SECOND TERM

SUBJECT: MATHEMATICS  CLASS: JSS 2

WEEK  TOPIC

1. Solving algebraic equations
2. Word problems on algebraic fraction
3. Linear inequalities
4. Linear inequalities in one variable/graphical representation
5. Graph of linear equation in two variables
6. Linear graphs in two variable
7. Plane figures and shapes
8. Scale drawing of lengths and distances
9. Solving Equation
10. Revision

WEEK ONE

TOPIC: SOLVING EQUATIONS

CONTENT

• Using directed number
• Unknown on both sides
• Equations with brackets
• Equations with fractions

Using Directed Numbers

Example

1.  Solve 25 – 9x = 2 and 12 = 9 – 3a

25 – 9x = 2

ecolebooks.com

Subtract 25 from both sides

25 – 25 – 9x = 2 – 25

-9x = -23.

Divide both sides by – 9

-9x = – 23

– 9 -9

X = 23/9 = 2 5/9

Check

when x = 23/9

LHS = 25 – 9x 23/9 = 25 – 23 = 2 RHS

2. 12 = 9 – 31

12 = 9 – 31

Subtract 9 – from both sides

12 – 9 = -9 – 9 – 3a

3 = -31

Divide both sides by -3

3/-3 = -31/-3

-1 = a

:. A = -1

Check when x = -1

LHS = 12 = 9 – 3 ( -1) = RHS

LHS = 12 = 9 + 3 + RHS

LHS = 12 = 12 = RHS

EVALUATION

Solve the following equations and check the Solutions.

1. 5 – 5n = 8

2. 12 + 5a = 23

3. 20 – 5t = 5

Unknown on both sides. If an equation has unknown terms on both sides of the equal sign, collect the unknown terms on one side and the number terms on the other side.

Worked Examples

Solve the equations.

1. 5x – 4 = 2x + 11

Solution

5x – 4 = 2x + 11

Subtract 2x from both sides of ( 1)

5x – 2x – 4 = 2x – 2x + 11

3x – 4 = 11 ………………. ( 2 )

Add 4 to both sides of …..(2)

3x – 4 + 4 = 11 + 4

3x = 15 ………………… ( 3)

3x/3 = 15/3

X = 5.

Check: when x = 5

LHS = 5 x 5 – 4 = 25 – 4 = 21

RHS = 2 x 5 + 11 = 10 + 11 = 21 = LHS

EVALUATION

Solve the following equations and check the Solution.

1. 18 – 5f = 2f + 4

2. 5a + 6 = 2a + 20

Equations with brackets

Always remove brackets before collecting terms.

Examples.

Solve 3 ( 3x – 1) = 4 ( x + 3 ) and

5 (x +11) + 2 (2x – 5) = 0

Solution

3 ( 3x -1) = 4 ( x + 3)…………..(1)

Remove brackets

9x – 3 = 4x + 12 …………….. ( 2)

Subtracts 4x from both sides and add 3 to both sides.

9x – 4x – 3 + 3 = 4x – 4x + 12 + 3

5x = 15 …………………………(3)

Divide both sides by 5

X = 3

Check : When x = 3

LHS = 3 ( 3 x 3 – 1 ) = 3 ( 9-1) = 3 x 8 = 24

LHS = 4 ( 3 + 3) = 4 x 6 =24 = LHS

( 2 ) .5 ( x + 11) + 2 ( 2x – 5 ) = 0

Remove brackets

5x + 55 + 4x – 10 = 0

Collect like terms

9x + 45 = 0

Subtract 45 from both sides

9x = -45

Divide both sides by 9

X = -5

Check : when x = -5

LHS = 5(-5 + 11) + 2 (2x (-5) – 5 )

= 5 x 6 + 2 ( -10 -5 )

=30 + 2x ( -15) = 30 – 30 = 0 = RHS

Evaluation

Solve the following equations and check the Solutions

1. 2 (x + 5) = 18 2. 5(a + 2) = 4 ( a -1) 3. 2(y – 2) + 3 ( y – 7 ) = 0

Equation with fractions

Always clear fractions before collecting terms. To clear fractions, multiply both sides of the equation by the LCM of the denomination of the fractions.

Example

Solve the equations

1. 4m – 2m = 4

5 3

2. 3x – 2 – 2x + 7 = 0

6 9

Solutions

4m – 2m = 4

5   3

The LCM of 5 and 3 is 15. Multiply every term by 15.

15 x 4m – 15 x 2m = 15 x 4

5 3

3 x 4m – 5 x 2m = 15 x 4

12m – 10m = 60

2m = 60

Divide both sides by 2.

m = 30

Check; When m = 30

LHS = 4 x 30 – 3 x 30 = 120/3 – 60/3

5 3

= 24 – 20 = 4 = RHS

2. 3x – 2 – 2x + 7 = 0

6 9

The LCM of 6 and 9 is 18

18 ( 3x – 2) – 18 ( 2x + 7 ) = 18 x 0

6 9

3(3x – 2) – 2 (2x + 7) = 0

Clear bracket

9x – 6 – 4x – 14 = 0

Collect terms

5x – 20 = 0

5x = 20

Divide both sides by 5

X = 4

Check : when x =4

LHS : 3 x 4 – 2 – 2 x 4 + 7

6 9

12 – 2 – 8 + 7

6 9

= 10 – 15 = 5 – 5 = 0 ( R.H.S)

6 9 3 3

EVALUATION

Solve the following equations.

1. x – 5 = x – 4

2 3

2. 3 ( 2a + 1 ) = 5 ( a + 5)

4 6

GENERAL EVALUATION

1.  solve 7a + 2 = 5 ( a – 4)

2.  Solve the equation 3 (2x + 1) = 4(x – 3 )

4 6

3.  Solve the equation x + 1 – 2 – x = 0

3 s

REVISION QUESTION

Solve the following equations

1. 3(x + 5) – 35 = -4(2x -6)

2. 4(2x + 5) -2 (x – 3 ) = 8(2x + 4 )

3. 2(3x – 2 ) = 4(x-2)

5 3

New General Mathematics, UBE Edition chapter 13 pgs. 110-113.

Essential Mathematics by AJS Oluwasanmi, chapter 19pgs 200-205

WEEKEND ASSIGNMENT

1.  Solve the equation 4b +24 = 0  A. -3 B. -4 C. -5 D. -6

2.  Solve 7c – 6 = c A. 1 B. 2 C. 3 D. 4

3.  Solve 15 = 3(x – 3)  A. 6 B. 7 C. 8 D. 4

4.  Solve 0 = 7 (x – 3 )  A. 1 B. 2 C. 3 D. 4

5.  Solve 3m + 8 = m  A. -1 B. -2 C. -3 D. -4

THEORY

Solve the following equations.

1.  3(2x-1) =39 – 2(y + 1)

2.   2a – 1 – a + 5 = 1

3 4 2

WEEK TWO

TOPIC: WORD PROBLEMS ON ALGEBRAIC FRACTIONS

CONTENTS

• From words to algebra
• Word problems with brackets
• Word problems with fraction.
• From words to Algebra

When solving word problems:

1.  Choose a letter for the unknown

2.  Write the information of the equation in algebraic form

3.  Make an equation

4.  Solve the equation

5.  Give the answer in written form.

6.  check the results against the information given in the question.

Example:

1. I think of a number. Multiply it by 5 and add 15. The result is 100. What is the number I thought of?

Let the number be n

I multiply n by 5:5n

I add 15: 5n + 15

The result is 100:5n + 15 = 100 ……….. ( i)

Subtract 15 from both sides

5n + 15 -15 = 100 – 15

5n = 85 ………………………………. 2

Divide both sides by 5

5n = 85

5 5

n = 17

Check: 17 x 5 = 85 + 15 = 100

EVALUATION

1. John thinks of a number, he doubles it. His result is 58. What number does he think of?

2. A number is multiplied by 6 and then 4 is added. The result is 34. Find the number.

Example

A rectangle is 8m long and its perimeter is 30m. Find the breadth of the rectangle.

Solution

Let the breadth of the rectangle be “b” meters.

Perimeter = 8 + b + 8 + b meters

= 16 + 2b = 30

Subtract 16 from both sides

2b = 30 -16 = 14

Divide both sides by 2

B = 14 = 7

2

The breadth of the rectangle is 7 meters

Check: 8m + 7m + 8 + 7m = 30m

EVALUATION

1.  A rectangle is 10m long and its perimeter is 26m. Find the breadth of the rectangle.

2.  A square has a perimeter of 32m. Find the length of one side of the square.

Word problems with brackets.

Example

I subtract 3 from a certain number, multiply the result by 5 and then add 9. If the final result is 54, find the original number.

Solution

Let the original number be x.

1 subtract 3, this gives x – 3.

I multiply by 5 this gives 5(x- 3)

I add 9 this gives 5 (x – 3) + 9

The result is 54.

So, 5 ( x – 3) + 9 = 54 ………..1

Clear brackets

5x – 15 + 9 = 54……………….2

Collect like terms

5x = 54 + 15 – 9 =60

x = 60 ÷ 5 = 12

The original number is 12

Evaluation

1.  I subtract 8 from a certain number. I then multiply the result by 3. The final answer is 21. Find the original number.

3.  I thought of a number, I multiplied it by 5. I then subtracted 19. Finally, I doubled the result, the result was 22. What number did I think of?

Worked Examples

Find two consecutive even number such that seven times the smaller number subtracted from nine times the greater number makes 46.

Solution

Note : 1, 2, 3, 4,5 ……………..are consecutive whole numbers

2,4,6,8,10 …………..are consecutive even numbers

1,3,4,7,9…………………….are consecutive odd numbers.

Let the number be x and (x + 2)

Multiply x by 7, gives 7x

Multiply (x + 2) by 9 gives 9 (x + 2)

The result is 46.

9(x+2) – 7x = 46 …………………….1

9x + 18 – 7x = 46 ……………………2

9x – 7x = 46 – 18 ……………………..3

2x = 28 …………………………………4

X = 28/2 = 14

If x = 14

(x+ 2) = 14 + 2 = 16

The numbers are 14 and 16.

EVALUATION

1.  Find the consecutive whole numbers such that five times the smaller number added to three times greater number makes 59. (Hint: let the numbers be x and (x +1).

2.  Find two consecutive odd numbers such that six times the smaller added to four times the greater comes to 138. (Hint: let the number is x and x + 2)

Word problems with Fractions

Examples: I add 55 to a certain number and then divide the sum by 3. The result is four times the first number, find the number.

Let the number be n

I add 55 to n this gives n + 55

I divide the sum by 3: this gives
n _ + 55

3

The result is 4n

So, n + 55 = 4n ………………..1

3

Multiply both sides by 3

3 ( n + 55) = 3 x 4n …………………..2

3

n + 55 = 12 n ………………………………3

Collect terms

55 = 12n – n

55 = 11n

So, n = 5

EVALUATION

I think of a number. I double it. I divide the result by 5. My answer is 6. What number did I think of?

New General Mathematics, UBE Edition chapter 13 pgs 114-118

Essential Mathematics by AJS Oluwasanmi, chapter 19pgs 205-207

GENERAL EVALUATION

1.  A woman is 7 times as old as her daughter in 2 years’ time, she will be 5 times as old as her daughter. How old are they now?

2.  If 60 is added to a number, and the sum is divided by 3, the result is 7 times the same number. Find the number.

REVISION QUESTION

1.  I think of a number. I add 7 to it and then divide the sum by 4.The result is 5.What is the number?

2.  A man spent two third of his monthly wages on his car and one quarter on food. If he spent #5500 altogether, how much does earn in a month?

3.  If 18 is added to a number to a number, and the sum is divided by 2,the result is 5 times the number. Find the number.

New General Mathematics pg. 209-211

WEEKEND ASSIGNMENT

1. What is the perimeter of an equilateral triangles of side 2b meters? A. 4b B. 5b  C. 6b D. 8b

2. How many altogether if a number is 3 less than a? A. a -3 B. – a + 3 C. 3 – a D. 3 + a

3.  How many altogether if a number is doubled and then 5 is taken away?  A. 5 – 2y B. 2y – 5 C. 5y – 2 D. 2 – 5y

4.  What is the perimeter of a rectangle of breadth b meters and length three times as long?  A. 4b meters  B. 5bmeters   C. 8b meters  D. 10bmeters

5.  What is the perimeter of a rectangle of length y meters and breath 4m less than the length? A. 4(y -2) meters  B. 2 (y- 4)meters  C. 4 (2-y)  D. 2 (4- y)meter

THEORY

1.  A rectangle is such that its breadth is 2m less than its length. Find the length of the rectangle if the perimeter is 8m.

2.  I am thinking of a number. I take away 5, the result is 14. What number did it think of?

WEEK THREE

TOPIC: BASIC SIGNS AND PROPERTIES OF LINEAR INEQUALITIES.

CONTENT

• Greater than and less than
• Properties of linear inequalities
• Not greater than and not less than
• Graphs of inequalities.

Greater than and less than

5 + 3 = 8 means equal to

X = 0 means x is not equal to 0

But 5 + 5 > 8, where 7 means greater than

Similarly, 3 x 2 < 8 where < means less than, > and < are inequality symbols.

Worked Examples

1.  Write the inequality symbols for the following

a. b is greater than 15

b. 9 x 3 20

2

2. (- 5)2 indicate < or > in the box

Solution

1a. b > 15

b. 9 x 3  20.

2.  (-2)2 = – 2 x 2 = 4

( -5)2 = -5 x -5 = 25

( -2) 2 ( -5)2

Evaluation

Write the inequality symbol for each of the following:

a.  -3 less than + 3

b.  y is less than – 2

c.  4 is greater than a

d.  (-5)2   (22) 2

e.  13  3 x 49

New General Mathematics UBE Edition, Chapter 22, pgs 209-211

Essential Mathematics by A. JS. Oluwasanmi Chapter 23 pgs 237-239

Properties of linear inequality. The symbol > and < can be used to change word statements into algebraic statements.

Worked Examples

1.  The distance between two villages is over 18km. write this as an inequality statement.

2.  I have x naira, I spend N20, the amount I have left is less than N5. Write inequality in x.

3.  The area of a square is less than 25cm2. What can be said about

a. the length of its sides b. its perimeters

Solution

1.  x > 18

2.  I spent N20 out of x naira

Amount left = N(x – 20)

Less than N5 N(x – 20 ) < N5 i.e. X - 20 < 5

3.  let the length be a , then a2 < 25

a < 25, a < 5

b.  perimeter = 4a since a = 4

then 4a < 4 x 5

4a < 20

a < 5cm

EVALUATION

1.  Write the inequality symbols places of the statements below:

a.  The car use more than 28 liters of petrol.

b.  The cost of the stamp was less than N25.

c.  The students got over 60% in the exam.

2.  A boy saved over N500. His father gave him N200, the boy now had altogether. Write an inequality in y.

3.  The perimeter of a square is less than 28cm what can be said about:

a. its length b. its area

New General Mathematics UBE Edition, Chapter 22, pgs 213-215

Essential Mathematics by A. JS. Oluwasanmi Chapter 23 pgs 237-239

NOT GREATER THAN AND NOT LESS THAN

When a particular variable say x does not exceeds a particular value, it means x is not greater than the given value.

For Example x < 50, it means x < 50 or x = 50; where < means less than or equal to. But when the variable x exceeds a given value for Example x > 50 or x = 50m it means less than or equal to. But when the variable x exceeds a given value for Example x > 50 or x = 50m it means x > 50 where > means greater than or equal to .

Worked Examples

1.   Note books cost N60 each Deborah has d naira it is not enough to buy a note book. Taiwo has t naira. He is able to buy a note book. What can be said about the value of d and it?

Solution

Deborah = d naira

Deborah amounts is less than N60 d < 60

Taiwo = t naira

In conclusion, t > d and d < t.

EVALUATION

1.  Write an inequality in terms of the unknown.

a. The number of goals n was five or more.

b. the temperature tc, was not greater than 38oC.

c. the number of students n was less than 36

2.  The pass mark in a test was 27, one person got x marks and failed. Another got y marks and passed, what can be said about x and y?

GRAPH OF INEQUALITIES

Consider the number line below:

-4 -3 -2 -1 0 +1 +2 +3 + 4

The inequality graph indicates an arrow head on a number line which show whether the values of x are greater than or less than a given value.

For Example:

(a) if x < 2, the inequality graph should show this

x < 2

2 -1 0 1 2 3

(b) if x >, -1, the inequality graph should show this

-2 -1 0 1 2 3

Note: included value ( ¸or > )
makes use of a close circle, ( * ) while not included value ( < or > ) makes use of an open circle ( o)

EVALUATION

1. Write down the inequality shown in the following graphs.

a).

b). x

8

c).

-3 -2 -1

GENERAL EVALUATION

1.  Say whether each of the following statement is true or false?

a. -20 is greater than – 5.

b. -3 x ( -2) > – 3 – 6

c. -18 <3 – 20

2.  Illustrate the following on a number line

a. x > 0 b. x ≤ – 1 c. x ≥ -2

REVISION QUESTION

1.  If x is a positive integer positive integer, for what range of values of x is 8 + 2x < 14.Draw number line to show your answer.

2.  If x is an integer, find the first three possible values of x in the inequality 6x – 5(x-2) ≤ 4 (2x-1)

3.  If x is a positive integer and 2x + 3 > -30 + 6p (a) solve for x (b) Find all the possible values of x and show them on a number line.

New General Mathematics UBE Edition, Chapter 22, pgs 213-215

Essential Mathematics by A. JS. Oluwasanmi Chapter 23 pgs 237-239

WEEKEND ASSIGNMENT

1.  The inequality symbol for -1 is greater than -5 is (i) – 1 < 5 ( b) -5 > -1   (c ) -1 > -5

2.  The time for a journey t mins was over 2hrs the inequality statement is ………………

A. t > 2hts B. 2hts > t C. t > 2hrs

3.  The graph below represents thus

-2 -1 0 1 2 ½ 3 4 5

A. n > 2 ½ B. x < 2 ½ C x < -2 ½

4. ( -3)2  22

The inequality symbol in the box is

A. < B. > C. ≥

5.  The cost of a stamp #x was not more than N20

A. x > N20 B. x = N20 C. x < N20

THEORY

1.  A square has an area of more than 36cm2. What can be said about the length of one of its sides.

b. the perimeter.

2.  Sketch the following inequality on a graph.

a. x ≤ – b b. x ≥ 3.

TOPIC: SOLUTION OF INEQUALITIES IN ONE VARIABLE

CONTENT

• Solution
of inequalities
• Multiplication and division of negative numbers
• Word problem involving inequalities

Solution of inequalities

Inequalities are solved in the same way as simple equation

For Example x = 23 is equation

While x < 3 is an inequalities.

Worked Examples

1.  Solve the inequality and show the Solution on a graph.

a.  x + 4 < 6

b.  6 < 2x - 1

c.  3x – 3 > 7

Solution

a.  x + 4 < 6

x < 6 - 4

x < 2

x

-2 -1 0 1 2 3

b. 6 < 2x - 14

7 < 2x

3 ½ < x

X > 3 ½

x > 3 ½

3 ½

c. 3x – 3 > 7

3x > 7 + 3

3x > 10

x > 10/3

x > 3

x > 3 ½

-2 -1 0 1 2 3 3 1/3 4 5 6 7

EVALUATION

1.  Solve the following inequalities and show the Solution on a graph:

a. 5x -2 > 8

b. 4x – 2 > 19

c. 3 < 3x X 5

d. x + 4 > 10

MULTIPLICATION AND DIVISION BY NEGATIVE NUMBER

When solving an inequality involving negative numbers, the inequality sign must be reversed. For Example if

-2x > 10 is true then on division throughout by -2.

X <- 5 will be true

Worked Examples

1. Solve 5 – x > 3

2. Solve 19 > 4 -5x

3. Solve 3 – 2x < 8

Solution

5 – x > 3

-x > 3 – 5

-x > -2

Dividing through by ( -1)

-x < – 2

-1 -1

X < 2

2. 19 > 4 – 5x

19 – 4 > – 5x

15 > – 5x

Dividing through by -5

15 < – 5x

-5 – 5

-3 < -5x

x > – 3

3. 3 – 2x < 8

-2x < 8 - 3

-2x X 5

-2 -2

X > -5/2

X > -2 ½

Evaluation

Solve the following inequalities

a. -2x + 5 > 16

b. 10 – 3x < - 11

c. 2r > 6r + 6

d. 9 < 3 - 4t

on a number line

New General Mathematics UBE Edition, Chapter 22, Nos 1- 2pgs 213-215

Essential Mathematics by A. JS. Oluwasanmi Chapter 23 pgs 40 – 243

WORD PROBLEMS INVOLVING INEQUALITIES

Worked Example

1.  A triangle has sides xcm, ( x + 4) cm and 11cm, where x is a whole number of cm, if the perimeter of the triangle is less than 32cm. find the possible values of x.

Solution

Perimeter of triangle = x + ( x + 4) + 11

But perimeter < 32cm

x+ 4 +11< 32

2x < 32 - 15

2x < 17

x< 17/2

x < 8 ½

Also in any triangle, the sum of the length of any two sides must be greater than the length of the third side.

Thus,

X + (x + 4) > 11

2x + 4 > 11

2x > 7

x > 7/2

x > 3 ½

Thus

x< 8 ½ and x > 3 ½ . but x is a whole number of cm therefore, the possible values of x are 4,5,6,7 or 8.

GENERAL EVALUATION

1.  If 9 is added to a number x, the result is greater than 17. Find the values of x

2.  Three times a certain number is not greater than 54. Find the range of values of the number .

REVISION QUESTION

1.  If 8 is subtracted from a number, the result is at most 15. Find the range of values of the number

2.  If x is subtracted from 5, and the result is greater than 15.Find the range of values of x.

WEEKEND ASSIGNMENT

1.  Find which symbol > or < goes in the box to make the statement 9 + 8 10 true.

A. <  B. >  C. none of the above

2.  Write the inequality of the statement ‘The cost of meal Nx was over N5. A. x > N5 B. x > N5 C. x < N 5

3.  Solve 5x – 7 > 9 A. x > 32.5  B. x > 3 1/5  C. x , 5 2/3

4.  Solve x -2 < 3 A. x < 5 B. x > 5 C. x > 5

5.  If 7.3 is subtracted from y, the result is less than 3.4. find the value of y. A. y < 10.7 B. y > 10.7 C. y < 10.7

THEORY

1.   A rectangle is 8cm long and bcm branch find the value of b if the perimeter of the rectangle is not greater than 50cm and not less than 18cm

2.  The sides of a rectangle are xcm (x+ 3cm) and 10cm. if xcm is a whole number, if the perimeter is less than 30cm, find the possible value of x.

Hint: The sum of the length of any two sides of a triangle must be greater than the length of the third side.

WEEK FIVE

TOPIC: THE CARTESIAN PLANE AND COORDINATES

CONTENT

• The position of points on a line
• Cartesian plane/coordinates
• Plotting points

The position of points on a line

The number line is a picture or graph of relative position of positive and negative numbers.

Worked Examples

1.  Find the values of the following variables in the number line below:

D B C A

-4 -3 -2 -1 0 +1 +2 + 3 + 4

Solution

i. A is 3 units to right of zero = A (3)

ii. B is 1 unit to left of zero = B (-1)

iii. C is 1 ½ units to the right of zero = 1 ½

iv. D is 2 units to the left of zero = (-2)

2. Draw a number line from -5 to 5 in the number line mark.

A. 2 B. (-4) C. (-2 ½) and D. (1)

Solution

D B C A

-4 -3 -2 -1 0 +1 +2 + 3 + 4

Evaluation

1.  Give the position of 1A ii. B iii C iv. D v. E in the number line below:

2.  Find the values of the following variables in the number line below:

D B C A

-4 -3 -2 -1 0 +1 +2 + 3 + 4

3.  Draw a line from -8 to 5 and indicate the following points x (-2), y(3) W(4) M (5) x (-6) and T(-8)

New General Mathematics UBE Edition, Chapter 12, Nos 1- 2pgs102-107

Essential Mathematics by A. JS. Oluwasanmi Chapter15 pgs 152 -155

Cartesian Plane/Coordinate

Cartesian graphs give a picture of the relative position of point on a plane. The Cartesian plane is a plane surface with two axes drawn on it. The axes are at right angle to each other . they are the horizontal ( x -axis) and the vertical (y-axis) crossing at the origin (the zero point for both axes).

Y – axis(positive )

x-axis (negative)   x- axis(positive)

y axis (negative)

COORDINATES

Every points on the Cartesian plane has a unique ordered pair of coordinates in the form ( a,b) where a is x – coordinates (distance of the point along the x – axis) and b is the y-coordinates ( the distance of the point along the y-axis).

Worked Examples

1.  Write down the coordinates of the point A, B, C, D, E and F in the figure below:

C  6-  A

5-

4-

3-

2-

1-

-5 -4 -3 -2 -1  1 2 3 4 5

-1

-2

-3

D   -4   B

2. Write down the coordinates of the vertices of triangle ABC and parrellogram PQRS in the figure below

Solution

1.  The coordinates of the points are

i. A (3, 5)

ii. B 9 2, -4)

iii. C (2, -3)

iv. D (-3,-4)

v. E (4, 0)

vi. F (1, -1)

2.  The vertices of triangle ABC are A. (-2,3) B. ( 2,1) C ( 0, -1)

The vertices of parallelogram PQRS are P (1,2)  Q ( 1,2)  R(-2,-2)  S (-3,0)

EVALUATION

1.  What are the coordinates of the points A B C, D, E, F, G, H in the figure below

2.  In the figure below name the points which have the following coordinates

a. (9,5) b. (5, – 8 ) c. (-15,10 ) d (-5, 8 ) e (12, 0 )

GENERAL EVALUATION

Show the position of the following points on the graph sheet. Using 1cm to 1 unit on both axes.

1.  P (2, 4) Q (-3, -4) R (2, -4) S (-12,9) T (8, -15 )

New General Mathematics UBE Edition, Chapter 12, Nos 1- 2pgs105-107

Essential Mathematics by A. JS. Oluwasanmi Chapter15 pgs 237-239

Plotting Point

Plotting points means to draw its positions on Cartesian plane. The easiest way to plot a graph is as follows:

Start at the origin

ii. Move along the x axis by amount and in a direction given by the x-coordinates of the point.

iii. move up or down parallel to the y-axis by an amount in a direction given by the y- coordinates.

Worked Example

1.  Plot the point (-1, 2) and ( 2,6-1.8) on a Cartesian plane .

1.  The vertices of quadrilateral PQRS have coordinates P (-3,18) Q ( 15,14) R (11,-4) and S ( -7,0) A and B are the points A(-3,-7) and B ( 3,0).

a. using a scale 2cm, represent 10 units on both axes plot points P.Q, R .S.A and B.

b. join the vertices of quadrilateral PQRS. What kind of quadrilateral is it?

c. Find the coordinates of the points where diagonal of PQRS cross.

Solution

a

b. PQRS is a square

c. the diagonals of PQRS cross at x(4,7)

d. B. A and Q lie on the same straight line

REVISION QUESTION

1. Draw the origin O, near the middle of a sheet of graph paper use a scale of 2cm represent 5 units on both sides. Plot the following points. Join each points to the next in order they are given start (-10,-5 ) (-5,10) (0,15) (5,17) (3,14) (3,12) (15,6) (14,3) (11,3),1 (13,2) (5,3) (6,-6) >

What pictures does your graph shows.

WEEKEND ASSIGNMENT

1. Use the diagram below to answer questions 1-3

A  C  B  D  E

-4 -3 -2 -1 0 1 2 3

2.  The value of a is A. 3 ½ B. -4 C. 3 D. 1 ½

3.  The value of D is A. -4 B. 1 ½ C. -3 ½ D. 3

4.  The value of C is A. -4 B. 1 ½ C. 3 ½ D. 3

5.  if the value of a point in the x axes are 2 and -4 respectively, it can be written in a

A. ( 2,-4) B. ( -4,2)  C. ( y ,2) D. ( x, 2)

5.  Point A in the diagram is A. (-5, 5) B.(5,5) C(-5,-5)  D. (0,5)

THEORY

1.  Draw the graph of the following coordinates and join all the points.

T(-2,1) U (-3,1) V(0,4) W(3,1) X (2,1) Y(2,-1) Z(-2,-1)

2.  Draw the origin () near the middle of a sheet of graph paper use scale of 2cm represents 1 unit on both

axes. Plot the following point, and then join each point to the next in alphabetical order.

A(0,1) B(1,2)C (1,1) D(2,1) E(1,0)F (2,-1)G (1,-1)H (1,-2) I(0,-1) J(-1,-2) K(-1,-1)L (-2,-1) M(-1,01) N(-2,1)P (-1,1) Q(-1,2) finally join Q to A.

WEEK SIX

TOPIC: LINEAR GRAPHS IN TWO VARIABLES, USING GRAPH TO SOLVE REAL LIFE SITUATIONS

CONTENT

• Continuous graphs
• Graph of real life situations
• Choosing scales.

Continuous Graph: Graphs are used to show the relationship between two quantities. A continuous graph is in the form of a continuous line and shows the relationship between the two quantities.

Worked Examples

1a.  A student walked along a road at a speed of 120m per minute.

Make a table of values showing how far the students has walked after 0, 1, 2,3,4,5, minutes

b.  using the scale of 1cm to 1min on the horizontal axis and 1cm to 100m on the vertical axis, draw a graph of the information .

2.  use the graph to find:

i. how far the students has walked after 2.6min.

ii. how long it takes the students to walk 500m

Solution

 Time(min) 0 1 2 3 4 5 Distance(m) 0 120 240 360 480 600

y

600

500

400

300

200

100

0  1 2 3 4 5 6  x

Time (mins)

c.  i. The student has walked 310cm after 2.6min.

ii. The student takes about 4.5 in walk 500m

EVALUATION

1.  A girl walks along a road at a speed of 100m per minute

A. Copy and complete the table

 Time(min) 0 1 2 3 4 5 6 Distance(m) 0 100 200

B. Using a scale of 2cm to 1min on the horizontal axis and

New General Mathematics Bk. 2 chapt.4,pg 119-123

Distance-time graphs

We use distance-and-time graphs to show journeys. It is always very important that you read all the information shown on these type of graphs.

A graph showing one vehicle’s journey

If we look at the graph shown below, you can see that the time in hours is along the horizontal, and the distance in miles is on the vertical axis. This graph represents a journey that Jan took, in travelling to Glasgow and back, from Aberdeen.

Important points to note are:

It took half an hour to travel a distance of miles

Between 1pm and 3pm there was no distance travelled. This means that the car had stopped.

The journey back, after 3pm, took one hour.

A graph showing two different journeys in the same direction

The next graph shows two different journeys. You can see that there is a difference with the steepness of the lines drawn. Remember that, the steeper the line, the faster the average speed. We can calculate the average speeds, by reading distances from the graph, and dividing by the time taken.

Line A: How long does journey A last, and what distance is travelled?

The journey takes 2 hours, and the distance travelled is 15km.

Line B : How long does journey B last, and what distance is travelled?

The journey takes 1 hour, and the distance travelled is also 15 km.

This means that the average speeds are:

A: 15/2 = 7.5 km per hour

B: 15/1 km per hour

You will also notice from the graph that the two lines cross. This means that, if the two vehicles were travelling along the same route, they would have met at that point, which was just after 10am. The vehicle on journey B overtook vehicle

A graph showing two journeys in the opposite direction

A different pair of journeys is shown below. It is important to note that one journey begins at a distance of , and the other at a distance of miles, from Perth. In fact, what is happening is that one journey is travelling away from, and the other is travelling towards Perth. Again, the two journeys meet. This time it is miles from Perth.

You will also see that the two journeys contain stops.

If we were to calculate the average speeds for each total journey we would have to include this time as well.

A graph showing a journey (or journeys) should have time on the horizontal axis, and distance from somewhere on the vertical axis.

A line moving up, as it goes from left to right, shows a journey moving away from a place, and a line moving down, as it goes from left to right, represents a journey towards a place.

A horizontal line is a break or rest.

Two lines, sloping the same way, cut: then an overtaking has taken place.

Two lines, sloping opposite ways, cut: a meeting has taken place.

Speed-time graphs

speed-time graph, velocity-time graph, shows how the speed of an object varies with time during a journey.

There are two very important things to remember about velocity – time graphs.

The distance traveled is the area under the graph.

The gradient or slope of the graph is equal to the acceleration. If the gradient is negative, then there is a deceleration. We may use the equations(1) or some rearrangement of this equation.

WEEKEND ASSESSMENT

Example. A car starts on a journey. It accelerates for 10 seconds at It then travels at a constant speed for 50 seconds before coming to rest in a further 4 seconds.

a. Sketch a velocity – time graph.

b. Find the total distance traveled.

c. Find the deceleration when the car is coming to a stop at the end.

d. Find the average speed.

WEEK SEVEN

Topic: PLANE FIGURES OR SHAPES

Any four-sided shape is a Quadrilateral.

But the sides have to be straight, and it has to be 2-dimensional.

Properties

Four sides (edges)

Four vertices (corners)

The interior angles add up to 360 degrees:

Try drawing a quadrilateral, and measure the angles. They should add to 360°

There are special types of quadrilateral:

Some types are also included in the definition of other types! For example a squarerhombus andrectangle are also parallelograms.

Let us look at each type in turn:

The Rectangle

means “right angle” | and || show equal side.

A rectangle is a four-sided shape where every angle is a right angle (90°).

Also opposite sides are parallel and of equal length.

The Rhombus

A rhombus is a four-sided shape where all sides have equal length.

Also opposite sides are parallel and opposite angles are equal.

Another interesting thing is that the diagonals (dashed lines in second figure) meet in the middle at a right angle. In other words they “bisect” (cut in half) each other at right angles.

A rhombus is sometimes called a rhombus or a diamond.

The Square

means “right angle” | shows equal side.

A square has equal sides and every angle is a right angle (90°)

Also opposite sides are parallel.

A square also fits the definition of a rectangle (all angles are 90°), and a rhombus (all sides are equal length).

The Parallelogram

A parallelogram has opposite sides parallel and equal in length. Also opposite angles are equal (angles “a” are the same, and angles “b” are the same).

NOTE: Squares, Rectangles and Rhombuses are all Parallelograms!

The Trapezoid

A trapezoid (called a trapezium in the UK) has a pair of opposite sides parallel.

It is called an Isosceles trapezoid if the sides that aren’t parallel are equal in length and both angles coming from a parallel side are equal, as shown.

And a trapezium (UK: trapezoid) is a quadrilateral with NO parallel sides:

The Kite

Hey, it looks like a kite. It has two pairs of sides. Each pair is made up of adjacent sides that are equal in length. The angles are equal where the pairs meet. Diagonals (dashed lines) meet at a right angle, and one of the diagonal bisects (cuts equally in half) the other.

Note: The only regular quadrilateral is a square. So all other quadrilaterals are irregular.

The definition shows that squares are special rectangles. Thus all squares are also rectangles. However, some rectangles are not squares.

Parallelograms have two pair sides parallel. This is also true of squares, rectangles and rhombuses. Thus squares, rectangles and rhombuses are special parallelograms. However, there are many parallelograms which are not squares, rectangle or rhombuses.

Weekend Assessment

1. Name four shapes which are special examples of trapezium

2. Name three shapes which are special examples of parallelograms

3. Name two shapes which are special examples of kites.

WEEK EIGHT

Topic: SCALE DRAWING

A drawing that shows a real object with accurate sizes except they have all been reduced or enlarged by a certain amount (called the scale).

The scale is shown as the length in the drawing, then a colon (“:”), then the matching length on the real thing.

Example: this drawing has a scale of “1:10”, so anything drawn with the size of “1” would have a size of “10” in the real world, so a measurement of 150mm on the drawing would be 1500mm on the real horse.

Since it is not always possible to draw on paper the actual size of real-life objects such as the real size of a car, an airplane, we need scale drawings to represent the size like the one you see below of a van.

In real-life, the length of this van may measure 240 inches. However, the length of a copy or print paper that you could use to draw this van is a little bit less than 12 inches

Since 240/12 = 20, you will need about 20 sheets of copy paper to draw the length of the actual size of the van

In order to use just one sheet, you could then use 1 inch on your drawing to represent 20 inches on the real-life object

You can write this situation as 1:20 or 1/20 or 1 to 20
Notice that the first number always refers to the length of the drawing on paper and the second number refers to the length of real-life object.

Example

Suppose a problem tells you that the length of a vehicle is drawn to scale. The scale of the drawing is 1:20 If the length of the drawing of the vehicle on paper is 12 inches, how long is the vehicle in real life?
Set up a proportion that will look like this:

Length of drawing/Real length = 1/20

Do a cross product by multiplying the numerator of one fraction by the denominator of the other fraction
We get :
Length of drawing × 20 = Real length × 1
Since length of drawing = 12, we get:
12 × 20 = Real length × 1
240 inches = Real length

Example
The scale drawing of this tree is 1:500
If the height of the tree on paper is 20 inches, what is the height of the tree in real life?

Set up a proportion like this:

Height of drawing/Real height = 1/500

Do a cross product by multiplying the numerator of one fraction by the denominator of the other fraction
We get :
Height of drawing × 500 = Real height × 1
Since height of drawing = 20, we get:
20 × 500 = Real length × 1
10000 inches = Real height

Scale Drawings

A map cannot be of the same size as the area it represents.  So, the measurements are scaled down to make the map of a size that can be conveniently used by users such as motorists, cyclists and bushwalkers.  A scale drawing of a building (or bridge) has the same shape as the real building (or bridge) that it represents but a different size.  Builders use scaled drawings to make buildings and bridges.

A ratio is used in scale drawings of maps and buildings.  That is:

The scale of a drawing = Drawing length: Actual length
Likewise, we have:

Map scale=Map distance: Actual distance
A scale is usually expressed in one of two ways:

Using units as in 1 cm to 1 km

Without explicitly mentioning units as in 1 : 100 000.

Note

A scale of 1 : 100 000 means that the real distance is 100 000 times the length of 1 unit on the map or drawing.

Example

Write the scale 1 cm to 1 m in ratio form.

Solution

1 cm to 1m = 1 cm : 1m

= 1 cm : 100 cm

= 1 : 100

Example

Simplify the scale 5 mm : 1 m.

Solution

5 mm : 1 m = 5 mm : 100 cm

= 5 mm : 1000 mm

= 5 : 1000

= 1 : 200

Example

Simplify the scale 5 cm : 2 km.

Solution

5 cm : 2 km = 5 cm : 2000 m

= 5 cm : 200 000 cm

= 5 : 200 000

= 1 : 40 000

Calculating the Actual Distance using the Scale

If the scale is 1 : x, then multiply the map distance by x to calculate the actual distance.
Example

A particular map shows a scale of 1 : 5000.  What is the actual distance if the map distance is 8 cm?

Solution

Scale = 1 : 5000 = 1 cm : 5000 cm

Map distance : Actual distance = 1 : 5000

Map distance = 8 cm

Let the actual distance be a cm.

8 : a = 1: 5000 {Units are in cm}

8/a = 1/5000 {Invert the fractions}

a/8 = 5000/1 {Multiply by 8}

8 X a/8 = 8 X 5000

a = 40 000

Actual distance = 40 000 cm

= 40 000/100 m

= 400 m

Alternative Way

Map distance = 8 cm

Scale = 1 : 5000 = 1 cm : 5000 cm

Map distance : Actual distance = 1: 5000

= 1 X 8 : 5000 X 8

= 8 : 40 000

Actual distance = 40 000 cm

=40 000/100 m

=400m

Calculating the Scaled Distance using the Actual Distance

If the scale is 1 : x, then divide the actual distance by x to calculate the map distance.

WEEKEND ASSESSMENT

A particular map shows a scale of 1 cm : 5 km.  What would the map distance (in cm) be if the actual distance is 14 km?

WEEK NINE

Topic: SOLVING EQUATIONS

Solving Equations (1)

2x – 9 = 15 is an equation in x. x is the unknown in the equation. 2x – 9 is on the left-hand side (LHS) of the equals sign and 15 is on the right-hand side (RHS) of the equal sign.

To solve an equation means to find the value of the unknown that makes the equation true.

The balance method (revision)

Think of the two sides of an equation as forming a balance. Keep the balance by doing the same operation to both sides of the equation.

Example

Solve 3x = 12

3x = 12

Divide both the LHS and RHS by 3, the coefficient of the unknown. This keeps the balance of the equation.

3x/3 = 12/3

x = 4

x = 4 is the solution of the equation 3x = 12

check: when x = 4, LHS = 3 X 4 = 12 = RHS

Example

Solve 2x – 9 = 15.

2x – 9 = 15

a. The LHS contains the unknown. Add 9 to 2x – 9. This leaves 2x. 9 must also be added to the RHS to keep the balance of the equation.

2x – 9 = 15

Add 9 to both sides (+9 is the additive inverse of -9)

Simplify 2x = 24

b. The equation is now simpler. Divide the LHS by 2 to leave x. The RHS must also be divided by 2to keep the balance of the equation.

2x = 24

Divide both sides by 2.

2x/2 = 24/2

x = 12

x = 12 is the solution of the equation 2x – 9 = 15.

Check: when x = 12, LHS = 2 x 12 – 9 = 24 – 9 = 15 RHS.

Exercise

Use the balance method to solve the following:

a. 3x – 8 = 10

b. 4x – 1 = 1

c. 27 =10x – 3

Solving Equations (2)

Using directed numbers

It is possible to use operations with directed numbers when solving equations.

Example

Solve 25 – 9x = 2

25 – 9x = 2

Subtract 25 from both sides.

25 – 25 – 9 = 2 – 25

– 9x = – 23

Divide both sides by -9.

– 9x/-9 = -23/-9

x = 23/9 = 2 5/9

check: when x = 23/9,

LHS = 25 – 9 X 23/9 = 25 – 23 = 2 = RHS

Unknowns on both sides

If an equation has unknown terms on both sides of the equal sign, collect the unknown terms on one side and the number terms on the side.

Example

Solve 5x – 4 = 2x + 11

5x – 4 = 2x + 11 (1)

Subtract 2x from both sides of (1).

5z – 2x – 4 = 2x – 2x + 11

3x – 4 = 11   (2)

Add 4 to both sides of (2).

3x – 4 + 4 = 11 + 4

3x = 15

Divide both sides of (3) by 3. (3)

x = 5

Check: x = 5,

LHS = 5 x 5 -4 25 – 4 =21

RHS = 2 x 5 + 11 = 10 + 11 = 21 = LHS

Note that equations (1), (2), and (3) are still equivalent.

Exercise

a. 13 – 6 = 1

b. 4b + 24 = 0

c. 12 + 5a = 23

Equations with brackets

Always remove brackets before collecting terms.

Solve 3(3x – 1) = 4(x + 3)

3(3x – 1) = 4(x + 3) (1)

Remove brackets.

9x – 3 = 4x + 12               (2)

Subtarct 4x from both sides and add 3 to both sides.

9x – 4x -3 + 3 = 4x – 4x + 12 + 3

5x = 15                           (3)

Divides both sides by 5.

x = 3

Check: when x = 3,

LHS = 3(3 x 3 -1) = 3(9 – 1) = 3 X 8 = 24

RHS = 4(3 + 3) = 4 X 6 = 24 = LHS

Example

Solve 5(x + 11) + 2(2x – 5) = 0.

5(x + 11) + 2(2x – 5) = 0.   (1)

5(x + 11) + 2(2x – 5) = 0.

Remove brackets.

5x + 55 + 4x – 10 = 0 (2)

Collect like terms.

9x + 5 = 0 (3)

Subtract 45 from both sides.

9x = -45  (4)

Divide both sides by 9.

x = -5

Check: when x = -5

LHS = 5(-5 + 11) + 2(2 X (-5) -5)

= 5 X 6 + 2(-10 -5)

= 30 + 2 X (-15) = 30 – 30 = 0 = RHS

EVALUATION

a. 5(x – 4) – 4(x + 1) = 0

b. 3(2x + 3) – 7(x + 2) = 0

c. 2(x + 5) = 18

Equations with fractions

Always clear fractions before collecting terms. To clear fractions multiply both sides of the equation by the LCM of the denominators of the fractions.

Example

Solve the equation 4m/5 – 2m/3 = 4.

4m/5 – 2m/3 = 4

The LCM of 5 and 3 is 15.

Multiply both sides of the equations by 15, i.e. multiply every term by 15.

15 X (4m/5) – 15 X (2m/3) = 15 X 4

3 X 4m – 5 X 2m = 15 X 4

12 – 10m = 60

2m = 60

Divide both sides by 2.

m =30

check: when m = 30,

LHS = 4 X 30/5 – 3 X 30/3 = 120/5 – 60/3

= 24 – 20 = 4 = LHS

Example

Solve the equation 3x – 2/6 – 2x + 7/9 = 0.

The LCM of 6 and 9 is `8.

18(3x – 2)/6 – 18(2x + 7)/9 = 18 X 0

3(3x – 2) – 2(2x + 7) = 0

Clear brackets.

9x – 6 – 4x – 14 = 0

Collect like terms.

5x – 20 = 0

5x = 20

Divide both sides by 5

x = 4

Check: when x = 4

LHS = 3 X 4 – 2/6 – 2 X 4 + 7/9

= 12 – 2/6 – 8 + 7/9

= 10/6 – 15/9 = 5/3 -5/3 = 0 = RHS

Exercise

a. x/3 = 5

b. x/5 = ½

c. 4/3 = 2z/15

d. x – 2/3 = 4

Word Problems

We can use equations to solve word problems, i.e. problems using everyday language instead of just numbers or algebra. There is always an unknown in a word problem. For example, if a question says what is the length of the room?. Then length is the unknown and the task is to find its numerical value.

From words to algebra

When solving a word problem:

1. Choose a letter for the unknown

2. Write down the information of the question in algebra form.

3. Make and equation.

4. Solve the equation

5. Give the answer in written form

6. Check the result against the information given in the question.

Example

I think of a number. I multiply it by 5. I add 15. The result is 100. What is the number I thought of.

Let the number be n

I multiply n by 5: 5n

I add 15: 5n + 15

The result is 100; 5n + 15 = 100 (1)

Subtract 15 from both sides of (1).

5n + 15 – 15 = 100 – 15

5n = 85 (2)

Divides both sides of (2) by 5.

5n/5 = 85/5

n = 17

The number is 17.

Check: 17 X 5 = 85; 85 + 15 = 100

Example

When 6 is added to four times a number, the result is 50. Find the number.

Step 1: What are we trying to find?

A number.

Step 2: Assign a variable for the number.

Let’s call it n.

Step 3: Write down what the variable represents.

Let n = a number

Step 4: Write an equation.

We are told 6 is added to 4 times a number. Since n represents the number, four times the number would be 4n. If 6 is added to that, we get 6 + 4n. We know that answer is 50, so now we have an equation 6 + 4n = 50

Step 5: Solve the equation.

6 + 4n = 50

4n = 44

n = 11

Step 6: Answer the question in the problem

The problem asks us to find a number. We decided that n would be the number, so we have n = 11. The number we are looking for is 11.

The answer makes sense and checks in our equation from Step 4.

6 + 4(11) = 6 + 44 = 50

Exercise

1. John thinks of a number. He doubles it. His result is 58. What numbe did John think of?

2. Six boys each have the same number of sweets. The total number of sweets is 78. How many sweets did each boy have?

3. A number is multiplied by 6 and then 4 is added. The result is 34. Find the first number.

Word problems with brackets

Example

1. If fish cost £2 and chips cost £1 and you went into the shop and asked for 2 fish and chips would you be expecting to pay £5 or £6?

Well it all depends on what you actually wanted.

Was it?

2 x (fish and chips) = 2 x (£2 + £1) = 2 x £3 = £6

or

(2 x fish) + chips = (2 x £2) + £1 = £4 + £1 = £5

2. I bought 3 boxes of eggs in the market. Each box contained 12 eggs. When I got home I found that 5 were broken and had to be thrown away. How many eggs did I have left?

(3 x 12) – 4 = 32

36 – 4 = 32

If I had not done the calculation in brackets first, I could have got 24 as an answer

3 x 12 – 4 = 24

3 x 8 = 24

and that would have been the wrong answer.

Exercise

1. The farmer has four chicken runs. In each run there are 67 brown and fourteen black hens. How many chicken are there altogether?

Hint:(67+14) x 4 = 224

1. 124 cakes were bought, but there wasn’t enough so they decided to buy 4 times more. Then there were too many so they took 10 away. How many did they have in the end?

Hint: (124×4) – 10 = 486

Word Problems with fractions

I add 55 to a certain number and then divide the sum by 3. The result is four times the first number. Find the number.

Let the number n.

I add 55 to n: this gives n + 55

I divide the sum by 3: this gives n + 55/3

The result is 4n.

So, n + 55/3 = 4n (1)

Multiply both sides by 3.

3(n + 55)/3 = 3 X 4n (2)

n + 55 = 12n (3)

Collect terms.

55 = 12n – n

55 = 11n (4)

2x – 9 = 15 is an equation in x. x is the unknown in the equation. 2x – 9 is on the left-hand side (LHS) of the equals sign and 15 is on the right-hand side (RHS) of the equal sign.

So, n = 5, the number is 5.

Weekend Assessment

1. I think of a number. I double it. I divide the result by 5. My answer is 6. What number did I think of?

2. I subtract 17 from a certain number and then divide the result by 5. My final answer is 3. What was the original number?

3. I add 9 to a certain number and then divide the sum by 16. Find the number if my final answer is 1.

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