Applications of Expressions, Equations and Inequalities

Hello learner! In this section, we will delve into the world of number patterns, where we will discover hidden structures and relationships between numbers. We will learn how to represent these patterns using algebraic expressions, and how to perform operations such as simplification, addition, subtraction, multiplication, and division. We will again delve into perfect squares; learning how to identify and use them in solving equations and inequalities, including difference of two squares of binomials. Perfect squares have unique properties and play a significant role in various mathematical operations such as factorising expressions, simplifying expressions and solving equations. Finally, we shall explore algebraic fractions by performing operations on simple fractions with monomial and binomial denominators and finally, investigate the conditions under which these fractions become zero or undefined. This will help us in mastering the art of analysing and simplifying expressions. These also have real-world applications in fields like geometry, trigonometry and calculus.

At the end of this section, you will be able to:

• Use number patterns and variables to formulate mathematical expressions and apply the algebraic order of the four operations to solve.

• Factorisation of algebraic expressions.

• Recognise perfect squares and apply the idea to solve problems including the difference of two squares of binomials.

• Analyse and apply operations on simple algebraic fractions including monomial and binomial denominators and determine the conditions under which algebraic fraction is zero or undefined.

Key ideas

• Number Patterns: These are sequences or patterns that help in recognising relationships and predicting outcomes. For example, recognising that the sequence 2, 4, 6, 8 follows a pattern of increasing by 2.

• Variables: These are unknown quantities that are represented by letters or symbols.

• Factorisation of algebraic expressions: Algebraic expressions can be broken down into factors. For example, x²– 4 can be factorised as (x – 2) (x + 2).

• A perfect square is an expression that can be written in the form (x)²or x², where x is a variable or an expression. For example, 4, 9, and 16 are perfect squares because they can be written as 2², 3², and 4²respectively.

• The difference of two squares formula is: a²− b²= (a + b)(a − b). This formula can be used to factorise expressions like x²− 4 or 9 − y².

• Algebraic Fractions: An algebraic fraction is an expression in the form a/b, where a and b are algebraic expressions. To perform operations on algebraic fractions, we can follow these rules:

• An algebraic fraction is zero if the numerator (a) is zero, and it is undefined if the denominator (b) is zero.

Number patterns are sequences of numbers that follow a certain rule or formula.

These patterns can be found all around us, from the natural world to man-made structures. Understanding number patterns helps us make predictions, solve problems, and uncover the underlying order in seemingly random sets of numbers.

Let us take a look at some examples of patterns in real life.

From the pictures below, we realise that patterns can be found in all aspect of our lives. From our clothes, home decorations, building designs and even animals on both land and sea, we see various beautiful patterns. Can you identify some patterns around you? Share with a friend.

In Junior High School, we also learnt about patterns from shapes and numbers.

Let us take a look at some examples.

Take a look at the pattern above, it is made up of circles. Can you determine how each successive group of circles came about? We realise that each next pattern is generated by adding the next counting number to the preceding pattern, right?

But assuming we are to determine the number of circles in the 100ᵗʰterm. Could that be easy? I’m sure you agree with me that it will be a challenging task. So, we will need to find a smart way to do that. This brings us to the need to find the rule or nth term rule. This will help us to be able to find any term in any given pattern.

Now in this lesson our focus will be on writing algebraic expressions for the rules of the patterns that we will explore. Then in SHS 2 we will turn our attention to exploring patterns and sequences including arithmetic and geometric patterns.

Dear Learner,

Now that we have explored Number Patterns, Algebraic Expressions, and their Operations, let’s apply what we’ve learned through a series of engaging activities.

Please follow the instructions below for each activity.

Activity 1

Let’s look at this pattern:

Pattern: 2, 5, 8, 11, 14, ...

Materials: Bottle tops.

Rule:

• Use bottle tops to create the above pattern.

• We start with 2 bottle tops.

• Each time, we add 3 bottle tops. This means that to go from ‘term to term’ to find the next term we + 3 Algebraic Expression: Let aₙ represent nth term in the pattern. The expression can be written as:

aₙ = 3n − 1 Where n is the position of the term in the pattern. Hence, the algebraic expression is 3n − 1 .

Activity 2

The pattern below is made up of squares, made from lines, as shown below.

Investigate the pattern, find the nth term rule and write an algebraic expression for the number of lines required to make the pattern, if the pattern continues.

Fig.1 Fig.2 Fig.3

From the pattern, we realise that, Fig 1 Fig 2 Fig 3 Nᵗʰ 4 7 10 ?

From the table let represent each term with n . Now, if we multiply n by 3 and add 1 we obtain the number of lines needed to create the shape in each term.

Therefore, our rule for the pattern is aₙ = 3n + 1 . Remember, 3n + 1 is an algebraic expression.

Activity 3

Ama is collecting seashells on the beach. On her first day, she collected 1 seashell. Each day after that she collected 2 more seashells than the previous day. How many seashells will she have on the fifth day? Let’s generate a pattern from the above and use it to formulate mathematical expression.

Step 1: Identify the pattern Day 1: 1 seashell Day 2: 1 + 2 = 3 seashells Day 3: 3 + 2 = 5 seashells Day 4: 5 + 2 = 7 seashells Day 5: 7 + 2 = 9 seashells

Step 2: Formulate the mathematical expression The pattern shows that each day, the number of seashells increases by 2.

Therefore, the mathematical expression is:

S = 1 + 2(n – 1) = 1 + 2n − 2 = 2n − 1 where S is the total number of seashells and n is the day number.

Step 3: Use the expression to find the number of seashells on the fifth day Substitute n = 5 into the expression:

S = 2(5) – 1 S = 10 – 1 S = 9 Therefore, Ama will have 9 seashells on the fifth day.

Have you ever heard of the term “Algebraic expression”? if you do that’s very good, if you do not let’s now look at the meaning of Algebraic expression.

An Algebraic expression in mathematics is an expression which is made up of variables, constants and arithmetic operations. (addition, subtraction, multiplication and division) Some examples include:

3x – 5 , n + 8 , x², 2 x²+ 3x – 5 and 2x + 5xy + 3 .

Definition of Key Concepts

1. Term: is the part of the expression, e.g. x²+ 5x – 4 (three terms)

2. Variable: is a symbol (usually a letter), used to represent one or more numbers in algebraic expression (e.g. from above, x, n , y are the variables)

3. Monomials: an algebraic expression with only one term (e.g.: y, 2 x², − 5a etc.)

4. Binomial: an algebraic expression with two terms (e.g.: 7 x²+ 3, y − 2, 4x + 9, n − m etc.)

5. Trinomial: algebraic expressions with three terms (e.g.: 2 x²+ 5x − 3, a − 5b + 2c, 8x + 5y − z )

6. Coefficient: is a number attached to a variable in an algebraic expression.

Consider the expressions below and their corresponding coefficients x + 5x – 3 9a – 3b + 2 1 5 9 -3

7. Constant term: is a number or symbol which is not attached to any variable in an expression. From the above expressions the constant terms are -3 and 2.

The diagram below gives a summary of the terms explained.

Formulation of Algebraic Expressions

Algebraic expressions can be created using models and variables.

Application of concepts in real world activities

1. a) Nine more than a certain number (x )

Solution: x + 9

b) Two less than one-third of a certain number (x ) is

Solution: 1__ 3x − 2

c) Twice the square of a number (y) minus the cube of another number (n).

Solution: 2 y²− n³d) The age of Mr. Mensah is thrice his son’s age (a) plus ten.

Solution: 3a + 10

e) There are 25 oranges in a bag. Write the algebraic expression for the number of oranges in x number of bags.

Solution: 25x Rules for the Use of the Operations of Algebraic Expression

2. Addition and Subtraction

In algebra, you can only add or subtract like terms, example 3x + 5x = 8x, 2y – 5y = − 3y , 7a + 5b = 7a + 5b

i. Like terms: Two or more terms are like terms if they have the same variables with the same exponent irrespective of their numerical coefficients. They can be added or subtracted to get a single term. For example: 3 x²and 7 x², 4xy and 2xy , 5 a³b and − 7 a³b

ii. Unlike terms: are terms that cannot be added or subtracted in an expression to get a single term. Example: 3 x²and a², 5y and 4xy

3. Multiplication and Division

Both like terms as well as unlike terms can be multiplied and divided. In multiplying, we make use of distributive property depending on the expression given. To divide, look for factors that are common to both numerator and denominator and this can be divided or cancelled. Factors that are common to all terms of an expression can be factored out.

Simplifying Algebraic Expression

Example 1

Simplify the following expressions.

a. x + 2y + 5x – y

b. 5p− c – 9c

c. 4x × 2y

d. x²( x³− 2y )

e. 10b ÷ 2b Solutions

a. 6x − y

b. 5p − 10c

c. 8xy

d. x⁵− 2 x²y

e. 5b

Activity 4

Expand (2x + 3)(x – 4)

Step 1: Multiply the two binomials using the distributive property:

(2x + 3)(x – 4) = 2x(x) – 2x(4) + 3(x) – 3(4)

Step 2: Multiply the terms:

2x × (x) = 2 x²2x × ( − 4) = − 8x 3 × (x) = 3x 3( − 4) = − 12

Step 3: Combine like terms:

2 x²– 8x + 3x – 12

Step 4: Simplify the expression by combining like terms:

2 x²– 5x – 12 Therefore, the expanded form of (2x + 3)(x – 4) is 2 x²– 5x – 12 We’ve explored number patterns, learned how to identify and describe the rules that govern them and also how to formulate algebraic expression from the number patterns. Let’s now look at factorisation of algebraic expressions.

In factorising algebraic expressions, we look for like terms, group them and find common factors. Factorising is the reverse process of expanding brackets in algebraic expressions. Algebraic expressions could be factorised in many ways depending on the given expression(s) and it includes the common factor approach, algebraic tiles, regrouping of terms approach, standard identity approach, splitting- the-middle-term approach and using the quadratic formula. For instance;

x²+ 6x + 5 (x + 5)(x + 1) Factorising Expanding brackets 3x + 6 3(x + 2) Factorising Expanding brackets Worked Examples Factorise the following

1. 4x + 4by

2. ac + bc + ad + bd Solutions

1. Look for the common factor(s) Factor the common term out (Highest common factor) 4x + 4by = 4(x + by)

2. First put the four terms in the expression into two groups of two and find the common factors from each. i.e., (ac + bc)+ (ad + bd) = c(a + b) + d(a + b) and add the outside terms and take one of the common terms to be the final answer. (a+ b)(c + d).

There are many ways to factorise algebraic expressions. Factorisation is an important skill in algebra that helps us solve equations. We’ll look at different methods to factorise expressions, and each method is useful in different situations.

By learning these methods, you’ll become good at factorising expressions and solving equations.

Factorising and Solving Equations Using Algebraic

Tiles

Activity 5

Solve x + ( − 2) = 5 using the algebraic tiles

Solution

Step 1

Let’s represent the equation with our algebraic tiles

Step 2:

Transpose -2 by adding 2 to both sides (yellow squares are positive while red squares are negative numbers)

Step 3:

I hope you understand factorisation by using algebraic tiles, Let’s explore another method of factorisation, called Factorisation by Regrouping of Terms.

Factorisation by Regrouping of Terms Approach

In some algebraic expressions, not every term may have a common factor. For instance, consider the algebraic expression, 12a + n – na –12. The terms of this expression do not have a particular factor in common but the first and last term has a common factor of ‘12’. Similarly, the second and third terms have n as a common factor. So, the terms can be regrouped as:

⇒ 12a + n – na – 12 = 12a – 12 + n – an ⇒ 12a – 12 – an + n = 12(a – 1) –n(a – 1) After regrouping, it can be seen that (a − 1) is a common factor in each term, ⇒ 12a + n – na – 12 = (a – 1) (12 – n) Thus, by regrouping terms we can factorise algebraic expressions.

Let’s now engage in an activity using the factorisation by regrouping of terms approach:

Activity 6

Example: Factorise the expression: x²+ 5x + 6 Follow these steps below:

Steps:

Step 1: Write down the expression: x²+ 5x + 6

Step 2: Look for two numbers whose product is 6 (the constant term) and whose sum is 5 (the coefficient of the linear (or x) term). These numbers are 2 and 3.

Step 3: Rewrite the expression by regrouping the terms: x²+ 2x + 3x + 6

Step 4: Factor out the common term from each group: x(x + 2) + 3(x + 2)

Step 5: Factor out the common binomial term: (x + 3)(x + 2) Therefore, the factorised form of the expression is (x + 3)(x + 2).

I hope you’re familiar with factorisation through regrouping terms. Now, let’s dive into another exciting method: factorising expressions using standard identities.

Factorising expressions using standard identities An equality relation is one which holds true for all the values of variables in mathematics. It is known as an identity and has this sign, ≡ , rather than the standard = sign. Consider the following identities:

(a + b)²≡ a²+ b²+ 2ab (a − b)²≡ a²+ b²− 2ab a²− b²≡ (a + b)(a − b) On substituting any value of a and b , both sides of the given equations remain the same. Therefore, these equations are called identities.

Hello learner! Let’s engage in this activity by applying some standard identities which suits our example.

Activity 7

Example: x²− 4 Follow these steps of Activities

Step 1: Identify the Expression:

Given the expression x²− 4

Step 2: Recognise the Pattern:

The expression fits the pattern of the difference of two squares: a²− b²Step 3: Rewrite the Expression:

Rewrite 4 as 2²so the expression becomes x²− 2²Step 4: Apply the Identity:

• Using the identity a²− b²≡ (a + b)(a − b)

• Here, a = x, b = 2

Step 5: Simplify and Combine:

Apply the identity: x²− 2²≡ (x + 2)(x − 2) Thus, the factorised form of x²− 4 is (x + 2)(x − 2) Rectangle A rectangle is a quadrilateral with opposite pair of sides equal The side AB = DC = Length(L) The side BC = AD = Width(W) The opposite sides are parallel and equal Area of a rectangle = Length(L) × Width(W) Area of a rectangle = L × W

Note that area is expressed in square units Let’s try these two examples on area of a rectangle

Example 2

The length of a rectangle is 4 cm and its width (breadth) is 3 cm. Find its area.

Solution

Area of a rectangle = Length × Width = 4 cm × 3 cm = 12 cm² Therefore, the area of the rectangle is 12 cm²Example 3 The picture below is Mr. Bluba’s rectangular garden, where he grows cabbage and Kontomire, measures 8m in length and 3m in width. What is the total area of the garden?

Solution

Area of the garden = Length of the garden × Width of the garden = 8m × 3 m = 24m² Thus, the area of the rectangular garden is 24m² Think of it like this! If you have a room with a length of 10 feet and a width of 5 feet, the area would be 10 x 5 = 50 square feet. That’s the amount of space you’d need to cover with flooring, paint, or carpet!

Now that we’ve learned about the area of a rectangle, let’s use this knowledge to explore how to factorise quadratic trinomials using algebraic tiles.

Factorise Quadratic Trinomials Using Algebraic

Tiles

Example 4

Factorise x²+ 7x + 12 using algebraic tiles

Solution

Use algebraic tiles Let’s undertake this activity on how to factorise quadratic trinomials using algebraic tiles

Activity 8

Let’s factorise x²+ x− 6 using trinomial algebraic tiles:

Step 1: Represent the expression as tiles:

x²= (x × x) x = (1 × x)

– 6 = (– 6 × 1)

Step 2: Arrange the tiles to form a rectangle:

(x × x) + (3 × x) + ( − 2 × x) + ( − 6 × 1)

Step 3: Identify the common factors:

(x × x) + (3 × x) = x(x + 3) ( − 2 × x) + ( − 6 × 1) = − 2(x + 3)

Step 4: Combine the factors:

x(x + 3) – 2(x + 3)

Step 5: Factor out the common binomial term:

(x – 2)(x + 3) Factorising Quadratic Trinomials Using the Standard Identities

Example 5

Factorise x²+ 6x + 9

Solution

x²+ 6x + 9 = 0 The identity is a²+ 2ab +b²≡ (a + b)²The LHS is of the form a²+ 2ab + b², ⇒ (x + 3)²Or (x + 3) (x + 3) Factors are (x + 3) and (x + 3) Let’s now look at how to factorise using quadratic equations using formula approach.

Factorising Quadratic Equation Using Formula

Approach This method is very similar to the method of splitting the middle term.

Step 1: Consider the quadratic equation ax²+ bx + c = 0

Step 2: Now, find two numbers such that their product is equal to ac and sum equals to b.

(number 1) (number 2) = ac (number 1) + (number 2) = b

Step 3: Substitute these two numbers in the formula given below:

(1 / a) [ax + (number 1)] [ax + (number 2)] = 0

Step 4: Finally simplify the equation

Example 6

Solve 3 x²+ 7x + 4 = 0 Follow the steps carefully

Solution

3 x²+ 7x + 4 = 0 Here, a = 3, b = 7, c = 4 ac = (3)(4) = 12 Let’s identify two numbers such that their sum is 7 and the product is 12.

Factors of 12: 1, 2, 3, 4, 6, 12 Sum of two factors = 7 Product of those two factors = 12 Number 1 = 3 and number 2 Now, substitute these two numbers in the formula (1/a) [ax + (number 1)] [ax + (number 2)] = 0.

(1/3) (3x + 3) (3x + 4) = 0 (3x + 3) (3x + 4) = 0 1/3 × 3 (x + 1) (3x + 4) = 0 (x + 1)(3x + 4) = 0 Thus, (x + 1) and (3x + 4) are the factors of the given quadratic equation.

Learners, now that we’ve mastered factoring quadratic equation using the formula approach. Let’s explore another fascinating method of factorisation of quadratic equation by splitting the middle term Get ready to dive into this exciting new technique.

Factorisation of Quadratic Equations by Splitting

the Middle Term

Step 1: Consider the quadratic equation a x²+ bx + c = 0

Step 2: Now, find two numbers such that their product is equal to ac and sum equals to b .

(number 1)(number 2) = ac (number 1) + (number 2) = b

Step 3: Now, split the middle term using these two numbers, a x²+ (number 1) x + (number 2) x + c = 0

Step 4: Take the common factors out and simplify.

Let’s have a look at the example problem given below:

Example 7

Factorise and then solve the quadratic equation x²+ 7x + 10 = 0 by splitting the middle term Given, x²+ 7x + 10 = 0 Here, a = 1, b = 7, c = 10 ac = (1) (10) = 10 Factors of 10: 1, 2, 5, 10 Let’s identify two of these factors such that their sum is 7 and their product is 10.

Sum of two factors = 7 = 2 + 5 Product of these two factors = (2)(5) = 10 Now, split the middle term.

x²+ 2x + 5x + 10 = 0 Take the common terms and simplify.

x(x + 2) + 5(x + 2) = 0 (x + 5)(x + 2) = 0 Thus, (x + 2) and (x + 5) are the factors of the given quadratic equation.

x + 2 = 0 x = 0 − 2 x = − 2 x + 5 = 0 x = 0 − 5 x = − 5 Solving these two linear factors, we get x = − 2, − 5 as the roots.

Factorising Quadratic Equation Using Quadratic

Formula In quadratic formula, to get the roots of quadratic equation a x²+ bx + c = 0 is given by x = − b ± √(b²− 4ac)_____________ 2a Substituting the values of a, b, c and simplifying the expression, we get the values of roots.

Great job, everyone! Now that we’ve covered how to factor quadratic equations using the quadratic formula, let’s put what we’ve learned into practice with a fun

activity.

Activity 9

Given:

x²+ 4x – 21 = 0 Here, a = 1, b = 4, c = − 21 b²− 4ac = (4)²– 4(1) ( − 21) = 16 + 84 = 100 Substituting these values in the quadratic formula, we get x = − 4 + 10/2(1) = − 4 ± 10/2 x = − 4 + 10/2 , or x = − 4 – 10/2 x = 6/2, or x = −14/2 x = 3, or x = − 7 Therefore, factors of the given quadratic equation are (x – 3) and (x + 7).

Can you think of integers that can be expressed as the square of another integer? I hope you have written numbers such as 1 = 1², 4 = 2², 16 = 4²and many others.

These numbers; 1, 4, 16, ... are what we call ‘Perfect Squares’. How would you describe ‘perfect squares’ in your own words?

What are Perfect Squares?

A perfect square is an integer that can be expressed as the square of another integer.

E.g. 4 is a perfect square of 2, because it can be expressed as 2²= 2 × 2 = 4. Other examples such as 1, 9, and 16 are perfect squares because they can be written as 1², 3²and 4²respectively.

It is important to note the following;

• any number that cannot be expressed as the square of another number (integer) is not a perfect square. For example; 2 is not a perfect square because it cannot be expressed as the square of another integer.

• negative numbers are not considered as perfect squares. This is because negative numbers cannot be expressed as the square of another integer. For

example, − 49 ≠ 7²or (− 7)²Let us go through the activity below on how to generate some perfect squares by adding consecutive odd numbers.

Activity 10

List the first 5 consecutive odd numbers.

Add consecutive odd numbers to generate perfect squares 1, 2, 3, 4, 5, 6, 7, 8, 9, 10… 1 =1 = 1² 1 + 3 = 4 = 2² 1 + 3 + 5 = 9 = 3² 1 + 3 + 5 + 7 = 16 = 4² 1 + 3 + 5 + 7 + 9 = 25 = 5² 1 hope you have observed the pattern and can follow through to generate more of these perfect squares. Write them down and compare it to that of your friend(s).

Let us now look us how to use the graph or squared paper to generate square numbers in the activity below.

Activity 11

1. Take a graph or squared paper and draw a square on it with a side length of 1 unit (centimetres will fit better than inches).

2. Draw another square to the right of the first one with a side length of 1 unit and extend it into a new square.

3. Shade the area of the new square obtained.

4. Count the number of grid squares in the shaded area.

5. Record the results:

- Side length of outer square

- Number of grid squares in shaded area (area of inner square)

6. Repeat steps 2-5 for different side lengths.

7. Look for patterns and relationships between the side lengths and areas.

Hello learner! I hope you have performed the activity, observed the pattern and can continue with that of 36, 49, 64, 81, .... etc and determine their respective differences. Very Good!

However, discuss your observations on the following with a classmate;

- How does the area of the larger square change as the side length increases?

- Can you predict the area of a larger square based on its side length?

- What is the pattern between the perfect squares and their areas?

Let us now investigate the difference between two perfect squares. I hope you have an idea.

Perfect squares can be used to illustrate the concept of difference of two squares.

Here’s how to go about it:

Activity 12

Let’s consider two perfect squares, say:

a²= a × a (a squared) b²= b × b (b squared) Now, let’s find the difference between these two squares:

a²− b²We can rewrite this expression as:

(a × a) - (b × b) Next, we can factorise the expression by grouping the terms:

(a × a) - (b × b) = (a + b)(a - b) Very Good! We’ve arrived at the difference of two squares formula:

a²− b²= (a + b)(a − b) This formula shows that the difference between two squares can be factorised into the sum and difference of the same two terms. This can be a powerful tool for simplifying and evaluating expressions and solving equations.

Also, click on the link to watch the video on the connection between Perfect squares and Difference of two squares ..\Downloads\videoplayback.mp4 Perfect squares and difference of two squares (square with expression as the sides and some taken out and the area of the rest) Fig. 1 Fig. 2 Fig. 3 Fig. 4 Figures are not drawn to scale.

We already know that to find the area of square = side(s) × side(s) Mathematically, Area of a Square = s²1. Fig. 1 above shows a square of side a and its Area= a × a = a²a²is the perfect square of a

2. Fig. 2 shows the result of taking a square of side b from one of the four corners of Fig. 1.

The result from Fig. 2 gives Fig. 3.

The Area of Fig. 3 = Area of region I + Area of region II ((a -b) × b)+(a × (a-b)) = a²− b²Area of Fig. 4 (a+b) (a-b) = Length x breadth Area of Fig. 3 = area of Fig. 4 (a + b) (a - b) = a²– b² Below is a pictorial representation on how to expand difference of two squares.

For example, x²− 4 , y²− 1 , a²− b²etc x²– 4 = (x +2)(x -2) , y²- 1 = (y -1)(y + 1), a²- b²= (a-b)(a+b)

NOTE: Anytime there is a binomial with each term being squared (i.e. having an exponent of 2) and subtraction (minus) as the middle sign then you are guaranteed to have the case of difference of two square.

Again, the exponents of each term must be a multiple of two (2) and the coefficients should be a perfect square.

Caution: In the sum of two squares, it is not possible to factor a sum of two squares using real numbers.

Thus; x²+ 9 ≠ (x + 3)(x + 3) or (x − 3)(x + 3) May I take you back to the previous lesson on expansion of binomials and consider the product, (a + b)(a − b).

Using the distributive property to expand gives us;

(a + b)(a − b) = a(a − b) + b(a − b) = a²− ab + ab − b²= a²− b²⇒ (a + b)(a − b) = a²− b²Let us now explore how to factorise difference of two squares.

Example 8

Factorise x²− 49

Solution

Since the first term (x²) has already been expressed as a perfect square, express 49 as a square of another number and that is 7 × 7 = 7²x²− 49 = x²− 7²= (x + 7)(x − 7)

Example 9

Factorise 25 a²− 16

Solution

25a²− 16 = 5²a²− 4²= 5(a)²− 4²= (5a − 4 )(5a + a)

Example 10

Factorise completely p⁴− 81 q⁴Solution p⁴− 81 q⁴= (p²)²− 9²(q²)²= (p²)²− (9 q²)²= (p²− 9 q²)(p²+ 9 q²) I hope you have observed that (p²− 9 q²) is also a difference of two squares and can be factorised as well. You can proceed to factorise the expression as this;

⟹ (p²− 3²q²)(p²+ 9 q²) = (p − 3)(p + 3)(p²+ 9 q²) Therefore, p⁴− 81 q⁴= (p − 3)(p + 3)(p²+ 9 q²) Can you now try this example?

Example 11

Factorise the expression (m²− n²) − k(m + n) completely

Solution

Since m²− n²is a difference of two squares, ⟹ (m²− n²) − k(m + n) = (m − n)(m + n) − k(m + n) It can also be observed that m + n is common to both terms so we can factorise it out as;

= (m + n)((m − n) − k) = (m + n)((m − n − k) Let us now apply the concept of the difference of two squares in solving real life problems.

Example 12

If the length of a rectangular mirror is (x + 2) units and the width is (x - 2) units, find the area of the mirror?

Solution

Area of a rectangle = L × B = (x + 2) × (x − 2) = (x²− 4) square units

Example 13

Without using calculator, evaluate 21²− 11²completely.

Solution

21²− 11²= (21 − 11)(21 + 11) = (10)(32) = 320

Example 14

Without using calculator, evaluate 9R + 48 = 75 hence, find the value(s) of R

Solution

9R = 75 − 48 9R = 3(25 − 16) 9R = 3(5²− 4²) 9R = 3(5 − 4)(5 + 4) 9R = 3(1)(9) 9R_ 9 = 27_ 9 R = 3 Do look for more examples, try your hands on them together with your classmates and compare your answers.

I hope we are making positive progress. Very Good!

Let us now consider how to operate on fractions that contain variable(s) either as the numerator or denominator or even both.

An algebraic fraction is a fraction that contains at least one variable. For example, x/2, x (the variable) is the numerator.

3/x + 1, the denominator is an expression in terms of x , x + 5_ 2x , both the numerator and the denominator contain an x (variable) term Just as ordinary fractions, you can add, subtract, multiply and divide them.

Application of Concepts

To simplify algebraic fractions, we do it the same way as numerical fractions. We cancel common factors from the numerator and denominator until no common factors remain.

Note the following:

1. Algebraic fractions can be added or subtracted if they have the same denominators. (a common denominator).

4__ 3x + 1__ 3x or 2x/5 − 3y/5 or x/x − 1 + 5/x − 1 and use the Lowest Common Multiple (LCM) approach when the denominators are not the same.

2. When the fraction is multiplying, such as m_ n × a_ b = ma_ nb where there are common factors, then cancel out if not, multiply the numerators and the denominators separately.

3. When a fraction is divided by another, multiply the first fraction by the reciprocal of the second fraction. Thus m_ n ÷ a_ b = m_ n × b_ a = mb_ na etc.

Operations on Algebraic Fractions Including

Monomial and Binomial Denominators

Hello Learner! In week 9, you have looked at monomials, binomials and trinomial algebraic expressions. Here, you are going to explore how to perform the four basic operations (+ , − , × , ÷) on Algebraic fractions involving monomial, binomial denominators. Let us begin with;

Addition and Subtraction of Algebraic Fractions with

Monomial Denominators

Can you do a quick recall of how to add and subtract simple fractions? Perfect.

Just as you learnt how to add and subtract ordinary fractions in Week 6, addition and subtraction of algebraic fractions also follow the same procedure. This means that, you need to first identify the denominators of the algebraic fractions and find their lowest common multiple (LCM). Use the LCM. as their common denominator and express the fractions as a single fraction.

Let us go through an activity to express 2/a +3/b as a single fraction. Here are the steps:

Activity 13

Step 1: Identify the denominators of the two fractions. They are a and b.

Step 2: Find the LCM of a and b. This will be the common denominator. The LCM is ab

Step 3: Rewrite each fraction with the LCM as the denominator:

2/a = 2 × b____ ab (b is the multiplier to make the denominator ab) 3/b = 3 × a____ ab (a is the multiplier to make the denominator ab)

Step 4: Add the fractions and simplify if possible ⇒ 2b_ ab + 3a_ ab = (2b + 3a)_ ab So, the final answer is: 2/a + 3/b = (2b + 3a)_______ ab I hope this activity has been helpful. Now take your jotter, try this example and compare your answers to that of your friend.

Example 15

Express 1/x − 3/x²as a single fraction

Solution

LCM = x²1_ x − 3_ x²= 1(x)_ x²− 3_ x²= x − 3_ x²Example 16 Express 4___ 3m + 2/m²− 1___ nm as a single fraction

Solution

The denominators are 3m, m²and nm. Their LCM is 3m²n 4___ 3m + 2/m²− 1___ nm = 4(mn)_____ 3 m²n + 2(3n)____ 3 m²n − 1(3m)_____ 3 m²n = 4mn + 6n − 3m/3 m²n Please look for more examples, try them and show your workings to a classmate.

Addition And Subtraction of Algebraic Fractions with

Binomial Denominators

Addition and subtraction of Algebraic Fractions with binomial denominators also go through the same steps as those with monomial denominators. Let us perform an activity with this example.

Activity 14

Work through the following examples with a classmate to make sure you fully understand what is happening.

Example 17

Simplify 10/x − 4 + 2/x + 1 as a single fraction.

Solution

To express 10/x − 4 + 2/x + 1 as a single fraction, we need to find a common denominator.

Here are some steps to guide you;

Step 1: Identify the denominators of the two fractions: x − 4 and x + 1 .

Step 2: Find the LCM of x − 4 and x + 1 . To do this, we need to find the LCM of the two expressions by multiplying both expressions together because they do not have common factors:

LCM = (x − 4)(x + 1) which can also be expressed as x²− 3x – 4

Step 3: Rewrite each fraction with the LCM as the denominator:

10/x − 4 = 10(x + 1)__________ (x − 4)(x + 1) = 10x + 10__________ (x − 4)(x + 1) 2/x + 1 = 2 × (x − 4)__________ (x + 1)(x − 4) = 2x − 8__________ (x − 4)(x + 1)

Step 4: Add the two fractions:

10x + 10__________ (x − 4)(x + 1) + 2x − 8__________ (x − 4)(x + 1) = 10x + 10 + 2x − 8_____________ (x − 4)(x + 1)

Step 5: Simplify the fraction, if possible:

Simplifying this gives us; 12x + 2__________ (x − 4)(x + 1) So, the final answer is:

10/x − 4 + 2/x + 1 = 12x + 2__________ (x − 4)(x + 1) or 12x + 2/x²− 3x − 4 There you have it!

I hope this activity has also been helpful. Please try the examples that follow and compare your answers after.

Example 18

Simplify the expression x/x + 1 − 2x/x + 2 as a single fraction

Solution

LCM= (x + 1)(x + 2) x_ x + 1 − 2x/x + 2 = x(x + 2) − 2x(x + 1)_______________ (x + 1)(x + 2) = x²+ 2x − 2x²− 2x_____________ (x + 1)(x + 2) = –x²__________ (x + 1)(x + 2)

Example 19

Express 5 b²______ a²− b²+ 2a/a − b − 3b/a + b as a single fraction

Solution

The denominators are a²− b², a − b, a + b Their LCM is (a − b)(a + b) = a²− b²– remember this is a difference of two squares.

5_ a²− b²+ 2a_ a − b − 3b_ a + b = 5 b²+ 2a(a + b) − 3b(a − b)___________________ (a − b)(a + b) = 5 b²+ 2a²+ 2ab − 3ab + 3b²____________________ (a − b)(a + b) = 8b²+ 2a²− ab___________ (a − b)(a + b)

Example 20

At a youth club there were k people present. Of those, 2/5 were playing football and 1/4 were playing other games.

i. How many people were playing games?

ii. How many were not playing games?

Solution

i. Those who were playing football= 2__ 5k Those playing other games = 1__ 4k Total number of those present who were playing games = 2__ 5k + 1__ 4k = 4(2k) + 5(k)_________ 20 = 13__ 20k Therefore, 13__ 20k of those present were playing games.

ii. Number who were not playing games ⟹ total number of those present (k)− total number of those playing games (13__ 20k) = k − 13_ 20k = 20(k) − 13k_ 20 = 7_ 20k Can you also recall how to multiply or divide ordinary fractions? Good.

Multiplication and division of algebraic fractions are also done in the same manner.

So, let us now look at how to multiply or divide Algebraic Fractions with monomial or binomial denominators.

Multiplication and Division of Algebraic Fractions

When multiplying algebraic fractions, first of all check whether there are common factors, cancel them out and when there are none, multiply the numerators and the denominators separately. For example; m/n × a/b = m × a/n × b = ma___ nb Also, when an algebraic fraction is divided by another, multiply the first fraction by the reciprocal of the second fraction. Thus m_ n ÷ a_ b = m_ n × b_ a = mb_ na etc.

Note: Always ensure that you leave your answer in its simplest form.

Now, let us consider an activity on how to perform these.

Activity 15

Simplify the expression 12xy/7 × 14x/20 in its simplest form.

Here are the steps:

Method 1 – with simplification happening at the end

Step 1: Multiply the numerators; 12xy × 14x = 168 x²y

Step 2: Multiply the denominators; 7 × 20 = 140

Step 3: Write the product as a fraction: 168x²y/140

Step 4: Simplify the fraction by dividing both the numerator and denominator by their common factor(s) ⇒ 168 x²y_ 140 = 6 x²y_ 5

Step 5: Write the final answer in its simplest form:

⇒ 12xy_ 7 × 14x_ 20 = 6 x²y_ 5 Method 2 – with simplification happening at the beginning To simplify the expression 12xy/7 × 14x/20 in its simplest form,

Step 1: Identify the common factors of the numerators and denominators and cancel them out. For instance, since 7 is a factor of 14, it can cancel out.

Again,12 and 20 have common factors that can also cancel out. This gives us 12xy_ 7 × 14x_ 20 = 3xy_ 1 × 2x_ 5

Step 2: Multiply the numerators; 3xy × 2x = 6 x²y

Step 3: Multiply the denominators; 1 × 5 = 5

Step 4: Write the product as a fraction: 6 x²y/5

Step 5: Write the final answer in its simplest form:

⇒ 12xy_ 7 × 14x_ 20 = 6 x²y_ 5 There you have it!

Let us consider another example.

Example 21

Simplify the expression 6x + 8/4 ÷ x²+ 3/5 x²as a single fraction.

Solution

Step 1: Simplify the numerator and denominator of the first fraction, 6x + 8/4 6x + 8_ 4 = 2(3x + 4)_ 4 = 2(3x + 4)_ 2 × 2 = 3x + 4_ 2

Step 2: Simplify the second fraction, x²+ 3/5 x²Since the numerator and denominator cannot be simplified further, we leave it as is.

Now, the expression becomes; 3x + 4/2 ÷ x²+ 3/5 x²Step 3: To divide fractions, we multiply by the reciprocal of the second fraction:

= 3x + 4_ 2 × 5x²_____ x²+ 3

Step 4: Multiply the numerators and denominators;

= (3x + 4) × 5x²_ 2 × (x²+ 3)

Step 5: Leave the answer in its simplest form.

⇒ 6x + 8/4 ÷ x²+ 3/5 x²= 15x³+ 20x²_________ 2 x²+ 6

Example 22

Simplify 2 m²___ 3 ÷ 4m/9

Solution

2 m²_ 3 ÷ 4m_ 9 Multiply the first fraction by the reciprocal of the second fraction ⇒ 2 m²_ 3 × 9_ 4m Cancel out those with common factors or multiply the numerators separately and the denominators also separately. Note, you can only cancel opposite numerators and denominators when we are at the multiplication stage, not while it is still in the division stage.

⇒ 18 m²____ 12m , here 6m divides both the numerator and the denominator to reduce it to;

3m/2 ⇒ 2 m²_ 3 ÷ 4m_ 9 = 3m_ 2 I encourage you to try your hands on more examples with your classmates and compare your answers.

Does it make sense when asked to divide an object into zero (0) equal parts and share it among a number of people? Obviously no!

What about dividing an object into equal number of parts and sharing them among nobody? Does this make a meaning? Of course, Yes!

This implies that when a number is divided by zero (0), we say the fraction is not defined or it is undefined. Also, when the numerator of a fraction is zero, the entire fraction is zero (0).

Therefore, an Algebraic Expression is said to be undefined or not defined if the variable at the denominator assumes a value (takes on a value) that makes the denominator zero.

Zero or Undefined Algebraic fractions Algebraic fractions are said to be undefined or have no meaning if the denominator is equal to zero.

For example; the fraction ²_ₓ is said to be undefined when x = 0 Likewise, an algebraic fraction is said to be zero if the numerator is equal to zero.

Let us do this activity to ascertain why a fraction is said to be undefined or zero.

Activity 16

Note:

Dividend_ Divisor = Quotient ⇒ Quotient × Divisor = Dividend Therefore, if 16_ 2 = 8 This implies 8 × 2 =16 Again, 0_ 13 = 0. This implies 13 × 0 = 0 How would you determine that a fraction is undefined? I hope you have an idea.

Below are some guidelines that help you 5_ 0 ≠ 0 , This implies 0 × 0 ≠ 5 (It cannot be defined) Have you now seen the reason why we say a fraction is undefined or not defined when the denominator is zero? Very well.

Let us now explore some conditions under which an algebraic fraction is considered undefined or zero.

Conditions Under Which an Algebraic Expression is Said to be Zero or Undefined To find the value(s) for which an algebraic fraction is undefined, here are some steps to guide you.

Step 1: Equate the denominator of the expression to zero

Step 2: Solve for the value(s) of the involving variable

Step 3: The value(s) of the variable makes the expression undefined.

This implies the variable at the denominator can assume or take all values except the value that makes the expression undefined. Let us consider the following examples.

Example 23

Find the value of x which makes the expressions undefined:

i. 3/x − 1

Solution

For the expression 3/x − 1 to be undefined, the denominator x −1 must be equal to zero: x−1=0 Solving for x:

x = 1 Therefore, the expression is undefined at x =1

ii. ⁽²x − 1)(x − 4)___________ 4x²− 1

Solution

For the expression ⁽²x − 1)(x − 4)___________ 4x²− 1 to be undefined, the denominator 4x²−1 must be equal to zero:

4x²− 1 = 0 This can be factorised as a difference of two squares:

(2x+1)(2x−1) = 0 Set each factor to zero:

2x +1 = 0 or 2x −1= 0 Solve for x in each case:

x = − 1_ 2 or x = 1_ 2 Therefore, the expression is undefined at x = −¹_ ₂ and x = ¹_ ₂.

Also, to find the value(s) for which an algebraic fraction is zero, here are some steps to guide you.

Step 1: Equate the fraction to zero.

Step 2: Solve for the value(s) of the involving variable.

Step 3: This value(s) of the variable makes the expression zero and is called the zeros of the expression. This means that, when the variable takes that value, the outcome of the entire expression is zero.

Let us consider an example.

Example 24

Determine the condition under which 3a _ a − 4 is undefined and determine the value of a and do the same for when the whole fraction is zero.

Condition: a – 4 = 0 a = 4, Therefore, when a = 4 the fraction is undefined.

Condition: under which the fraction is zero, put the whole fraction to zero.

3a_ a − 4 = 0 , 3a = 0, a = 0, Therefore, when a is 0 the fraction is zero.

Example 25

For what values of x is the expression zero.

Solution

x²+ x − 6_ x Equate the fraction to zero x²+ x − 6_ x = 0 x²+ x − 6 = 0 Solve the quadratic x²+ 3x − 2x − 6 = 0 x(x + 3) − 2(x + 3) = 0 (x − 2)(x + 3) = 0 x − 2 = 0 or x + 3 = 0 x = 2 or x = − 3 This implies that the expression x²+ x − 6/x is zero when x = 2 or x = − 3 .

I encourage you to try more examples from the extended reading materials and discuss your findings with a classmate.

Review Questions 3.1

1. Identify patterns from the following sets of numbers and create a mathematical expression for each sequence

a. 1, 3, 5, 7, 9…

b. -1, 2, 5, 8, 11…

2. In a math competition, participants are asked to solve a series of problems.

In the first round, each contestant solves 1 problem. In each subsequent round, they solve 2 more problems than in the previous round. How many problems will they need to solve in the 7th round?

3. Expand the algebraic expressions.

a. (x + 5 ) (x + 3)

b. (x + 2) (x - 4)

c. (y +7) (y + 1)

d. (z + 2) (z – 5)

4. Factorise completely the following

a. xy + 2xz

b. 4(4n – 12x)

c. 3 x²– 9xy

d. 6ab − 4pb − 2pq + 3aq

5. Factorise completely the following quadratic expressions

a. x²+ 2x − 3

b. 6 – 5x – x²c. 3 x²– 17x + 10

d. p²− 1

6. The length of a rectangular field is 5 meters more than its width. Write down an expression for the perimeter of the field.

7. The length of a rectangular garden is 4 meters longer than its width. Write an expression to find the area of the garden.

Review Questions 3.2

1. Express the following numbers as perfect squares

a. 121

b. 81

c. 64

d. 169

e. 100

2. Simplify the following as single fractions

a. x/x²− 5x + 6 + 1/x − 2 + 3/x − 3

b. 2x − 1_ 3 − x + 3_ 2

c. x − 2_ 3 + x + 3_ 5

d. 5x_ 3 × x + 3_ 2

e. 2x − 1_____ 2x ÷ 4 x²− 1/6

3. Identify with reasons which of the following fractions have monomial or binomial denominators.

a. 2__ 5x

b. y/x + 1

c. x − 3/x + 6

d. 7y − 1/x²− 4 ,

e. 3z + 4/z²4. Given that p = 2x/1 − x²and q = 2x/1 + x , simplify 3p – 2q

5. Simplify the following and state the value of the variable that makes the expression undefined or zero.

i. 4m²− 9_______ (m − 1)²÷ 2m + 3/m²− m

i. 6x²+ 12x/x²+ x × x²− 1/x²+ 10x + 16

ii. y²− 7y + 12/y³− 9y

6. Fully factorise the following:

i. 4x²– 9

ii. 16y²-25a²iii. a²b²-121

iv. 4m²– 100

7. Find the value(s) for the unknown using the idea of perfect squares.

a. 3z + 128 =162

Review Questions 3.1

1. a. 1, 3, 5, 7, 9… Pattern: Adding 2 to the previous term Algebraic expression: aₙ = 2n − 1 (where n is the term number)

b. -1, 2, 5, 8, 11… Pattern: Adding 3 to the previous term Algebraic expression: aₙ = 3n – 4

2. Therefore, in the 7th round, contestants will need to solve 13 problems.

3. a. (x + 5 ) (x + 3) = x²+ 5x + 3x + 15 = x²+ 8x + 15

b. (x + 2) (x - 4) = x²+ 2x − 4x − 8 = x²− 2x − 8

c. (y +7) (y + 1) = y²+ 7y + y + 7 = y²+ 8y + 7

d. (z + 2) (z – 5) = z²+ 2z − 5z − 10 = z²− 3z − 10

4. a. xy + 2xz = x(y + 2z)

b. 4(4n − 12x) = 4(4(n − 3x)) = 16(n − 3x)

c. 3 x²− 9xy = 3x(x − 3y)

d. 6ab − 4pb − 2pq + 3aq = (3a − 2p)(2b + q)

5. a. x²+ 2x − 3 = (x + 3)(x − 1)

b. 6 – 5x – x²= -(x²+ 5x − 6) = −(x + 6)(x − 1)

c. 3 x²– 17x + 10 = 3x²− 15x − 2x + 10 = 3x(x − 5) − 2(x − 5) = (3x − 2)(x − 5)

d. p²− 1 = (p + 1)(p − 1)

6. Let w = the width of the field, so the expression for the perimeter of the field is:

w + w + w + 5 + w + 5 = 4w + 10 = 2(2w + 5).

OR Let l = length of the field, so the expression for the perimeter of the field is:

l + l + l - 5 + l - 5 = 4l – 10 = 2(2l – 5)

7. Let the width of the garden = w, so the expression for the area of the garden is w(w + 4).

OR Let the length of the garden = l, so the expression for the area of the garden is l(l – 4)

Review Questions 3.2

1. a) 11²b) 9²c) 8²d) 13²e) 10²2. a. 5x − 9__________ (x − 3)(x − 2)

b. x − 11/6

c. 8x − 1/15

d. 5 x²+ 15x/6

e. 3/2 x²− x

3. Monomial denominators; a, e Binomial denominators; b, c and d

4. 2x + 4 x²__________ (1 − x)(1 + x)

5. i. 2 m²− 3m/m − 1 ∴ undefined when m = 1 and zero when m = 0 or m = 3/2

ii. 6(x − 1)______ x + 8 ∴ undefined when x = − 8 and it is zero when x = 1

iii. y − 4/y(y + 3) ∴ undefined when y = 0 or y = − 3 and zero when y = 4

6. i. (2x − 3)(2x + 3)

ii. (4y − 5a)(4y + 5a)

iii. (ab − 11)(ab + 11)

iv. (2m − 10)(2m + 10)

7. z = 34/3

Proficiency in algebraic manipulation and problem-solving lays the foundation for tackling a multitude of mathematical challenges and real-world situations. A bedrock of algebraic proficiency involves the ability to formulate equations that represent real-world situations and to interpret these equations in context. This skill empowers individuals to manipulate formulae to solve problems, including situations necessitating a change of subjects within the formula. Linear equations and inequalities are highly relevant in day-to-day activities, including commerce and industry. For example, they are applied in the following areas: Budgeting and Finance, Supply Chain Management, Production and Manufacturing, Marketing and Sales, Operations Management, Economics and Business Analysis, Risk Assessment and Decision Making, and Optimisation. Relations, mappings, and functions play significant roles in various aspects of daily life, commerce, and industry. These concepts are applied in areas such as: Communication and Relationships, Commerce and Business, Technology and Industry, Data Analysis and Decision Making, Education and Learning, Healthcare and Medicine. In conclusion, relations, mappings, and functions are foundational concepts with wide-ranging applications in daily life, commerce and industry. They help in understanding, analysing, and optimising relationships, processes, and systems across diverse fields, contributing to efficiency, innovation, and informed decision- making. We’ll also learn how to find the gradient, equation of a straight line and calculate the distance between two points. Imagine being able to calculate the distance between two cities on a map or the length of a bridge. It’s all about understanding the relationships between points and lines.

At the end of this section, you will be able to:

• Construct and interpret formulae for a given task and apply to problems involving a change of subjects.

• Solve linear equations in one and two variable(s); and brackets and fractions for given problems and relate it to real life situations.

• Find solution set or truth set of linear inequalities and illustrate on the number line.

• Identify relations from functions and differentiate between the types of relations and functions using models such as graphs.

• Investigate relationships between two number sets and determine the rules of given mappings or functions.

• Extend the knowledge of coordinates of two points to find the gradient and equation of a straight line

• Recognise and interpret two points on a straight line and use it to find the distance between them.

Key Ideas

• A formula is a rule which gives the relationship between things or quantities.

• A Linear equation is of the form, ax + by + c. where a, b, and c are constants, and x and y are variables. This form is known as the standard form of a linear equation. Or Linear equation can be in slope intercept form as y = mx + c where m is the slope of the line and c is the y-intercept.

• A linear inequality in two variables is typically expressed in the following general form:

ax + by ≥ c or ax + by ≤ c, where a, b and c are constants, x and y are variables.

• Folding the human right arm resembles the idea of greater than symbol.

Folding the human left arm resembles the idea of less than symbol.

• A relation between two sets is a collection of ordered pairs where each element of the first set is related to one or more elements of the second set. A function is a special type of relation where each input (from the domain) is related to exactly one output (from the range).

• There are various types of relations. Some of which are: One-to-One relations, One-to-many relations, many-to-one relations and many-to- many relations. One-to-one and Many-to-one relations are functions.

• A mapping can either be linear, quadratic or exponential. In this section, we shall look at Linear and Exponential mappings as well as determining the rules of mappings. We shall also identify and interpret linear graphs.

This refers to the process of rearranging an equation to isolate a different variable or parameter. This technique is often used in solving equations or formulae where one needs to manipulate the equation to express a different variable as the subject of the formula.

For example, if you have an equation A = 1/2 bh and you want to solve for the height, you perform a “change of subject” by rearranging the equation to isolate h: h = 2A/b In this case, the process involved changing the subject from A = 1/2 bh to h = 2A/b , by rearranging the terms of the equation. This method is fundamental in algebraic manipulation and problem-solving in mathematics.

Definition of Key Concepts

A formula is a rule which gives the relationship between things or quantities.

The letters in the formula always stand for something specific, like cost, speed, number of books. There is always more than one unknown in a formula. For

example, the perimeter of a rectangle P = 2l + 2b, the area of a triangle, A = 1/2 bh .

Activity 1

Let’s break down the process involving a change of subject.

Steps to Construct and Interpret Formulae

Step 1: Identify Variables

First, identify the variables involved in the problem. For example, let’s consider a problem involving the formula for the area of a rectangle, which is given by: A = L × W. where A represents the area, L represents the length and W represents the width of the rectangle.

Step 2: Formula Manipulation

Solving for a Specific Variable: Sometimes, you may need to solve the formula for a specific variable. For instance, if you want to solve for W in terms of A and L, you would rearrange the formula:

W = A/L Here, you isolate W on one side of the equation.

Step 3: Applying to Problems

Consider a scenario where you are given the area, A, and length, L, of a rectangle and need to find the width, W. Using the formula A = L x W and knowing A and L, you can substitute these values to find W:

W = A/L This formula allows you to calculate W directly once you know A and L.

I trust you can now discuss and manipulate the area, length and breadth of your classroom. Thumbs up!

Example 1

Study the example below.

The area of a rectangular field is 24 square units, and its length is 6 units. Find the width of the field.

Solution

1. Given:

A = 24 square units L = 6 units

2. Use the formula, A = L × W to find W:

W = 24/6

3. Therefore, the width, W of the rectangle is 4 units.

Interpretation In this example, by constructing and interpreting the formula A = L × W, and applying it correctly to the problem where values of A and L are known, we were able to find the value of W, which represents the width of the rectangle.

Application of concepts and examples Hello Learner! Let’s now look at the following examples on change of subject.

Example 2

1. Make c the subject of the formula y = mx + c

Solution

To make c the subject, subtract mx from both sides of the equation to isolate c.

mx − mx + c = y − mx ∴ c = y − mx

Example 3

From the formula 3c + 2r = md + k, make r the subject.

Solution

Given 3c + 2r = md + k, Make 2r the subject, ie isolate the r term on one side of the equation:

2r = md + k – 3c Divide both sides by 2 so we only have a single r left.

r = md + k − 3c/2

Example 4

The relation between energy E, mass m, and velocity of light v, is given by E = mv².

Find the value v, when E = 20 and m = 5

Solution

Given that E = mv², let us rearrange to isolate the v term.

v²= E/m ∴ v = √E_ m v = √20_ 5 ∴ v = 2

An equation of the form ax + b = c, where a, b and c are real numbers, and a ≠ 0.

A Linear equation in one variable has exactly one solution.

Activity 2

Let’s engage in an activity to solve the age problem together.

Steps:

1. Consider the age problem: “John is 5 years older than twice Maria’s age.

If John is 25 years old, how old is Maria?”

Define the variable: Let’s denote Maria’s age to be x.

2. Setting up the Equation According to the problem statement, John’s age is 25 years, and he is 5 years older than twice Maria’s age. So, we can write the equation:

25 = 2x + 5

3. Solving the Equation:

Now, let’s solve the equation to find Maria’s age (x).

25 = 2x + 5 Subtract 5 from both sides to isolate the x term 25 – 5 = 2x 20 = 2x Divide both sides by 2 to solve for a single x.

x = 20/2 x = 10 Hello Learner! I hope you enjoyed the above activity. Let’s now consider the following examples.

Linear Equations Involving one and Two variables

Example 5

Solve for the variable indicated in the following equations.

a) 3x − 12 = 21

b) 5 − 3y = 3y + 7

Solution

a) To solve 3x − 12 = 21, You need to isolate the x term on one side, so add 12 to both sides of the equation, 3x = 33 Then divide both sides of the equation by 3, to make a single x the subject x = 33/3 = 11

b) To solve 5 − 3y = 3y + 7 First group the like terms (ie isolate the ys onto one side and the constants (the numbers) onto the other) 5 – 7 = 3y + 3y Simplify the terms -2 = 6y Make a single y the subject by dividing both sides by the coefficient of y.

Therefore y = −1/3

Example 6

A company produces and sells handmade pottery. The total cost C (in Ghana cedis) to produce x units of pottery is given by the equation C = 100x - 500. The company sells each unit of pottery for p Ghana cedis. The total revenue R (in Ghana cedis) from selling x units of pottery is R = px

1. Express the company’s profit, P (in GH€) as a function of x and p.

2. If the company sells each unit of pottery for GH€20.00 how many units of pottery must they sell to break even (i.e., make zero profit)?

Solution

Step 1: Expressing Profit as a Function

The profit, P is given by subtracting the total cost C from the total revenue, R.

P = R – C Substitute the given equations for C and R:

1. P = px - (100x - 500) Simplify the expression:

P = px – 100x + 500 So, the profit, P as a function of x and p is P(x) = px – 100x + 500.

Step 2: Finding the Break-even Point

To find the break-even point, we set the profit P(x) equal to zero (break-even condition):

px – 100x + 500 = 0 Now, substitute p = 20 (since the company sells each unit for GH€20):

20x – 100x + 500 = 0 Combine like terms:

– 80x + 500 = 0 Subtract 500 from both sides:

– 80x = – 500 Divide both sides by –80 − 80x_ − 80 = −500_ − 80 x = 50/8 Simplify the fraction:

x = 6.25 Since x represents the number of units of pottery, and it must be a whole number, we round up to the nearest whole number because the company cannot sell a fraction of a pottery unit:

x = 7 Therefore, the company must sell at least 7 units of pottery to break even (make zero profit) when selling each unit for GH€20.00 Linear Equations Involving Brackets Here, we shall be applying the concept of distributive property in the expansion of brackets.

Example 7

Solve the equation 7(x − 6 ) = 3(x + 9)

Solution

To solve the equation 7(x − 6 ) = 3(x + 9) , First multiply the brackets (expansion) on both sides of the equation, 7x − 42 = 3x + 27 Then subtract 3x from both sides (so you are isolating your xs on one side of the equation) 4x − 42 = 27 Then add 42 to both sides of the equation (so the xs are isolated) 4x = 69.

Divide both sides of the equation by 4 to find the value for a single x.

x = 69/4 = 171/4 Linear Equations Involving Fractions This concept was treated in week 10 of section 3. Let’s look at the following

example to refresh our minds.

Example 8

Solve 3x/2 − 2 = 1/2

Solution

3x/2 − 2 = 1/2

1. Eliminate the fraction by multiplying each term on both sides of the equation by the LCM 2 × 3x/2 − 2 × 2 = 1/2 × 2 3x − 4 = 1

2. Group the like terms and simplify the equation 3x = 1 + 4 3x = 5 Divide both sides by 3, x = 5/3

Linear inequality in one variable is of the form ax + b < c, ax + b ≤ c, ax + b > c, ax + b ≥ c.

Folding the human right arm resembles the idea of greater than (>) symbol and that of the left arm also resembles the idea of less than (<) symbol.

4 < 5 (4 is less than 5) 6 > 4 (6 is greater than 4) x ≤ 4 (Depending on the values of x, which is an unknown variable, but it must be less than or equal to 4) Suppose we are given an inequality of the form x > -5, the solution set for an inequality (as it is for an equation) is the set of all values for the variable that make the inequality a true statement.

An appropriate way to picture the solution set is by a graph on a number line or the use of Geodot to generate conjectures. A sample of the geodot is shown here.

Figure 1: Geodot

Graph the set; {x: x < 4} Explanations: We want to include all real numbers less than 4, that is, to the left of 4 on the number line. An open circle is used to indicate that the point corresponding to 4 is not included in the graph. It is called an open half line; it extends to the left and not including 4.

Two other symbols as shown in the introduction, ≤ and ≥ , are also used in writing inequalities. In each case, they combine the inequality symbols for less than or greater than with the symbol for equality.

The following explain the use of these symbols. The expression a ≤ b is read as “a is less than or equal to b”

Note that this combines the symbol ‘< ‘and ‘=’ and means that either a < b or a = b. Similarly, a ≥ b reads “a is greater than or equal to b”. Implying, either a > b or a = b. etc.

Activity 3

Solve the inequality: 2x – 5 > 3 Steps

1. Isolate the variable:

Start by isolating the x term on one side of the inequality.

2x – 5 > 3 Add 5 to both sides to remove the constant term on the left side:

2x – 5 + 5 > 3 + 5 Simplify:

2x > 8

2. Divide by the coefficient of x Divide both sides by 2 to solve for a single x 2x_ 2 > 8 __ Simplify:

x > 4

3. Write the solution:

The solution to the inequality 2x - 5 > 3 is x > 4.

This activity shows the basic steps involved in solving a simple linear inequality. Always remember to perform operations on both sides of the inequality to maintain its validity. However, do avoid multiplying or dividing by a negative number as this reverses the inequality sign.

Review Questions 4.1

1. Solve 1/2 x − 1/3(x + 4) > 4x + 2/3

2. If Kofi’s age is 30 years now, what is his age in 5 years’ time?

3. If Ama is 40 years now, what was her age 4 years ago?

4. Solve the following equations:

i. 6 + y – 2 = 12

ii. 14y – 5 = 2y

iii. 10(y + 2) = 14

5. Find the solution set of the following inequalities and illustrate your answer on the number line;

i. 3x − 2 ≥ 12 − x

ii. x − 2/3(x + 1) ≤ 1/2(4 − x) − 5

6. Answer the following questions;

i. The sum of four consecutive even numbers is 36. Find the numbers.

ii. The perimeter of a football field in a rectangular form of a certain school is 296m. If the breadth is 2/3 of the length, find the length.

iii. Find the number N, such that when ¹_ ₃ of it is added to 8, the result is the same as 18 from 1/2 of it.

Review Questions 4.2

1. Identify whether each of the mappings below are functions. Give a reason for each answer.

i. ( x y z) ( 1 2

3) ii. ( x y z) ( A D T)

iii. ( x y z) ( 1 2 3)

2. Which of the following relations defined on the set of real numbers are functions? Give a reason for your answers.

A = (x, y): y = 3x + 1 B = (x, y) : y = 2x²C = (x, y) : y²= x

3. Draw mapping diagrams to represent the following

i. y = 2 x²− 1 , where {x:x = 0, 1, 2, 3, 4}

ii. y = 3x + 2 , where {x:x = 0, 1, 2, 3, 4, 5}

4. Find the rule of the following mapping P 5 6 7 8 q 17 21 25 29

5. Draw the following linear graphs and interpret them

i. y = 2x + 1

ii. y = − 3x + 1

iii. y = − 5x + 1

6. A Bakery sells a total of 250 loaves of bread per day. The number of whole wheat loaves (x) and white loaves (y) sold are related by the equation y = 150 − x . How many whole wheat loaves are sold if 75 white loaves are sold?

7. A person’s height in inches (h) is related to their shoe size (s) by the equation, h = 5s + 20 .

What is the person’s height if their shoe size is 8 ?

8. Find the rule for the mapping below.

9. Find the rule for the mapping below.

Review Questions 4.3

1. Suppose A is the point (3, 4) and B is the point (8, 14). What is the gradient/ straight line joining these points?

2. Calculate the gradient of the straight line given the coordinates A (2,6) and B (8,24)

3. Find the midpoint of the line joining the points or with end points of (1,3) and (5,6).

4. Show that the line segment joining the points (1, 4) and (3, 10) is parallel to the line segment joining the points (−5, −10) and (−2, −1).

5. Are the lines L1 through (2, 3) and (4, 6) and L2 through (-4, 2) and (0, 8) parallel, or do they intersect? Give reason for your answer.

6. Determine the equation of the line perpendicular to 2y + 3x = 6 which goes through the point (5, 2)