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WEEK 2
OPEN SENTENCE
(a). If 40 note are to be shared Among 5 pupils, how many books Will be given to a pupil?
5 pupils 40 notes
1 pupils (40 + 5) notes
= 8 notes
(b). Find the letters e.g.
2y + 6 = 30
2y = 30 –6
2y = 24 divide both sidesBy 2
2Y/2=24/2
= y = 12
Closed and open sentences
Study the following mathematical statements:
13 + 6 = 19 23 + 12 = 35
42 − 20 = 22 63 − 49 = 14
7 × 5 = 35 11 × 12 = 132
40 ÷ 5 = 8 120 ÷ 10 = 12
The mathematical statements above are called closed number sentences.
Closed number sentences can either be true or false.
Examples
15 + 7 = 22 (True mathematical statement) 18 + 3 = 19 (False mathematical statement)
3 × 6 = 12 (False mathematical statement) 42 ÷ 6 = 7 (True mathematical statement)
Study each of the following mathematical statements:
[]+ 9 = 13 11 +[] = 25 [] − 4 = 11 20 − = 7
[]× 5 = 15 4 ×[] = 24 [] ÷ 6 = 5 48 ÷ = 12
In each of the statement above, there is a missing number called unknown represented by
.[] They are called open sentences.
An open sentence is a mathematical statement that involves equality signs and a missing
quantity represented by[] that the four arithmetic operations of addition, subtraction,
multiplication and division can be applied to solve.
Open sentences can either be true or false depending on the value [].
Exercise
1.[] + 2 = 9 2[]. + 3 = 7 3.[] + 7 = 12 4.[] − 3 = 1
9.[] ÷ 2 = 2
Operation of addition and subtraction involving open sentences (Revision)
Examples
Here the number represented by in each of the following has been found.
1.[] + 14 = 36 2. 12 +[] = 8 3[]. − 4 = 30 4. 15 –[] = 9
Solution
1.[] + 14 = 36 can be interpreted as “what can be added to 14 to get 36?”
[]+ 14 = 20 + 16
[]+ 14 = 20 + 2 + 14
[]+ 14 = 22 + 14
[]= 22
Check:
22 + 14 = 36
Short method
If[] + 14 = 36
then []= 36 − 14
= 22
= 22
Check:
22 + 14 = 36
12 +[] = 12 + 8 + 10
12 + []= 12 + 18
= 18
Check:
12 + 18 = 30
Short method
If 12 +[] = 30
Then[] = 30 – 12
= 18
= 18
Check:
12 + 18 = 30
Note: Since the problem is addition, the number is subtracted from each other to find .
3 [] . − 4 = 8 can be interpreted as “what number minus 4 gives 8?”
[]− 4 = 12 – 4
[]= 12
Check:
12 − 4 = 8
Short method
If []− 4 = 8
Then[] = 8 + 4
= 12
Check:
12 − 4 = 8
Note: The numbers 8 and 4 are added to get the number represented by[] .
15 –[] = 9
15 –[] = 15 – 6 [15 = 9 + 6]
[]= 6
Check:
15 − 6 = 9
Short method
If 15 –[] = 9
Then[] = 15 – 9
= 6
Check:
15 − 6 = 9
Note: 9 is subtracted from 15 to get the number represented by [] .
Exercise
1.[] − 16 = 13 2[]. − 7 = 23 3. 19 –[] = 11
Operation of multiplication and division involving open sentences (Revision)
Examples
Find the number represented by in each of the following:
Solution
7 ×[] = 7 × 8
= 8
Check:
7 × 8 = 56
Short method
If 7 × = 56
then =
56/7
=8 × 7/7 = 8
Check:
7 × 8 = 56
= 12
Check:
12 × 4 = 48
Short method
If × 4 = 48
then =
48/4
= 12 × 4/4 = 12
Check:
12 × 4 = 48
‘what number divides 60 gives 12?’
60 ÷ = 12
60 = 5 × 12
60 ÷ 5 = 12, 60 ÷ 12 = 5
60 ÷ = 60 ÷ 5
= 5
Check:
60 ÷ 5 = 12
a number is divided by 8, the answer is 9’
÷ 8 = 9
÷ 8 = 72 ÷ 8
= 72
9 × 8 = 72
72 ÷ 8 = 9
72 ÷ 9 = 8 Check: 72 ÷ 8 = 9
Exercise
Find the number represented by in each of the following.
of = 16 11. 12
of = 18 12. 3 × = 18
10 of = 9 15. 680 ÷ = 34 16. 13
of = 12
10 of = 42
Use of letters to find the unknown
Activity
Study the following mathematical statements.
+ 5 = 11, a + 5 = 11 6 + = 15, 6 + y = 11 − 3 = 2, x − 3 = 2
× 2 = 12, 2 × m = 12 32 ÷ = 8, 32 ÷ n = 8
By comparing each statement, you will discover that the box is replaced with a letter of
the alphabet. That is;
+ 5 = 11 is the same as a + 5 − 3 = 2 is the same as x − 3 = 2 and so on.
Mathematical statements containing simple letters and numbers are called simple equations.
When the value of the letter is solved, the equation is solved.
Examples
= 6
Hint: Write a sentence to show the meaning of each equation.
Solution
2 × m = 2 × 7
m = 7
Check:
2m = 2 × m = 2 × 7 = 14
Short method
If 2m = 14
then m = 14
2
= 7
Check: 2m = 2 × m = 2 × 7 = 14
a5
= 6 5 × 6 = 30
a5
= 30
5 30 ÷ 5 = 6, 30 ÷ 6 = 5 a = 30
Check: a5
= 30
5 = 6 5 × 6 = 30
Short method
If a5
= 6
then a = 5 × 6 = 30
x + 5 = 7 + 5
x = 7
Check:
7 + 5 = 12
Short method
If x + 5 = 12
then x = 12 − 5
= 7
Check:
x + 5 = 7 + 5 = 12
y − 12 = 3
y − 12 = 15 − 12
y = 15
Check:
15 − 12 = 3
Short method
If y − 12 = 3
then y = 3 + 12
= 15
Check:
y – 12 = 15 − 12 = 3
Check: a5
= 30
5 = 6
= 6 can be interpreted as ‘when a number is divided by 5 we get 6’
Exercise
Solve the following equations.
Examples
Word problems
Solution
The number I think of + 7 = 21
Let m stand for the unknown number then,
m + 7 = 21
m + 7 = 10 + 10 + 1
m + 7 = 11 + 3 + 7
m + 7 = 14 + 7 m = 14
Short method
m + 7 = 21
m = 21 − 7
= 14
Check:
m + 7 = 14 + 7
= 21
Solution
Unknown number − 43 = 38
Let x stand for the unknown number, then
x − 43 = 38
x − 43 = 81 − 43
x = 81
Short method
x – 43 = 38
x = 38 + 43 = 81
Check:
x – 43 = 8 1
− 4 3
3 8
Solution
Unknown number × 3 = 36
Let y be the unknown number, then
y × 3 = 36
y × 3 = 12 × 3
y = 12
Solution
Unknown number ÷ 7 = 9
Let q be the unknown number
q ÷ 7 = 9
q ÷ 7 = 63 ÷ 7 q = 63
7 × 9 = 63
63 ÷ 7 = 9
63 ÷ 9 = 7
Check: q ÷ 7 = 63 ÷ 7 = 9
Exercise
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