 2 x = 2
y 12 10 6 y -6 -2 2 x = -1

Exercise
2x-3y=8 (b) x+y=7
3x+ 2y=7 x-y=1
SUBSTITUTION METHOD
Examples (a) x-2y=4 , 2X – Y = 52 (b) 4x+y=8
3x+ 2y=7
SOLUTION
x-2y=4 …………1
2x- y=5 ………….2
Step 1: Rearrange one of the equations so that are variable is made the subject
That is from eqn I x=2y+4………..3
Step 2: Substitute into the second equation. That is substitute and solves the resulting equation.
x=2y+4 Into eqn…………2
2(2y+4) – y = 5 4y+8-y=2
3y3 = -63
Y = -1

Step 3: Substitute your answer into 3 to find the other variable
That is x=2y+4
x=2(-1)+4
x=-2+4
x=2
∴x, y=(2,-1)
4x+y=8 ,3x+ 2y=7
From eqn……..1 y=8-4x ……… 3
Substitute eqn3 into eqn 2 gives
3x-8-4x= 6
3x-8+4x= 6
7×7= 147
x = 2
From Eqn 3
y=8- 4x
=8- 4x
y =8- 4 (2)
y =0
∴x, y=(2, 0)
EXERCISE: 2x+y=10, 4x-2y=4 (b) y=5x+2, x+2y=15

ELIMINATION METHOD
Examples: (a) 5x+2y=10, 3x-2y=6 b 3a+3b=15, 4a+3b=20
When one of the unknown has equal coefficient
SOLUTION
5x+2y=10 …………. 1 3a+3b=15
+3x-2y=6…………… 2 4a+3b=20
8×8 = 168 – a = -5
x =2 a = 5
From eqn 1 , 5x+2y=10 from Eqn 1 ,3 5+ 3b=15
10+2y=10 15+8b=15
2y2 = 08
y=0 b=0
x, y=(2, 0) a, b=(5, 0)

Example 2 (When none of the unknown has equal coefficient)
Example: x-2y=4 (b) 3c+4d= -9
2x-y=5 4c+5d= -1
SOLUTION
To make the coefficients of x equal multiply eqn 1 by 2 and eqn 2 by 1
2 x 1 2x-4y=8……….. 3
1 x 2 2x-y=5…………. 4
-3y-3 = 33
y=-1

From Eqn 1 x=2y+4
x=2 -1+ 4
x= -2+4
x=2
3c+4d= -9 ——-1
4c+5d= -11 ——-2
4 x 1 12c+16d= -36 ——–3
3 x 2 12c+15d= -33 d= -3 …………….4
From Eqn (1) 3c+4d= -9
3c+4-3= -9
3c-12= -9
3c= -9+12
3c 3 = 33
c=1
c, d=(1, -3)
EXERCISE: 3x+2y=13, 2x+3y=12; 4x-4y=44 ,5x+2y=34 ; 6x+9y=31 , 4x+3y=6
WORD PROBLEMS
EXAMPLES
The sum of two numbers is 30 and their difference is 15. Find the two numbers
3 boxes and 2 packages weigh 1240g while 5 boxes and 7 packages weigh 2800g. What is the weight of a box and a package?
SOLUTION
x+y=30 (2) 3x+2y=1240 —–(1)
-x-y=15 5x+7y=2800 – – – -(2)
2y=15 5 x 1 15x+10y=6200
y=7.5 3 x 2 15x+21y=8400
x=30-7.5 11y 11 = 220011
x-22.5 y=200
From (1) 3x=1240-2200
3x=1240-400
3x=840
x=260
Exercise:
The sum of two numbers is 18 and their difference is 12. Find the two numbers
This shape is an equilateral triangle with dimension show finds its perimeter.

2x+5y-2 4x – y + 1

x+y+5
Andre has more money than Bob. If Andre gave Bob \$20, they would have the same amount. While if Bob gave Andre \$22, Andre would then have twice as much as Bob. How much does each one actually have?
In a two digit number. The units digit is thrice the tens digit. If 36 is added to the number, the digits interchange their place. Find the number.
If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions.
ASSIGNMENT:
EXERCISE 15.5; NO 2 – 5.PAGE 127

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WEEK 11
REVISION and EXAMINATION

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