- WHOLE NUMBERS
- WHOLE NUMBERS CONT’D
- ADDITION AND SUBTRACTION IN BASE 2
- MULTIPLICATION AND DIVISION IN BASE 2
- RATIONAL AND NON-RATIONAL NUMBERS
- FACTORIZATION
- REVISION OF FIRST HALF TERM’S LESSON AND PERIODIC TEST
- FORMULAE: SUBSTITUTION AND CHANGE OF SUBJECT
- SIMPLE EQUATIONS INVOLVING FRACTIONS
- CONVERTING NUMBERS TO BASES
WHOLE NUMBERS
JSS 3 Mathematics First Term
Week 2
Topic: Whole Numbers
Contents
- Binary Number System
- Using Computer
- Word Problems
A. Number bases
Most people count in tens. For instance, the place value of the digits in number 4956 in this system
Thousands hundreds tens units
↓ ↓ ↓ ↓
4 9 5 6
The place values are in powers of ten. It is called the base ten system.
4956 = 4 X 1000 + 9 X 100 + 5 X 10 + 6 X 1
= 4 X103 + 9 X 102 + 5 X 101 + 6 X 10
Some people traditionally count in 5s, other in 20s. Using the method of the base ten system, a base five system is in powers of five. So, for example, in base five, 342 would be
342five = 3 X 25 + 4 X 5 + 2 X 1
= 3 X 52 + 4 X 51+ 2 X 50
Notice that 342five is short for 342 in base five.
Converting base ten numbers to other bases
To convert from base ten to another base, express the given number in powers of the new base.
Example
Convert 37ten a. to base eight, b. to base five.
a. Since 37 < 64, there are no sixty-fours in 37. To find the number of eights in 37, divide by 8.
37 ÷ 8 = 4, remainder 5
37 = 4 eights + 5 units
37ten = 45eight
Check: 45eight = 4 X 8 + 5 X 1
= 32 + 5
= 37
Since 37 > 25, there must be a twenty-five in 37.
37 ÷ 25 = 1, remainder 12
37 = 1 twenty-five + 12 units.
Consider the 12 units. Since 12 > 5, there must be some fives in 12.
12 ÷ 5 = 2, remainder 2
12 = 2 fives + 2 units
∴ 37 = 1 twenty-five + 2 fives + 2 units
= 1 x 52 + 2 X 51 + 2 X 1
37ten = 122five
Check: 122five = 1 X 25 + 2 X 5 + 2 X 1
= 25 + 10 + 2
= 37
The method in part b of Example 2 can be shortened as follows:
5 | 37
5 | 7 + 2 (i.e 7 X 5 + 2 X 1)
5 | 1 + 2 ( i.e. 1 X 52 + 2 X 51)
5 | 0 + 1 ( i.e. 0 X 53 + 1 X 5 2)
↑ ↑
Continued division by 5 gives remainders. Reading the remainders upwards gives 37ten = 122five
(see the arrows above).
To change from base ten to another base:
1. Divide the base ten number by the new base number.
2. Continue dividing until 0 (zero) is reached, writing down the remainder each time.
3. Start at the last remainder and read upwards to get the answer.
Convert from other bases to base ten
Convert an Octal into a Decimal
This is a little bit of the same as before except our base is now 8 instead of 16. Try it with the number 7238 also written 0723 in computer programming notation. Go ahead and convert it. I’ll outline the steps below again.
7238 = (7 X 82) + (2 X 81) + (3 X 80)
= 448 + 16 + 3
= 46710
723 octal is 467 decimal. See, that’s not so tough. Now, the astute reader may have noticed another way this can be accomplished using the same information from the original formula. Take the previous octal number as an example here. Since you know what the base and exponent will be for all digits’ place, you can go ahead and write that out: 82 = 64, 81 = 8, 80 = 1. Write your number beneath and multiply out, then add the products.
Operations with binary numbers
To add, subtract and multiply binary number, use the method that you use with base ten numbers. However, you must remember that you are working with powers of two, not powers of ten. The following identities are very useful:
Addition
0 + 0 = 0 1 + 0 = 1
0 + 1 = 2 1 + 1 = 10
Multiplication
0 X 0 = 0 1 X 0 = 0
0 X 1 = 0 1 X 1 = 1
The example below shows how to use these identities when operating with binary numbers. Follow the notes carefully.
Example
Calculate the following.
a. 1 001 b. 1 110 c. 110
+ 1 011 – 101 X 101
———– ———– ———–
1 001
+ 1 011
————
10 100
Note: 1st column: 1 + 1 = 10; write down 0, carry 1
2nd column: as above
3rd column: 1 + 0 + 0 = 1
4th column: 1 + 1 = 10; write down 0, carry 1
b. 1 110
– 101
———–
Note: 1 st column: 1 from 0’won’t go’. Move the 1 in the 2nd column to the 1st column: 10 – 1 = 1; write down 1
2nd column: the 1 has been removed, leaving 0; 0 – 0 = 0; write down 0
3rd column: 1 – 1 = 0; write down 0
4th column: 1 – 0 = 1; write down 1
c. 110
X 101
———–
110
110..
———–
11110
Note: Set out as in a normal long multiplication, multiplying by 1 or 0 as necessary. Take care with placing the digits. Add as explained in part a.
All calculations may be checked by converting to base ten. E.g. part c of example 10 is equivalent to 6 X 5 = 30.
B. Using Computer – ICT and Computer
In the past few decades of Information and Communications Technology (ICT) on many people’s personal and professional lives has been immense. Mobile phones, computers and computer programs have played a central role in ICT development and will play an even greater role in future.
Although computers were originally developed to speed up calculation, a huge leap forward in ICT development took place when it became possible to connect them to the internet on a world wide web (www) via telephone, satellites and radio systems.
Think of the internet as the global communication network that provides national and international access to information and to other people. Millions of people use the internet on a daily basis to access information, to purchase and provide goods and services, and to communicate with each other by electronic mail (email).
Nowadays, to meet ICT demands, computers have been smaller, faster and more versatile, more reliable and more affordable. Computer chips are used in cars mobile phones, television sets, music players, cameras and are carried by many in the form of bank cards. The tendency for greater portability, greater connectivity and more power for less cost will continue throughout this century. For example, many mobile phones now have considerable computing, photographic , entertainment and internet capability.
ICT is mainly used for connecting people ‘usually be the spoken or written word.
Desktop computer
Laptop computer
Mobile phone with in-built computer
Typical in-built programs – word processing, spreadsheet, database, overhead presentation, media player and artwork.
Computers and mathematics
This section will only be meaningful if you have access to a computer that has a spreadsheet program, preferably Microsoft Excel as used here.
Follow through the Class Activities on your computer.
As already mentioned, computers had a historically important role in speeding up calculation and handling numerical information. Of the software listed above, spreadsheet programs currently have the greatest everyday application to mathematics, statistics and economics.
A spreadsheet has the appearance of an extensive matrix of cells. Data either written or numerical are entered into each cell.
There are many spreadsheet programs. The one used here is the most common: Microsoft Excel. Here we identify each cell by an ordered pair: column letter, row number. The cell highlighted in the figure below is C7. This is similar to an ordered pair, (x, y), on the Cartesian plane. Use the arrow keys on the keyboard, or the computer mouse to locate a cell.
Entering data into a spreadsheet
To begin this part of the course, we’ll do something really simply: we’ll enter some text and numbers into some cells. When we’re finished, our spreadsheet will look like this:
So, to enter something into a cell, do the following:
Click on cell A1 with your left hand mouse button. Type the text “Numbers” (without the quotation marks). Press the Return key on your keyboard. The darker border will jump down one cell to A2. Type a 3 and then press the Return Key on your keyboard. The darker border will jump down one cell to A3. Enter a 6 and a 9 in exactly the same way
When you’re finished, your spreadsheet will look like the one in the image above.
The word “Numbers” was our heading. We’re not going to do anything to the heading. It is there purely for our benefit, in order to serve as an explanation for what the numbers are.
Except “Numbers” is not very descriptive. Let’s change it to something else. We’ll change it to “Add these numbers”.
Editing data in a cell
When we wanted to enter some data in a cell, we simply clicked on an individual cell and started typing. But you can’t edit the data in a cell using that method. If there is already something in a cell, and you tried to type something else, the old contents would be entirely erased. Try it for yourself.
Click on the cell A1. Type the letter “A” of “Add”. The word “Numbers” is erased. Click on Edit from the menu bar. From the menu that drops down, click on “Undo Typing”. The word “Numbers” will be restored
So how do you Edit the data in a cell? You have to do it from the Formula bar. The formula bar is the thin white text area running right across the top of the spreadsheet. Notice where it says “Formula Bar”. The thin white text area just above the label is the “Formula Bar”. Notice, too, that there is some thin lines in the shape of an I. This is actually the mouse pointer. It has changed shape, and is now an I-bar.
Click on the cell A1. Then click inside the formula bar. You will see your cursor flashing away. The spreadsheet should now look like the one in the next picture.
The formula bar is displaying the contents of cell A1. To edit the contents, you can use the backspace key on your keyboard to erase anything you don’t want. And you can just type something new in the formula bar. When you’ve finished editing, press the Return key on your keyboard.
In the next image, the text “Numbers” has been changed to “Add these numbers”. Make the changes on your own spreadsheet so that it look like the one below. Notice how the formula bar now reads “3” when the Return key is pressed. The cell A2 is showing in the Name Bar.
Using formulae functions
Sum
To find the sum of the numbers in cells B4 to F4 and place the total in Cell G4:
1. First click in cell G4
2. Then type = SUM (B4:F4) and press Enter.
Notice that = SUM (B4:F4) is short for ‘the sum of the values in cells B4 to F4’. Another way to find the sum is to select cells B4 to F4 and press the ∑ icon (∑ is the greek letter S, which is short for ‘sum’). Try this for cells B4 to F5. Use both of these methods to check the data in the Totals (1) row and Totals (2) column.
Average
To find the average of the numbers from B4 to F4 on the spreadsheet:
Click the cell where you want the average to go (H4)
Type = AVERAGE (B4:F4), then presenter.
C. Word Problems – Sum and difference
The sum of a set of numbers is the result when the numbers are added together.
Example
The sum of four consecutive numbers is 58. Find the numbers.
Let the numbers be n, n+ 1, n + 2, n + 3.
n + (n + 1) + (n + 2) + (n + 3) = 58
4n + 6 = 58
Subtract 6 from both sides.
4n = 58 – 6
4n = 52
Divide both sides by 4.
4n = 52/4
n = 13
The numbers are 13, 14, 15, 16
The difference between two numbers is the result of subtracting one from the other. It is usual to subtract the smaller number from the larger. This gives a positive difference.
Example
The difference between 8 and another number is 7. Find two possible values for the number.
Let the number be x.
i. Assuming x > 8, then
x – 8 = 17
Add 8 to both sides
x = 17 + 8 = 25
ii. Assuming x < 8, then 8 – x = 17
Add x to both sides
8 = 17 + x
Subtract 17 from both sides
8 – 17 = x
X = -9
Thus, the number could be 25 or -9.
Product
The product of two or more numbers is the result when the numbers are multiplied together.
Example
Find the product of -6, 0.7 and 6(2/3).
Product = -6 X 0.7 X 6(2/3)
= -6 X 7/10 X 20/3
= -6 X 7 X 20/10 X 3
= -2 X 7 X 2
= -28
Example
The product of two numbers is 8 4/9 if one of the numbers is 1/4 , find the other.
Let the number x.
¼ X x = 8 4/9
Multiply both sides by 4.
x = 8 4/9 X 4
= 33 7/9
Combining products with sums and differences
Example
Find the positive difference between 31 and the product of 4 and 14
Product of 4 and 14 = 4 X 14
= 56
Difference between 31 and 56 = 56 – 31
= 25
Notice that the problem is to find the difference between 31 and a product. Therefore, find the product first. (4 X 14) – 31 is equivalent to ‘the positive difference between 31 and the product of 4 and 14’.
Expressions with fractions
Example
Find one-quarter of the positive difference between 29 and 11.
Required value = ¼ (29 – 11)
= ¼(18)
= 18/4
= 4 ½
Example
Divide 40 by the sum of 3 and 5.
Required value = 40/3 + 5
= 40/8
= 5
Notice that the dividing line of a fraction acts like a bracket on the expressions above or below the line. For example,
40/3 + 5 = 40/(3 + 5)
Always simplify the expressions above or below the line before dividing out the fraction.
Example
Find the product of 10 and 6, subtract 23; then divide the result by 4.
Required value = (10 X 6) – 24/4
= 60 – 24/4
= 36/4
= 9
From numbers to words
Change the following numerical expressions into word statements
a. (2 + 7) – 3
b. 5(9 -6)
Solution
a. (2 + 7) is ‘the sum of 2 and 7’.
(2 + 7) – 3 is ‘the positive difference between the sum of 2 and 7 and the number 3’ or ‘from the sum of 2 and 7, subtract 3’.
b. 5(9 -6) is ‘the product of 5 and (9 – 6)’.
But (9 -6) is ‘the positive difference between 9 and 6’.
5(9 -6) is ‘the product of 5 and the positive difference between 9 and 6’.
Problems involving equations
The product of a certain number and 5 is equal to twice the number subtracted from 20. Find the number.
Let the number be x.
The product of x and 5 is 5x
Twice x is subtracted from 20 is 20 – 2x
Thus, 5x = 20 – 2x
Add 2x to both sides.
7x = 20
Divide both sides by 7.
x = 20/7 = 2 6/7
The number is 2 6/7.
Example
The sum of 35 and a certain is number is divided by 4. The result is equal to double the number. Find the number.
Let the number be n.
the sum 35 and nis the 35 + n
the sum divide by 4 is 35 + n/4
double the number is 2n
thus, 35 + n/4 = 2n
Multiply both sides by 4.
35 + n = 8n
Subtract n from both sides.
35 = 7n
Divide both sides by 7.
5 = n
The number is 5.
Assessment
1. Find the sum of 12 and 9
2. Find the sum of 82 and 148.
3. Find the positive difference between 19 and 8.
4. Find the difference between 63 and the product of `10 and 5.
5. Find the difference between 27 and the product of 8 and 9.
6. The sum of 8 and a certain number is equal to the product of the number and 3. Find the number.
7. Four times a certain number is equal to the number subtracted from 40. Find the number.
Whole Numbers Cont’d
JSS 3 Mathematics First Term
Week 3
Topic: Whole Numbers Cont’d
Contents
- Expression Involving Fractions
- Direct and Inverse Proportion
- Compound Interest
By an equation with fractions, I’ll mean an equation to solve in which the variable appears in the denominator of one or more fractions. As you’ve seen with equations involving number fractions, the natural approach is to multiply to clear denominators. To do this, you should:
Factor any denominator that can be factored.
Multiply both sides of the equation by the least common multiple of the denominators to clear the fractions.
Fractions with unknowns in the denominator
Example
2 ¾ + 33/2x = 0
2 ¾ + 33/2x = 0
Express 2 ¾ as an imporper fraction.
11/4 + 33/2x = 0
The denominators are 4 and 2x. Their LCM is 4x. Multiply each term in the equation by 4x.
4x(11/4) + 4x(33/2x) = 4x X 0
11x + 66 = 0
11x = -66
X = -6
Check: when x = -6,
LHS = 2 ¾ + 33/-12 = 2 ¾ – 11/4 = 0 = RHS
Example
Solve 1/3a + ½ = 1/2a
1/3a + ½ = 1/2a
The denominators are 3a, 2 and 2a. Their LCM is 6a. Multiply each term in the equation by 6a.
6a X (1/3a) + 6a X ½ = 6a X (1/2a)
2 + 3a =3
3a = 1
a= 1/3
Check: when a = 1/3
LHS = 1/3 X 1/3 + ½ = 1 + ½ = 1 ½
RHS = 1/3 X 1/3 = 3/2 = 1 ½ = LHS
The two examples above show that when unknowns, such as x or a, appear in the denominator, they are treated like numbers.
Clear fractions by multiplying each term of the equation by the LCM of the denominators of the fractions. Then solve the equations in the usual way.
Exercise
1. 1/x = 1/5
2. 1/9 = 1/r
3. 1/m – ¼ = 0
Factors with binomials in the denominator
Note: You should always check your answers to equations with fractions, since it’s possible to produce bogus solutions.
Example
Solve 2/x – 1 + 3 = 4x /x – 1.
When you have an equation with fractions, it’s often good to start by clearing denominators:
(x – 1)( 2/x – 1 + 3) = (x – 1) . 4x /x – 1.
I can cancel x – 1 ‘s provided that x ≠ 1. Assuming that this is true,
2 + 3(x – 1) = 4x
2 + 3x – 3 = 4
3x – 1 = 4
Then
3x – 1 = 4x
– 3x 3x
———————
– 1 = x
Check: If x = -1,
2/x – 1 + 3 = 2/ -2 + 3 = -1 + 3 = 2,
4x/x – 1= -4/-2 = 2
Example
Solve 4 + 5/x – 1= 5x/x – 1.
Clear denominators:
(x – 1)(4 + 5/x – 1) = (x – 1).5x/x – 1
4(x – 1) + 5 = 5x
4x – 4 + 5 = 5x
4x + 1 = 5x
Then
4x + 1 = 5x
– 4x 4x
——————-
1 = x
Check: If x = 1, 4 + 5/x – 1 is undefined.
Therefore, there are no solutions.
Word problems involving fractions
In this lesson, we will learn how to solve fraction word problems that deal with fractions i.e. parts of a whole. Remember to read the question carefully to determine the numerator and denominator of the fraction.
We will also learn how to solve word problems that involve comparing fractions, adding mixed numbers, subtracting mixed numbers, multiplying fractions and dividing fractions.
Example
A class has 20 girls and 30 boys. What part of the class are boys?
Solution:
Step 1: Numerator: boys = 30
Step 2: Denominator: class = 20 + 30 = 50
Step 3: Part or fraction
30/50 = 3/5
Answer: 3/5 of the class are boys
Example
If John earns $x in a week and spend $y, what part of his weekly salary did he save?
Solution
Step 1: Numerator: amount saved= x – y
Step 2: Denominator: salary = x
Step 3: Part or fraction
x – y/x
B. Direct and Inverse Proportion
Direct Proportion
If two quantities are in direct proportion, as one increases, the other increases by the same percentage.
If y is directly proportional to x, this can be written as y ∝x
A simple example of two things that are in the same proportion is the amount of apples you might buy and the amount you pay for them. If you buy twice as many apples as your friend, you pay twice as much.
We can write the connection between the cost and the amount as an equation:
Cost of apples = price per apple × number of apples bought.
This can also be written as y = kx, where k is the cost (the price per apple).
This means that, for some constant k, y = kx for all values of x and k is called theconstant of proportionality.
Example
If y is directly proportional to x.
When x = 12 then y = 3
Find the constant of proportionality and the value of x when y = 8.
We know that y is proportional to x so y = kx
We also know that when x = 12 then y = 3
To find the value of k substitute the values y = 3 and x = 12 into y = kx
3 = k × 12
So k = 3/12 = 1/4
To find the value of x , when y = 8 substitute y = 8 and k = 1/4 into y = kx
8 = (1/4) x
So x = 32 when y = 8
Direct Proportion to Powers
y can be directly proportional to x2 , x3 and other powers of x.
They can always form an equation with k, a constant multiplier (the constant of proportionality), at the start.
eg y = kx2
Example
y ∝ x 3
If y = 1 when x = 2, find the value of y when x = 4
Solution
y ∝ x 3
So y = kx3
Substitute the value y = 1 and x = 2 into y = kx3 to find the value of k.
1 = k × 23
So k = 1/8
Now use the values k = 1/8 and x = 4. y = 1/8 × 64
Gives the answer y = 8
Inverse Proportion
Inverse proportion is when one value increases at the other value decreases.
A simple example of inversely proportional quantities is the lengths and widths of rectangles with the same area. As the length of one side doubles, the width has to be halved for the area to stay the same.
Example
y is inversely proportional to x. When y = 3, x = 12 .
Find the constant of proportionality, and the value of x when y = 8.
y ∝ 1/x
y = k/x
So xy = k
Substitute the values x = 12 and y = 3 into xy = k
3 × 12 = 36
So k = 36
To find the value of x when y = 8, substitute k = 36 and y = 8 into xy = k
8x = 36
So x = 4.5
Again, you can have questions involving squares, cubes or other powers of the variables.
Exercise
v is inversely proportional to r3. When r = 2, v = 25. Find r when v = 60.
Answer
v = ∝ so v =
Re-arrange the above to get k on its own.
k = vr3
k = 25 × 23
So k = 200
When v = 60
60r3 = 200
r3 = 200/60
r3 = 3.333
So r equals the cube root of 3.333
So r = 1.494
Graphical Representation
When two variables are related in such a way that the ratio of their values always remains the same, the two variables are said to be in direct variation.
In simpler terms, that means if A is always twice as much as B, then they directly vary. If a gallon of milk costs $2, and I buy 1 gallon, the total cost is $2. If I buy 10 gallons, the price is $20. In this example the total cost of milk and the number of gallons purchased are subject to direct variation — the ratio of the cost to the number of gallons is always 2.
To be more “mathematical” about it, if y varies directly as x, then the graph of all points that describe this relationship is a line going through the origin (0, 0) whose slope is called the constant of the variation. That’s because each of the variables is a constant multiple of the other, like in the graph shown below:
Inverse Variation
(The Opposite of Direct Variation)
In an inverse variation, the values of the two variables change in an opposite manner – as one value increases, the other decreases.
For instance, a biker traveling at 8 mph can cover 8 miles in 1 hour. If the biker’s speed decreases to 4 mph, it will take the biker 2 hours (an increase of one hour), to cover the same distance.
Inverse variation: when one variable increases,
the other variable decreases.
Notice the shape of the graph of inverse variation.
If the value of x is increased, then y decreases.
If x decreases, the y value increases. We say that y varies inversely as the value of x.
An inverse variation between 2 variables, y and x, is a relationship that is expressed as:
Y = k/x
where the variable k is called the constant of proportionality.
As with the direct variation problems, the k value needs to be found using the first set of data.
The Reciprocal of a Number
Clearly, 3 X 1/3 = 1
1/3 is called the reciprocal of 3
3 is called the reciprocal of 1/3
One number is the reciprocal of another if their product is 1.
Example, the reciprocal of 5/6 is 6/5 since 5/6 X 6/5 = 1.
In general:
The reciprocal of a fraction is obtained by interchanging the numerator and the denominator, i.e. by inverting the fraction.
Example
Find the reciprocal of 20.
Solution:
Reciprocal of 20 is 1/20
Example
Find the reciprocal of 3/7.
Solution:
Example 9
Solution:
Note:
To find the reciprocal of a mixed number, change it into an improper fraction and then invert it.
Additive Inverse
If we add 0 (zero) to any number, the result is the same as the given number.
For example,
3 + 0 = 3, 0 + 8 = 8
We say that 0 is the identity for addition. If the sum of two numbers is 0, we say that each number is the additive inverse of the other.
For example, (+3) + (-3) = 0.
(-3) is the additive inverse of (+3).
(+3) is the additive inverse of (-3).
(-8) + (+8) = 0.
(+8) is the additive inverse of (-8).
(-8) is the additive inverse of (8+).
Example 1
State the additive inverse of:
a. -19
b. 0.32
c. -7/8
d. 6 x 107
e. – 3.1 x 10-5
Given number Additive inverse
a. -19 + 19
b. 0.32 -0.32
c. -7/8 +7/8
d. 6 x 107 -6 x 107
e. – 3.1 x 10-5 +3.1 x 10-5
In parts d and e of example 1, remember that the power of 10 in a number in standard form places the decimal the decimal point. It is not significant in deciding whether the number is positive or negative.
Solve the following equations:
a. x + 7 = 2
Solution
x + 7 = 2 is the same as x + (+7) = 2
Add (-7) to both sides.
X + (+7) + (-7) = 2 + (-7)
X + 0 = 2 – 7
X = – 5
Multiplicative Inverse
If we multiply any number by 1 the result is the same as the given number. For example,
1 x 9 = 9, -5 x 1 = 15, 1 x /34 = ¾
We say that 1 is the identity for multiplication. If the product of two numbers is the multiplicative inverse of the other. For example,
9 x 1/9 = 1
1/9 is the multiplicative inverse of 9.
9 is the multiplicative inverse of 1/9.
(-5) x(-1/5) = 1
-1.5 is the multiplicative inverse of 3/4.
-5 is the multiplicative inverse of -1/5.
¾ x 4/3 = 1
4/3 is the multiplicative inverse of ¾.
¾ is the multiplicative inverse of 4/3.
You have already used multiplicative inverses. In Book 1 you used reciprocals. The reciprocal of a fraction is that fraction turned upside down. The reciprocal of 2/3 is 3/2. Thus the multiplicative inverse of a number is the same as its reciprocal. 1/8 is the reciprocal of 8/1 or 8. 1/8 is the multiplicative inverse of 8.
Example
Find the multiplicative inverses of the following:
- -32/ b. 0.3 c. 2 ½ d. n
Reciprocal of -3/2 = -2/3
-2/3 is the multiplicative inverse of -3/2.
Check: (-3/2) x (-2/3) = + (3/2 x 2/3) = 1.
- 0.3 = 3/10
Reciprocal of 3/10 = 10/3.
10/3 (or 3 1/3) is the multiplicative inverse of 0.3.
Check: 0.3 x 3 1/3 = 3/10 x 10/3 = 1.
c. 2 ½ = 5/2
2/5 is the multiplicative inverse of 2 ½.
The check is left as an exercise.
d. i/n is the multiplicative inverse of n.
n x 1/n = 1.
Example
Solve -5x = 20.
Method I:
Notice that -5 is the multiplying x. Multiply both sides by the multiplicative inverse of -5.
Multiply both sides by -1/5.
(-1/5) x (-5) X x = (-1/5) x (+20)
1 X x = -(1/5 x 20)
x = -4
Method II:
Notice that multiplying by -1/5 is equivalent to dividing by -5. The example can be solved as follows.
-5x = 20
Divide both sides by -5.
(-5) X x/(-5) = +20/-5
1 X x = -(20/5)
x = -4
The second method is usually quicker.
Inverse Operation
Do the following:
1. Stand up. Sit down.
2. Add 3 to 15. Subtract 3 from the result.
3. Multiply 7 by 2. Divide the result by 2.
In each case you should end where you start. When this happens, we say that the two actions are inverse operations.
Sitting down is the inverse operation of standing up. Adding a number is the inverse operation of subtracting the same number. Multiplying a number and dividing by the same number are inverse operations.
| operation | Inverse operation |
| Shut the door | Open the door |
| Add 20 | Subtract 20 |
| Subtract -3 | Add -3 |
| Multiply by 4 | Divide by 4 |
| Divide by 0.3 | Multiply by 0.3 |
C. Compound Interest
Simple Interest
Interest is the payment given for saving money. It can also be the price paid for borrowing money. When interest is calculated on the basic sum of money saved (or borrowed) it called simple interest.
To find the simple interest, use this formula: Interest = Principal × Rate of interest × Time
The principal is the amount of money you borrow or invest.
The rate of interest is the percent charged for the use of money, the percentage charged will be divided by hundred to get the actual value for application in solving a problem.
Exercises
Compute the interest if the principal is 2000 dollars at a rate of interest of 5% for 4 years.
Using a calculator,
Interest = 2000 × 5% × 4 = 2000 × 0.05 × 4
Interest = 100 × 4 = 400
Exercises
Compute the interest if the principal is 2,000,000 dollars at a rate of interest of 4% for a year
Using a calculator,
Interest = 2,000,000 × 4% × 1
Interest = 2,000,000 × 0.04 × 1
Interest = 80,000 × 1 = 80,000
If you have 2 million dollars and your bank pay you 4% interest every year, you will earn 80,000 dollars every year.
Great, you can quit your day time job!
Exercises
Compute the interest if the principal is 100 dollars at a rate of interest of 2% for 10 year
Using a calculator,
Interest = 100 × 2% × 10
Interest = 100 × 0.02 × 10
Interest = 2 × 10 = 20
With little money invested and low interest, 10 years investment gives you a mere 20 dollars
This might be a waste of time!
Compound Interest
When money is saved with simple interest, the interest is paid at regular intervals and the principal remains the same.
With compound interest the interest is added to the principal at the end of each interval.
Thus, the principal increases and so the interest becomes greater for each interval. Most savings schemes give compound interest, not simple interest.
Example
Find the compound interest on N60 000 for 2 years at 8% per annum.
Note: ‘per annum’ means ‘each year’.
The interest is added at 1 year intervals.
1st year: I1 = N 60 000 X 8 X 1/100 = N 4 800
Amount at end of 1st year = N60 000 + N4 800 = N64 800
2nd year: The principal is now N64 800
= N648 X 8 = N5 148
Amount at end of 2nd year = N64 800 + N5 184
= N69 984
Compound interest = N69 984 – N60 000 = N9 984
The working is easier if arranged in a table. The annual interest can be calculated by inspection. For example, 6% of N21 000 is found by multiplying N21 000 by 6, and moving the digits two places to the right (to divide by 100:; i.e.
6% of N21 000 = N210 X 6 = N1 260
The example below shows how to arrange the working
Example
Find the amount that N5 000 becomes if saved for 3 years at 6% per annum compound interest.
1st year: Principal N5 000
6% Interest + 300 (6/100 X 5 000)
—————-
2nd year: Principal N5 300
6% Interest + 318 (6/100 X 5 300)
—————-
3rd year: Principal 5 618
6% Interest + 337.08 (6/100 X 5 618)
——————
Amount N 5 955.08
Assessment
Find the a. amount b. the compound interest, for each of the following
1. N40 000 for 2 years at 8% per annum
2. N60 00 for 2 years at 7% per annum
3. N50 000 for 2 years at 6% per annum
When calculating compound interest, the arithmetic often gives final answers to many decimal places. Final answers should be rounded to the nearest naira. Such rounding should be left to the last line of the working. If possible, use a calculator to calculate interest.
When money is borrowed, interest must be paid back as well as the principal. When a large sum of money is paid back over a number of years, the principal gradually reduces.
Depreciation
Many items, such as cars, clothes, electrical goods, lose values and time passes. This loss in value is called depreciation. Depreciation is usually given as a percentage of the item’s value at the beginning of the year. For example, if a radio costing N10 000 depreciates by 20% per annum, then its value will be N8 000 at the end of the first year. At the end of the second year, its value will be N8 000 less 20% of N8 000, i.e. N8 000 – N1 6000 = N 6 400.
Example
A car costing N680 000 depreciates by 25% in its first year and 20% in its second year, Find its avalue after 2 years.
1st year:
Value of car N680 000
25% depreciation – 170 000 (1/4 of 680 000)
——————–
2nd year:
Value of car 510 000
20% depreciation – 102 000 (1/5 of 510 000)
Value after 2 yr = N408 000
Inflation
Due to rising prices, money loses its value as time passes. Loss in value of money is called inflation. Inflation is usually given as the percentage increase in the cost of buying things from one year to the next. For example, if the rate of inflation 15% per annum, then a CD player which cost N10 000 a year ago will now cost N11 500. Money has lost it’s a value since it now costs more to buy the same thing.
Example
How long will it take for prices to double if the rate of inflation is 20% per annum?
Start with an initial cost of 100 units.
Initially, cost = 100
rise = 20
—————-
after 1 year, cost = 120
rise = 24 (i.e. 20% of 120)
—————
after 2 years, cost = 144
rise = 28.8 (20% of 144)
—————-
after 3 years, cost = 172.8
rise = 34.56 (20% of 172.8)
The cost after 4 years is a little more than double the initial cost. Hence prices will double in just under 4 years.
Assessment
1. 6 times a number is 48. What is the number?
2. Find the number which, when multiplied by 10, gives 70.
3. A number divide by 5 gives 9, what is the number?
4. 12/ x – 1 = 3
5. 4/ 1 + 4 = 1
6. 2 = 7/ y + 2
Solve the following equations:
7. 4x = 28
8. 3x = 18
9. -4d = 20
10. 7x = 4 2/3
Addition and Subtraction in Base 2
JSS 3 Mathematics First Term
Week 4
Topic: Addition and Subtraction in Base 2
Let’s first take a look at decimal addition.
As an example we have 26 plus 36,
26
+36
To add these two numbers, we first consider the “ones” column and calculate 6 plus 6, which results in 12. Since 12 is greater than 9 (remembering that base 10 operates with digits 0-9), we “carry” the 1 from the “ones” column to the “tens column” and leave the 2 in the “ones” column.
Considering the “tens” column, we calculate 1 + (2 + 3), which results in 6. Since 6 is less than 9, there is nothing to “carry” and we leave 6 in the “tens” column.
26
+36
62
Binary addition
Works in the same way, except that only 0’s and 1’s can be used, instead of the whole spectrum of 0-9. This actually makes binary addition much simpler than decimal addition, as we only need to remember the following:
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10
As an example of binary addition we have,
101
+101
a) To add these two numbers, we first consider the “ones” column and calculate 1 + 1, which (in binary) results in 10. We “carry” the 1 to the “tens” column, and the leave the 0 in the “ones” column.
b) Moving on to the “tens” column, we calculate 1 + (0 + 0), which gives 1. Nothing “carries” to the “hundreds” column, and we leave the 1 in the “tens” column.
c) Moving on to the “hundreds” column, we calculate 1 + 1, which gives 10. We “carry” the 1 to the “thousands” column, leaving the 0 in the “hundreds” column.
101
+101
1010
Another example of binary addition:
1011
+1011
10110
Note that in the “tens” column, we have 1 + (1 + 1), where the first 1 is “carried” from the “ones” column. Recall that in binary,
1 + 1 + 1 = 10 + 1
= 11
Binary subtraction
is simplified as well, as long as we remember how subtraction and the base 2 number system. Let’s first look at an easy example.
111
– 10
101
Note that the difference is the same if this was decimal subtraction. Also similar to decimal subtraction is the concept of “borrowing.” Watch as “borrowing” occurs when a larger digit, say 8, is subtracted from a smaller digit, say 5, as shown below in decimal subtraction.
35
– 8
27
For 10 minus 1, 1 is borrowed from the “tens” column for use in the “ones” column, leaving the “tens” column with only 2. The following examples show “borrowing” in binary subtraction.
10 100 1010
– 1 – 10 – 110
1 10 100
Addition in base two
We will follow the exact same pattern above to show how to add in base 2
Base 2 uses 0 and 1
We show a place value for base 2 below:
Notice that the twos place in base 2 is the tens place in base 10
| Thirty-twos | Sixteens | Eights | Fours | Twos | Ones |
| 0 | 1 | 1 | 0 | 1 | 0 |
Let’s practice now with some examples
Example #1:Addition in base two with carry
To avoid confusion with base 10, we put a 2 next to each number
However, it is clear to you that the addition is being done in base 2, there is no need to write down the 2
Add: 1012 + 1012
1 1 1
0 1 12
1 0 12
___________________________
1 0 0 02
Explanation:
Add the numbers in the ones place: 1 + 1 = 2 = 2 + 0.
Write 0 in the ones place and carry the 2 to the twos place. You can just use a 1 to represent the 2 (shown in red)
Add the numbers in the twos place: 1 two + 1 two + 0 two = 2 twos = 4 = 4 + 0
Write 0 in the twos place and carry the 4 to the fours place. You can just a 1 to represent the 4 (shown in green)
Add the numbers in the fours place: 1 four + 0 four + 1 four = 2 fours = 8 = 8 + 0
Put 0 in the fours place and carry the 8 to the eights place. You can just put a 1 to represent the 8 (shown in black)
Bring down the 1 shown in black et voila!
Example #2:
Add: 11012 + 11012
1 1 1
1 1 0 12
1 1 0 12
________________________________
1 1 0 1 02
Explanation:
Add the numbers in the ones place: 1 + 1 = 2 = 2 + 0.
Write 0 in the ones place and carry the 2 to the twos place. You can just use a 1 to represent the 2 (shown in red)
Add the numbers in the twos place: 1 two + 0 two + 0 two = 1 two
Write 1 two in the twos place.
Add the numbers in the fours place: 1 four + 1 four = 2 fours = 8 = 8 + 0
Put 0 in the fours place and carry the 8 to the eights place. You can just put a 1 to represent the 8 (shown in green)
Add the numbers in the eights place: 1 eight + 1 eight + eight = 2 eights + 1 eight = 16 + 1 eight
Write 1 eight in the eights place and carry the the 16 to the sixteens place. You can just use a 1 to represent the 16 (shown in black)
Bring down the 1 in the sixteens place shown in black et voila!
If you struggle to understand addition in base two, you may need to go back to the example I gave about base 10 and try your best to relate this to base two
Subtraction in Base 2
Base 2 uses only 0 and 1
With base 2, you will borrow a 2,4,8,16, etc depending on the place value, not a 10 when needed. Why is that?
Some explanations:
In base 10, the number 739 can mean everything you see below:
7 × 102 + 3 × 10 + 9
7 groups of 100, 3 groups of 10, and 9
7 hundreds + 3 tens + 9
In base 2, the number 1012 could also mean everything you see below:
1 × 22 + 0 × 2 + 1
1 groups of 4, 0 groups of 2, and 1
1 four + 0 two + 1
Simply put, it is because you are in base 2, so any borrowing is done with 2, 4, 8, etc
Say you want to perform the following subtraction in base two
1 1 0
– 1 0 1
_____________
From the twos place, borrow 2 from 1 two. 1 two in now 0 two
Then, add that 1 two to the 0 in the ones place to make it 1 two or 2
Rewrite the problem and subtract
0 2
1 1 0
– 1 0 1
_____________
0 0 1
Now you are ready to do some more subtraction in base two
Example #2:No carry
To avoid confusion with base 10, we put a 2 next to each number
However, it is clear to you that the subtraction is being done in base two, there is no need to write down the 2
1 1 12
– 0 1 02
_____________________
1 0 12
The subtraction above is easy since there was no carry
Assessment
Add: 10112 + 10112
Subtract: 1102 – 1012
Multiplication and Division in Base 2
JSS 3 Mathematics First Term
Week 5
Topic: Multiplication and Division in Base 2
Binary multiplication
Is actually much simpler than decimal multiplication. In the case of decimal multiplication, we need to remember 3 x 9 = 27, 7 x 8 = 56, and so on. In binary multiplication, we only need to remember the following,
0 x 0 = 0
0 x 1 = 0
1 x 0 = 0
1 x 1 = 1
Note that since binary operates in base 2, the multiplication rules we need to remember are those that involve 0 and 1 only. As an example of binary multiplication we have 101 times 11,
101
x11
First we multiply 101 by 1, which produces 101. Then we put a 0 as a placeholder as we would in decimal multiplication, and multiply 101 by 1, which produces 101.
101
x11
101
1010 <– the 0 here is the placeholder
The next step, as with decimal multiplication, is to add. The results from our previous step indicates that we must add 101 and 1010, the sum of which is 1111.
101
x11
101
1010
1111
Binary division
Is almost as easy, and involves our knowledge of binary multiplication. Take for example the division of 1011 into 11.
11 R=10
11 )1011
-11
101
-11
10 <– remainder, R
To check our answer, we first multiply our divisor 11 by our quotient 11. Then we add its’ product to the remainder 10, and compare it to our dividend of 1011.
11
x 11
11
11
1001 <– product of 11 and 11
1001
+ 10
1011 <– sum of product and remainder
The sum is equal to our initial dividend, therefore our solution is correct.
Assessment
Multiply: 111 x 101
Divide: 1101 / 11
Rational and Non-rational Numbers
JSS 3 Mathematics First Term
Week 6
Topic: Rational and Non-rational Numbers
Rational Numbers
We can write numbers such as 8, 4½, 1/5, 0.211, √49/16, 0.3 as exact fractions or ratios:
8/1, 9/2, 1/5, 211/1 000, 7/4, 1/3.
Such numbers are called rational numbers.
Numbers which cannot be written as exact fractions are called non-fractional numbers, or irrational numbers. √7 is an example of a non-rational number. √7 = 2.645 751 …., the decimals extending without end and without recurring.
π is another example of a non-rational number. Π = 3.141 592 …., again extending forever without repetition. The fraction 22/7 is often used for the value of π. However, 22/7 is a rational number and is only an approximate value of π.
All recurring decimals are rational numbers. Read the following example carefully.
Example
Write 3.17 as a rational number
Let n = 3/17
i.e. n =3.17 17 17 ………… (1)
Subtract (1) from (2),
99n = (317.17 17 . . .) – (3.17 17 . . .)
99n = 314
Thus, 3.17 = 314/99, a rational number.
A non-rational number extends forever and is non-recurring.
Assessment
Which of the following are rational and which are non-rational?
a. 9
b. 1/9
c. √9
d. 0.9
e. 2 2/3
Square Roots
Some square roots are rational:
√4 = 2, √6.25 = 2.5 = 5/2
Other square roots are non-rational :
√11 = 3.316 624 . . ., √3.6 = 1.897 366 . . .
The fact that many square roots are non-rational, was first discovered by Pythagoras around 500 BC. He tried to find the length of a diagonal of a ‘unit square’. The fig below is a unit square, a square with side 1 unit.
In ∆ABC, using Pythagoras’ rule,
BC2 = AB2 + AC2
BC2 = 12 + 12 = 2
BC = √2
Pythagoras was unable to find a rational value for √2. Thus, although it is possible to draw the diagonal of a unit square, it is possible to measure its length accurately! This troubled Pythagoras so much that he called non-rational numbers ‘unspeakables’.
It is possible to find the approximate value of non-rational square roots by using a ‘trial and improvement’ method. The example immediately below shows:
Example
Find the value of √2 correct to 2 significant figures.
Since 2 lies between 1 and 4, √2 lies between √1 and √4, i.e. √2 lies between 1 and 2:
Try 1.5 : 1.52 = 2.25 (too large)
Try 1.4: 1.42 1.96 (too small)
Thus, √2 lies between 1.4 and 1.5. Since 1.96 is much closer to 2 than 2.25.:
Try 1.41: 1.412 = 1.9881∗ (too small)
Try 1.42: 1.422 = 2.0146∗ (too large)
Thus, √2 lies between 1.41 an 1.42.
Thus, √2 1.4 to 2 s.f.
(∗ Check the calculation of these squares.)
Pi (π)
As we have earlier in the course, Pi, or π, is the ratio of the circumference of a circle to its diameter:
π = length of circumference of circle/length of diameter of circle = c/d = c/2r
where r is the radius of the circle.
The problem of finding the value of V has occupied mathematicians through the ages. The most famous attempt to find π was by Archimedes, around 250 BC. His method, using the fact that the area of a circle is πr2, was as follows.
The area of the circle is about halfway between the area of the inner square and that of the outer square. Let the top of the inner square be labeled A, the left side labeled B, while the centre be point O.
Area of inner square = 4 X ∆AOB
= 4 X ½r2 = 2r2
Area of outer square = 8 X ∆AOB
= 8 X ½r2 = 4r2
Thus, the area of the circle lies between 2r2 and 4r2. It follows that the value of π must lie between 2 and 4, probably around 3.
Archimedes worked in this way, using regular polygons with more and more sides.
In the figure above it can be seen that the greater the number of sides of the polygon, the greater there are is to that of the circle. Using polygons of 96 sides, Archimedes showed that the value of π lies between 310/71 and 31/7.
Both of these values are correct to 2 decimal places.
Practice
a. Try to collect some tins and bottles of various diameters.
b. Measure the diameter, d, of each subject. (An easy way is to place the object on a ruler; the take readings at opposite ends of a diameter.)
c. Use the piece of string or a strip to measure the circumference, C, of each object.
d. Make a table of values of d and C.
e. Draw a graph of d (on the horizontal axis) against C (on the vertical axis).
Assessment
Write down the first digit of the square roots of the following
- 22
- 6
- 14
- 88
- 71
- 42
Use the method of Examples 2 and 3 to find the value of the following correct to 2 significant figures
- √8
- √52
- √69
- √3
- √23
Factorization
JSS 3 Mathematics First Term
Week 8
Topic: Factorization
Expression of Algebraic Expressions
The expression (x + 2)(x – 5) means (x + 2) X (x – 5). The product of the two binomials (x + 2) and (x – 5) is found by multiplying each term in the first binomial by each term in the second binomial. Read the following examples carefully.
Example 1
Find the product of (x + 2) and (x – 5).
= (x + 2)(x – 5) = x(x + 2) – 5(x + 2)
= x2 + 2x – 5x – 10
= x2 – 3x – 10
Example 2
Expand (2c – 3m)(c – 4m)
(2c – 3m)(c – 4m)
= c(2c – 3m) – 4m(2c -3m)
= 2c2 – 3cm – 8cm + 12m2
= 2c2 – 11cm + 12m2
Example 3
Expand (3a + 2)2
(3a + 2)2
= (3a + 2)(3a + 2)
= 3a(3a + 2) + 2(3a + 2)
= 9a2 + 6a + 6a + 4
= 9a2 + 12a + 4
Assessment
Expand each expression
- (2x + 1)2
- (d – 6)(d + 3)
- (x – 1)(x + 2)
- (3y – 5)(2y + 1)
- (5x + 2)(2x – 3)
Factorisation of Quadratic Expressions
A quadratic expression is one in which 2 is the highest power of the unknown in the expression. For example x2 – 4x – 12, 16 – a2, 3x2 + 17xy + 10y2.
Since (x + 2)(x – 6) = x2 – 4x – 12
x + 2 and x – 6 are the factors of x2 – 4x – 12.
To factorise quadratic expressions, it is to express it as a product of its factors
Example 5
Factorise x2 + 7x + 10
The problem is to fill the bracket in the statement x2 + 7x + 10 = ( )( )
1st step: Look at the first term in the given expression x2 . From work done in expanding brackets, when the first term is x2, it appears in each bracket: x2 + 7x + 10 = (x )(x )
2nd step: Look at the last term given in the given expression, +10. The product of the last terms in the two brackets must be +10. Number pairs which have a product of +10 are
- +10 and +1
- +5 and +2
- -10 and -1
- -5 and -2
These give four possible answers
- (x + 10)(x + 1)
- (x + 5)(x + 2)
- (x – 10) (x – 1)
- (x – 5)(x – 2)
3rd step: Look at the coefficient of the middle term in the given expression, +7. The sum of the last terms in the two brackets must be +7. Adding the number pairs in turn:
- (+10) + (+1) = +11
- (+5) + (+2) = +7
- (-10) + (-1) = -10
- (-5) + (-2) = -7
Of these, only b gives +7. Thus,
x2 + 7x + 10 = (x + 5)(x + 2)
Note:
- The answer can be checked by expanding the brackets
- The order of the brackets is not important
(x + 5)(x + 2) = (x + 2)(x + 5)
Assessment
- x2 + 12x + 11
- c2 + 8c + 15
- x2 + 6x + 5
- s2 + 10s + 16
Revision of First Half Term’s Lesson and Periodic Test
JSS 3 Mathematics First Term
Week 7
Topic: Revision of First Half Term’s Lesson and Periodic Test
Teachers and Students are expected to do a review of the first 6 weeks. Students will also have their continuous assessment test.
FORMULAE: SUBSTITUTION AND CHANGE OF SUBJECT
JSS 3 Mathematics Third Term
Week 10
Topic: FORMULAE: SUBSTITUTION AND CHANGE OF SUBJECT
Formulae and Substitution
A formula is an equation with letters which stands for quantities. For example
C = 2πr
Is the formula which gives the circumference, c, of a circle of radius r.
In science,
I = V/R
Is the formula which shows the relationship between the current I amps, voltage, V volts, and resistance, R ohms, in an electrical circuit. In arithmetic,
I = PRT/100
Is the formula which gives interest, I, gained on a principal, P, invested at R% per annum for T years. Sometimes the same letter can stand for different quantities in different formulae. For example, I stands for current in the science formula and I stands for interest in the arithmetic formula. Formulae is the plural of formula.
Substitution
To substitute in a formula means to replace letters by their values. This makes it possible to calculate other values.
Example
A gas at a temperature of 00C has an absolute temperature of T K, where T = θ + 273.
a. Find the absolute temperature of a gas at a temperature of 68 0C.
b. If the absolute temperature of a gas is 380 K, find its temperature in 0C.
Solution
a. T = θ + 273
when θ = 68
T = 68 + 273
= 341
The absolute temperature is 341 K.
b. T = θ + 273
when T = 380,
380 = θ +273
Subtract 273 from both sides.
380 – 273 = θ
107 = θ
The temperature of the gas is 107 0C.
Example
The formula W = VI gives the power, W watts, used by an electrical item when a current of I amps flows through a circuit of V volts.
a. An air conditioner on maximum power needs a current of 25 amps in a 120 volt circuit. Find the power of being used.
b. An electric light bulb is marked 100 watts, 240 volts. Find the current required to light the bulb.
Solution
a. W = VI
when V = 120 and I = 25
W = 120 X 25
= 3 000
The maximum power is 3 000 watts.
b. W = VI
when W = 100 and V = 240,
100 = 240I
Divide both sides by 240.
100/240 = I
I = 10/24 = 5/12
The current required is 5/12 amp.
Example
If y = 5x2 – 1, find
a. the value of y when x = -2
b. the values of x when y = 79.
Solution
a. y = 5x2 – 1
when x = -3
y = 5 X (-3)2 – 1
= 5 X (+9) – 1
= 45 – 1
= 44
b. y = 5x2 – 1
when y = 79
79 = 5x2 -1
Add 1 to both sides.
80 = 5x2
Divide both sides by 5.
16 = x2
Take the square root of both sides
√16 = x
x = +4 or -4
Notice that there are two possible values for x. We can shorten this to x = ±4 where ± is short for ‘+ or –‘.
Change of Subject
Formula means
Relationship between two or more variables
Example y = x + 5 where x and y are variables.
Subject of a Formula means
The variable on its own, usually on the left hand side.
Example y is the subject of the formula y = x + 5
Changing The Subject Of A Formula means rearranging the formula so that a different variable is on its own.
Making x the subject of the formula y = x + 5 gives x = y – 5
Example
Make x the subject of
y=x+3
We require x to be the subject of the formula. The subject is written on the left, so we switch the sides to get x on the left
Switch sides
x+3=y
We require x by itself on the left hand side. But we have x + 3. The inverse of addition is subtraction
We need to subtract 3 from the left side. But, to keep the equality true, we need to subtract 3 from the right side as well.
So subtract 3 from both sides
Subtract 3 from both sides
x + 3 − 3 = y − 3
Simplify
x = y − 3
Example
Make x the subject of
y = x + 3
Switch sides
x + 3 = y
Subtract 3 from both sides
x + 3 − 3 = y − 3
Simplify
x = y − 3
Example
Make x the subject of
y = x + m
Switch sides
x + m = y
Subtract m from both sides
x + m − m = y − m
Simplify
x=y−m
When you do a question yourself it is often helpful to write in these key points before you do the actual algebra. It gets you to think of the logic of the process
Example
Make x the subject of
y = x − 5
Switch sides
x − 5 = y
Add 5 to both sides
x − 5 + 5 = y + 5
Simplify
x = y + 5
Example
Make x the subject of
y = x − m
Switch sides
x − m = y
Add m to both sides
x − m + m = y + m
Simplify
x = y + m
Example
Make x the subject of
y = 8x
Switch sides
8x = y
Divide both sides by 8
8 x 8 = y8
Simplify
x = y8
Example
Make x the subject of
y = mx
Switch sides
mx = y
Divide both sides by m
mxm = ym
Simplify
x=ym
Example
Make x the subject of
y= x8
Switch sides
x8 = y
Multiply both sides by 8
8 x 8 = 8y
Simplify
x = 8y
Changing The Subject of A Formula
Example
Make x the subject of
y = xm
Switch sides
x m =y
Multiply both sides by m
m x m = my
Simplify
x = m y
Example
Make x the subject of
y = 2x + 5
Switch sides
2x + 5 = y
Subtract 5 from both sides
2x + 5 − 5 = y − 5
Simplify
2x = y − 5
Divide both sides by 2
2 x 2 = y − 52
Simplify
x = y − 52
Example
Make x the subject of
y = m x + c
Switch sides
mx + c = y
Subtract c from both sides
mx + c − c = y − c
Simplify
mx = y − c
Divide both sides by m
m x m = y − cm
Simplify
x = y − cm
Example
Make x the subject of
y = 3x − 7
Switch sides
3x − 7 = y
Add 7 to both sides
3x − 7 + 7 = y + 7
Simplify
3x = y + 7
Divide both sides by 3
3 x 3 = y + 73
Simplify
x = y + 73
Example
Make x the subject of
y=mx − c
Switch sides
mx − c = y
Add c to both sides
Mx – c + c = y + c
Simplify
mx =y + c
Divide both sides by m
mxm=y+cm
Simplify
x=y+cm
Example
Make x the subject of
y=x2+5
Switch sides
x2+5=y
Fractions are more difficult to work with so to make work easier and errors less likely get rid of the fraction first
Multiply EVERYTHING on both sides by 2
2(x2)+2(5)=2(y)
Simplify
x+10=2y
Subtract 10 from both sides
x+10−10=2y−10
Simplify
x=2y−10
Example
Make x the subject of
y = xm + c
Switch sides
xm + c = y
Fractions are more difficult to work with so to make work easier and errors less likely get rid of the fraction first
Multiply EVERYTHING on both sides by m
m(xm) + m(c) = m(y)
Simplify
x + cm = my
Subtract cm from both sides
x + cm – cm = my − cm
Simplify
x = my − cm
Example
Make x the subject of
y=x4−7
Switch sides
x4−7=y
Fractions are more difficult to work with so to make work easier and errors less likely get rid of the fraction first
Multiply EVERYTHING on both sides by 4
4(x4) − 4(7) = 4(y)
Simplify
x−28=4y
Add 28 to both sides
x−28+28=4y+28
Simplify
x=4y+28
Example
Make x the subject of
y=xm − c
Switch sides
xm – c = y
Fractions are more difficult to work with so to make work easier and errors less likely get rid of the fraction first
Multiply EVERYTHING on both sides by m
m (xm)−m(c)=m(y)
Simplify
x − cm = my
Add cm to both sides
x − cm + cm = my + cm
Simplify
x = my + cm
Example
Make x the subject of
y=2×3
Switch sides
2×3=y
Multiply both sides by 3
3(2×3)=3(y)
Simplify
2x=3y
Divide both sides by 2
2×2=3y2
Simplify
x=3y2
Example
Make x the subject of
y = axb
Switch sides
axb = y
Multiply both sides by b
b(axb) = b(y)
Simplify
ax=by
Divide both sides by a
axa = bya
Simplify
x = bya
Assessment
- If y = 2x – 9. Express x in terms of y. Find x when y = 5
- Make P the subject of the simple interest formula I = PRT/100.
Hence find the principal which earns an interest of N29750 in 7yrs at a rate of 5% per annum - If p = c/d
express d in terms of p and c
find d when p = 3 and c = 5.7 - The wage, w naira of a person who works r hours overtime is given by the formula w = 200r + 5900
Make r the subject of the formula. Hence find the number of hours of overtime worked by someone whose total wage is N8200.
Simple Equations Involving Fractions
JSS 3 Mathematics First Term
Week 9
Topic: Simple Equations Involving Fractions
Do you go blank when you see x, y and z in mathematics? Well, this is your abc to solving equations.
Solving simple equations
In an equation, letters stand for a missing number. To solve an equation, find the values of missing numbers. A typical exam question is:
Solve the equation 2a + 3 = 7
This means we need to find the value of a. The answer is a = 2
There are two methods you can use when solving this type of problem:
Trial and improvement
Using inverses
Trial and improvement
This method involves trying different values until we find one that works.
Look at the equation 2a + 3 = 7
To solve it:
Write down the equation: 2a +3 = 7
Then, choose a value for ‘a’ that looks about right and work out the equation. Try ‘3’.
a = 3, so 2 ×3 + 3 = 9.
Using ‘3’ to represent ‘a’ makes the calculation more than 7, so choose a smaller number for ‘a’.
Try a = 2
2 × 2 + 3 = 7
Which gives the right answer. So a = 2
Be systematic in your approach:
- choose a number
- work it out
- then move the number up or down
However, sometimes the answers are negatives or decimal fractions, and the trialand improvement method will take a long time. Luckily, there is a better method.
Using inverses
The best way to solve an equation is by using ‘inverses’, or undoing what the equation is doing.
To use this method to solve equations remember that:
Adding and subtracting are the inverse (or opposite) of each other.
Multiplying and dividing are the inverse of each other.
This method is explained in the following pages. But for now, here is how to solve the question in the above example using inverses:
First, write down the expression:
2a + 3 = 7
Then, undo the + 3 by subtracting 3. Remember, you need to do it to BOTH sides!
2a + 3 – 3 = 7 – 3,
so 2a = 4.
Undo the multiply by 2 by dividing by 2, again on both sides:
2a ÷ 2 = 4 ÷ 2
The answer is: a = 2
Solution of an equation
It is usually possible to find the value of the unknown which makes an equation true. We call this value the solution of the equation.
x = 6 is the solution of 3x = 18.
To solve an equation means to find its solution.
Example
Solve the equation 18 – x = 7
The problem is to find a number which when taken from 18 gives 7. The number is 11.
x = 11 is the solution
Find the solution of 3x = 15.
Which number multiplied by 3 gives 15? The number is 5. Thus
x = 5.
Exercise
Solve the following equations
1. 20 + x = 28
2. 14 – x = 11
3. x – 2 = 15
4. 4x = 20
5. 3x + 4 = 17
1. Read the problem carefully and figure out what it is asking you to find.
Usually, but not always, you can find this information at the end of the problem.
2. Assign a variable to the quantity you are trying to find.
Most people choose to use x, but feel free to use any variable you like. For example, if you are being asked to find a number, some students like to use the variable n. It is your choice.
3. Write down what the variable represents.
At the time you decide what the variable will represent, you may think there is no need to write that down in words. However, by the time you read the problem several more times and solve the equation, it is easy to forget where you started.
4. Re-read the problem and write an equation for the quantities given in the problem.
This is where most students feel they have the most trouble. The only way to truly master this step is through lots of practice. Be prepared to do a lot of problems.
5. Solve the equation.
The examples done in this lesson will be linear equations. Solutions will be shown, but may not be as detailed as you would like. If you need to see additional examples of linear equations worked out completely, click here. (link to linear equations solving.doc)
6. Answer the question in the problem.
Just because you found an answer to your equation does not necessarily mean you are finished with the problem. Many times you will need to take the answer you get from the equation and use it in some other way to answer the question originally given in the problem.
7. Check your solution.
Your answer should not only make sense logically, but it should also make the equation true. If you are asked for a time value and end up with a negative number, this should indicate that you’ve made an error somewhere. If you are asked how fast a person is running and give an answer of 700 miles per hour, again you should be worried that there is an error. If you substitute these unreasonable answers into the equation you used in step 4 and it makes the equation true, then you should re-think the validity of your equation.
Let’s Practice:
1. 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 = 44
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.
Step 7: Check the answer.
The answer makes sense and checks in our equation from Step 4.
6 + 4(11) = 6 + 44 = 50
2. The sum of a number and 9 is multiplied by -2 and the answer is -8. 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 know that we have the sum of a number and 9 which will give us n + 9. We are then told to multiply that by -2, so we have -2n (n+9). Be very careful with your parentheses here. The way this is worded indicates that we find the sum first and then multiply. We also know the answer is -8. So we will solve -2(n+9) = -8
Step 5: Solve the equation.
-2(n+9) = -8
-2n – 18 = -8
-2n=10
n=-5
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 = -5. The number we are looking for is -5.
Step 7: Check the answer.
The answer makes sense and checks in our equation from Step 4.
-2(n+9)=-2(-5+9)=-2(4)=-8
- On an algebra test, the highest grade was 42 points higher than the lowest grade. The sum of the two grades was 138. Find the lowest grade.
Step 1: What are we trying to find?
The lowest grade on an algebra test.
Step 2: Assign a variable for the lowest test grade.
Let’s call it l.
Step 3: Write down what the variable represents.
Let l = the lowest grade
Step 4: Write an equation.
Whatever the lowest grade is, we are told that the highest grade is 42 points higher than that. That means we need to add 42 to the lowest grade. This tells us the highest grade is l+42. We also know that the highest grade added to the lowest grade is 138. So, (highest grade) + (lowest grade) = 142. In terms of our variable, (l+42)+(l)=138
Step 5: Solve the equation.
(l+42)+(l)=138
2l+42=138
2l=96
l=48
Step 6: Answer the question in the problem
The problem asks us to find the lowest grade. We decided that l would be the number, so we have l = 48. The lowest grade on the algebra test was 48.
Step 7: Check the answer.
The answer makes sense and checks in our equation from Step 4.
(48+42)+(48)=90+48=138
Solve the Equation x – 4 = 2 – x
5 2
x – 4 = 2 – x
5 2
There are two denominators , 5 and 2. Their LCM is 10. Multiply each term in the equation by 10.
10x – 4 = 10 x 2 – 10 x x
5 2
2(x – 4) = 20 – 5x
Clear brackets
2x – 8 = 20 – 5x
Add 5x to both sides
2x + 5x – 8 = 20 – 5x + 5x
7x – 8 = 20
Add 8 to both sides
7x -8 + 8 = 20 + 8
7x = 28
Divide both sides by 7
7x/7 = 28/7
x = 4
Assessment
- x/9 = 2
- 2(4x – 1) = 9(x + 1)
3 4 - 2x = 5x + 1 = 3x – 5
7 2 - 3/4 = a/20
- h = 18 + 5h
7
Converting Numbers to Bases
JSS 3 Mathematics First Term
Week 10
Topic: PROBLEM SOLVING ON NUMBER BASES EXPANSION, CONVERSION AND RELATIONSHIP
Converting from base b to base 10
The next natural question is: how do we convert a number from another base into base 10? For example, what does 42015 mean? Just like base 10, the first digit to the left of the decimal place tells us how many 50’s we have, the second tells us how many 51’s we have, and so forth. Therefore:
42015 = (4.53 + 2.52 + 0.51 + 1.50)10
= 4.125 + 2.25 + 1
=55110
From here, we can generalize. Let x =(anan-1 … a1a0)b be an n + 1 -digit number in base b. In our example (274610) a3 = 2, a2 = 7, a1 = 4 and a0 = 6. We convert this to base 10 as follows:
x = (anan-1 … a1a0)b
= (bn.an + bn-1 . an-1 + … + b.a1 + a0)10
Converting from base 10 to base b
It turns out that converting from base 10 to other bases is far harder for us than converting from other bases to base 10. This shouldn’t be a suprise, though. We work in base 10 all the time so we are naturally less comfortable with other bases. Nonetheless, it is important to understand how to convert from base 10 into other bases.
We’ll look at two methods for converting from base 10 to other bases.
Method 1
Let’s try converting 1000 base 10 into base 7. Basically, we are trying to find the solution to the equation 1000 = a0 + 7a1 + 49a2 + 343a3 + 2401a4 + …
where all the are digits from 0 to 6. Obviously, all the ai from a4 and up are 0 since otherwise they will add in a number greater than 1000, and all the terms in the sum are nonnegative. Then, we wish to find the largest a3 such that 343a3 does not exceed 1000. Thus, a3 = 2 since 2a3 = 686 and 3a3 = 1029.This leaves us with 1000 = a0 + 7a1 + 7a1 + 49a2 + 343(2) ↔ 314 = a0 +7a1 + 49a2.
Using similar reasoning, we find that a2 = 6, leaving us with 20 = a0 + 7a1.
We use the same procedure twice more to get that a1 = 2 and a0 = 6.
Finally, we have that 100010 = 26267.
An alternative version of method 1 is to find the “digits” a0, a1,… starting with a0. Note that is just the remainder of division of 1000 by 7. So, to find it, all we need to do is to carry out one division with remainder. We have 1000:7 = 142(R6). How do we find a1, now? It turns out that all we need to do is to find the remainder of the division of the quotient 142 by 7: 142:7 = 20(R2), so a1 = 2. Now, 20:7 = 2(R6), so a3 = 6. Finally, 2:7 = 0(R2), so . We may continue to divide beyond this point, of course, but it is clear that we will just get 0:7 = 0(R0) during each step.
Note that both versions of this method use computations in base 10.
It’s often a good idea to double check by converting your answer back into base 10, since this conversion is easier to do. We know that 26267 = 343.2 + 6.49 + 2.7 = 1000, so we can rest assured we got the right answer.
Method 2
We’ll exhibit the second method with the same problem used to exhibit the first method.
The second method is just like how we converted from other bases into base 10. To do this, we pretend that our standard number system is base 7. In base 7, however, there is no digit 7. So 7 is actually represented as “10.” Also, the multiplication rules we know do not hold. For example, 3.3 ≠ 9 (in base 7). For one, there is no 9 in base 7. Second, we need to go back to the definition of multiplication to fully understand what’s happening. Multiplication is a shorthand for repeated addition. So, 3.3 = 3 + 3 + 3 = 127.
In base 7, we have that 10 (the decimal number 10) is 13. Thus, if we view everything from base 7, we are actually converting 100013 to base 10. So, this is just 133. Remember that we aren’t doing this in our regular decimal system, so 133 ≠ 2197. Instead, we have to compute 13 x 13 x 13 as (13 x 13) x 13 = 202 x 13 = 2626.
This method can be very confusing unless you have a very firm grasp on the notion of number systems.
Binary and Hexadecimal
Two of the most common number bases are binary (base 2) and hexadecimal (base 16). Both of these are used in computer science. A binary number consists entirely of 0s and 1s. Each binary digit is called a bit. A base 16 number requires additional symbols after 9 since any positive integer less than 16 is a single digit. The standard notation is to use the letters a=10,b=11,c=12,d=13,e=14,f=15. In order to indicate that a number is written in hexadecimal, the prefix 0x is used. For instance, 0x95 is the same as 9516 and is equivalent to 9 x 161 + 5 = 149.
Examples
1. Convert the binary number 1111 1110 to decimal. (A space is often inserted every 4 digits to improve readability of binary numbers. A group of 4 digits (bits) is called a nible and a group of 8 bits is called a byte)
We could simply add all of the digit values: 0 x 1 + 1 x 2 + 1 x 4 + 1 x 8 + 1 x 16 + 1 x 32 + 1 x 64 + 1 x 128. But, there is a much shorter way. Note that in base 10, 99 = 102 – 1 and 999 = 103 – 1. Similarly, in base 2, 11 = 22 – 1, 111 = 23 – 1, and so on. Therefore, 1111 1111 = 28 – 1 = 255. The number we are trying to convert is one less, so it must be 254.
2. Convert the hexadecimal number 0xA7 into decimal.
Since A means ten and the A is in the “16s place,” we have 0x A7 = 10 x 16 + 7 = 167
3. Convert 700 into hexadecimal. 162 = 256, so 700 will require three digits. Using Method 1 above, we want to solve:
700 = a0 + 16a1 + 256a2
700/256 = 2 remainder 188, so a2 = 2
188 = a0 + 16a1
188/16 = 11 remainder 12, so a1 = 11 and a0 = 12 we have
700 = 12 + 16 x 11 + 256 x 2
Using standard hexadecimal notation, we can write
700 = 0x2BC
Numeral systems conversion table
| DecimalBase-10 | BinaryBase-2 | OctalBase-8 | HexadecimalBase-16 |
| 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 |
| 2 | 10 | 2 | 2 |
| 3 | 11 | 3 | 3 |
| 4 | 100 | 4 | 4 |
| 5 | 101 | 5 | 5 |
| 6 | 110 | 6 | 6 |
| 7 | 111 | 7 | 7 |
| 8 | 1000 | 10 | 8 |
| 9 | 1001 | 11 | 9 |
| 10 | 1010 | 12 | A |
| 11 | 1011 | 13 | B |
| 12 | 1100 | 14 | C |
| 13 | 1101 | 15 | D |
| 14 | 1110 | 16 | E |
| 15 | 1111 | 17 | F |
| 16 | 10000 | 20 | 10 |
| 17 | 10001 | 21 | 11 |
| 18 | 10010 | 22 | 12 |
| 19 | 10011 | 23 | 13 |
| 20 | 10100 | 24 | 14 |
| 21 | 10101 | 25 | 15 |
| 22 | 10110 | 26 | 16 |
| 23 | 10111 | 27 | 17 |
| 24 | 11000 | 30 | 18 |
| 25 | 11001 | 31 | 19 |
| 26 | 11010 | 32 | 1A |
| 27 | 11011 | 33 | 1B |
| 28 | 11100 | 34 | 1C |
| 29 | 11101 | 35 | 1D |
| 30 | 11110 | 36 | 1E |
| 31 | 11111 | 37 | 1F |
| 32 | 100000 | 40 | 20 |
Assessment
Convert the following number into hexadecimal:
- 500
- 800
- 40
- 950