Welcome to the World of Algebra!
Hello! Get ready to dive into one of the most powerful tools in mathematics: Algebra. You might have heard it's tricky, but don't worry! We're going to break it down into simple, easy-to-understand pieces.
Think of algebra as a secret code that helps us solve puzzles and real-world problems, like figuring out the price of something on sale, planning a trip, or even building a website. In this chapter, you'll learn how to speak this 'language' of letters and numbers. Let's get started!
Part 1: The Building Blocks - Algebraic Expressions
This is where it all begins. We learn how to turn words into maths and understand the basic parts of an algebraic expression.
1.1 Using Letters to Represent Numbers
In algebra, we use letters to represent unknown numbers. These letters are called variables. It's like having a box where you can put any number you want.
Why is this useful? Imagine you get some money for your birthday, but you don't know the exact amount yet. You can just call it 'm' for money!
Key Skill: Translating Words into Algebra
This is the most important first step! Let's use the letter 'n' to stand for "a number".
- "5 more than a number" becomes $$n + 5$$
- "A number decreased by 10" becomes $$n - 10$$
- "The product of a number and 3" becomes $$3n$$ (In algebra, we usually write the number before the letter and we don't need the 'x' sign for multiplication. $$3 \times n$$ is just $$3n$$).
- "A number divided by 4" becomes $$ \frac{n}{4} $$ (We use fractions for division).
Key Skill: Translating Algebra into Words
Let's go backwards!
- $$x + 2$$ could be "2 more than x" or "the sum of x and 2".
- $$y - 7$$ could be "7 less than y" or "y decreased by 7".
- $$5p$$ could be "5 times p" or "the product of 5 and p".
1.2 Understanding the Parts of an Expression
Let's look at the expression: $$ 5x + 3y - 7 $$
- Terms: These are the parts separated by + or - signs. The terms here are $$5x$$, $$3y$$, and $$-7$$.
- Variable: A letter representing an unknown value. The variables here are x and y.
- Coefficient: The number multiplied by a variable. The coefficient of x is 5. The coefficient of y is 3.
- Constant Term: A number on its own, without a variable. The constant here is -7.
Quick Review
In the term $$8k$$, the coefficient is 8 and the variable is k.
1.3 Polynomials: Naming the Expressions
"Poly" means "many". A polynomial is an expression with one or more terms. We have special names for polynomials with a few terms:
- Monomial: An expression with just ONE term (e.g., $$7x$$, $$-5$$, $$2y^2$$).
- Binomial: An expression with TWO terms (e.g., $$x+4$$, $$3a-2b$$).
- Trinomial: An expression with THREE terms (e.g., $$x^2+2x+1$$).
1.4 Simplifying Expressions by Combining Like Terms
Think about fruit. If you have 3 apples and you get 2 more apples, you have 5 apples. But if you have 3 apples and 2 bananas, you can't combine them to get "5 apple-bananas". It's the same in algebra!
Like Terms are terms that have the exact same variable part (including the power).
- $$4x$$ and $$7x$$ are like terms.
- $$5y^2$$ and $$-2y^2$$ are like terms.
- $$3a$$ and $$6b$$ are unlike terms (different variables).
- $$2x$$ and $$9x^2$$ are unlike terms (different powers).
To simplify, we just add or subtract the coefficients of the like terms.
Example: Simplify $$ \bf{7x} + \bf{2y} - \bf{3x} + \bf{4y} $$
1. Group the like terms together: $$ (7x - 3x) + (2y + 4y) $$
2. Combine them: $$ 4x + 6y $$
1.5 Operations with Polynomials
You can add, subtract, and multiply polynomials just like you do with numbers!
Addition: Remove the brackets and combine like terms.
Example: $$(2a + 5b) + (6a - 3b)$$
$$= 2a + 5b + 6a - 3b$$
$$= (2a+6a) + (5b-3b)$$
$$= 8a + 2b$$
Subtraction: Flip the sign of every term in the second bracket, then combine like terms.
Example: $$(4x + 7) - (3x - 2)$$
$$= 4x + 7 - 3x + 2$$ (Notice the -2 became +2!)
$$= (4x - 3x) + (7 + 2)$$
$$= x + 9$$
Multiplication: Multiply the term outside the bracket by every term inside the bracket.
Example: $$3(2x + 5)$$
$$= (3 \times 2x) + (3 \times 5)$$
$$= 6x + 15$$
Key Takeaway for Part 1
Algebraic expressions use letters (variables) to represent numbers. We can simplify them by combining like terms, and we can perform operations like addition, subtraction, and multiplication on them.
Part 2: Finding the Balance - Equations
Now that we know what expressions are, let's learn about equations. An equation is all about balance!
2.1 What is an Equation?
An equation says that two expressions are equal. It will always have an equals sign (=).
Analogy: A Balanced Seesaw
Think of an equation as a perfectly balanced seesaw. If you add or remove something from one side, you must do the exact same thing to the other side to keep it balanced.
Example: $$2x + 1 = 7$$
This is a linear equation in one unknown because it has only one variable (x) and the highest power of the variable is 1.
2.2 Solving Linear Equations in One Unknown
"Solving" an equation means finding the value of the unknown variable that makes the equation true. Our goal is to get the variable by itself on one side of the equals sign.
The Golden Rule: Whatever you do to one side of the equation, you MUST do to the other side.
Step-by-Step Guide to Solving $$3x - 5 = 16$$
- Goal: Get the 'x' term alone. The problem is the "-5".
- Action: Do the opposite of "-5", which is "+5", to both sides to keep it balanced. $$3x - 5 + 5 = 16 + 5$$ $$3x = 21$$
- Goal: Get 'x' completely alone. The problem is the "3" which is multiplying the x.
- Action: Do the opposite of "multiply by 3", which is "divide by 3", on both sides. $$\frac{3x}{3} = \frac{21}{3}$$ $$x = 7$$
2.3 Formulating Equations from Word Problems
This is where algebra becomes a real-life tool!
Problem: Sam is twice as old as his brother, Tom. Their total age is 18. How old is Tom?
- Define the unknown: Let Tom's age be 'a'.
- Write expressions for other values: Sam is twice as old, so Sam's age is $$2a$$.
- Form the equation: Their total age is 18. So, Tom's age + Sam's age = 18. $$a + 2a = 18$$
- Solve the equation: $$3a = 18$$ $$\frac{3a}{3} = \frac{18}{3}$$ $$a = 6$$
- Answer the question: Tom is 6 years old. (And Sam is 2 x 6 = 12 years old).
2.4 Solving Simultaneous Linear Equations
What if you have TWO unknowns (like x and y) and TWO equations? These are called simultaneous equations. You need two equations to find the two unknown values.
Real-world example: You buy 2 apples and 1 orange for $7. Your friend buys 1 apple and 1 orange for $5. What is the price of one apple and one orange?
Let apple price = a and orange price = o.
Equation 1: $$2a + o = 7$$
Equation 2: $$a + o = 5$$
There are two main algebraic ways to solve this:
Method 1: Substitution
- Rearrange one equation to get a variable by itself. Let's use Equation 2: $$a + o = 5 \implies o = 5 - a$$
- Substitute this expression into the OTHER equation. We'll put $$(5-a)$$ where 'o' is in Equation 1: $$2a + (5-a) = 7$$
- Solve this new equation which only has one variable: $$2a + 5 - a = 7$$ $$a + 5 = 7$$ $$a = 2$$
- Find the other variable by putting $$a=2$$ back into any of the original equations. $$o = 5 - a \implies o = 5 - 2 \implies o = 3$$
Solution: An apple costs $2 and an orange costs $3.
Method 2: Elimination
The goal is to add or subtract the equations to eliminate one variable.
Our equations are:
1: $$2a + o = 7$$
2: $$a + o = 5$$
- Line them up. Notice that both equations have '+o'.
- Subtract Equation 2 from Equation 1 to eliminate 'o'. $$(2a + o) - (a + o) = 7 - 5$$ $$2a + o - a - o = 2$$ $$a = 2$$
- Find the other variable just like in the substitution method. Put $$a=2$$ into Equation 2: $$2 + o = 5 \implies o = 3$$
We get the same answer! You can choose whichever method you find easier.
Did you know?
The graph of a linear equation in two unknowns (like $$y = 2x + 1$$) is always a straight line. The solution to a pair of simultaneous equations is the point where their two lines intersect on a graph!
Key Takeaway for Part 2
Equations are about balance. To solve them, you must always do the same thing to both sides. For two unknowns, you need two equations, which you can solve using substitution or elimination.
Part 3: Special Tools - Identities & Factorisation
Let's learn some special algebraic tricks that make solving harder problems much faster!
3.1 Equations vs. Identities
An equation is only true for a specific value. For example, $$x+2=5$$ is only true when $$x=3$$.
An identity is true for ALL possible values. It's like two different ways of writing the same thing. We sometimes use the $$ \equiv $$ symbol.
Example: $$2(x+1) \equiv 2x+2$$. No matter what number you pick for x, this will always be true!
3.2 Important Identities to Memorise
These are super useful shortcuts for expanding brackets.
- Perfect Square (sum): $$(a+b)^2 = a^2 + 2ab + b^2$$
- Perfect Square (difference): $$(a-b)^2 = a^2 - 2ab + b^2$$
- Difference of Two Squares: $$(a+b)(a-b) = a^2 - b^2$$
Example using an identity: Expand $$(x+3)^2$$
Using the formula $$(a+b)^2 = a^2 + 2ab + b^2$$, where $$a=x$$ and $$b=3$$:
$$(x)^2 + 2(x)(3) + (3)^2$$
$$= x^2 + 6x + 9$$
This is much faster than multiplying $$(x+3)(x+3)$$ by hand!
3.3 Factorisation: The Reverse of Expanding
Factorising means taking an expression and putting it back into brackets. It's like finding the original ingredients that were multiplied together.
Method 1: Extracting a Common Factor
Look for the highest common factor (HCF) of all the terms.
Example: Factorise $$6x + 9$$
The HCF of 6 and 9 is 3. So, we 'take out' the 3.
$$= 3(2x + 3)$$
You can check your answer by expanding it: $$3 \times 2x = 6x$$ and $$3 \times 3 = 9$$. It works!
Method 2: Using the Identities
If you see an expression that looks like the right side of one of our identities, you can use it to factorise!
Example 1: Factorise $$x^2 - 25$$
This looks like the 'Difference of Two Squares' ($$a^2 - b^2$$), where $$a=x$$ and $$b=5$$.
So, it factorises to $$(a+b)(a-b)$$, which is $$(x+5)(x-5)$$.
Example 2: Factorise $$x^2 + 10x + 25$$
This looks like the 'Perfect Square' ($$a^2 + 2ab + b^2$$), where $$a=x$$ and $$b=5$$.
Let's check the middle term: $$2ab = 2(x)(5) = 10x$$. It matches!
So, it factorises to $$(a+b)^2$$, which is $$(x+5)^2$$.
Method 3: The Cross-Method
This is used for trinomials like $$x^2 + bx + c$$.
Example: Factorise $$x^2 + 7x + 12$$
1. We need two numbers that multiply to give the last term (12).
2. Those same two numbers must add to give the middle coefficient (7).
3. Let's list factors of 12: (1, 12), (2, 6), (3, 4).
4. Which pair adds up to 7? It's 3 and 4! ($$3 \times 4 = 12$$ and $$3+4=7$$).
5. So the factors are $$(x+3)$$ and $$(x+4)$$.
Key Takeaway for Part 3
Identities are shortcuts for expanding expressions. Factorising is the reverse process, where we put expressions back into brackets. Learning to recognise these patterns is a key algebra skill!
Part 4: Advanced Skills - Formulae & Inequalities
Let's use our new skills to work with scientific formulas and compare quantities.
4.1 Working with Formulae
A formula is an equation that describes a relationship between different variables, like the area of a triangle: $$A = \frac{bh}{2}$$.
Substitution
If you know the values of some variables, you can 'substitute' them into the formula to find the value of the unknown.
Example: Find the area (A) of a triangle with base (b) = 10 cm and height (h) = 4 cm.
$$A = \frac{bh}{2}$$
$$A = \frac{(10)(4)}{2}$$
$$A = \frac{40}{2} = 20$$
The area is 20 cm².
Changing the Subject of a Formula
This means rearranging the formula to get a different variable on its own. We use the same balancing rules as for solving equations.
Example: Make 'h' the subject of the formula $$A = \frac{bh}{2}$$.
1. Goal: Get 'h' alone. First, let's get rid of the 'divide by 2'.
2. Action: Multiply both sides by 2.
$$2A = bh$$
3. Goal: Get 'h' alone. Now, let's get rid of the 'b' which is multiplying it.
4. Action: Divide both sides by 'b'.
$$\frac{2A}{b} = h$$
So, our rearranged formula is $$h = \frac{2A}{b}$$.
4.2 Algebraic Fractions
These are fractions that contain variables. We can simplify and operate on them just like normal fractions.
Adding/Subtracting: You need a common denominator!
Example: $$ \frac{3}{x+1} + \frac{2}{x+2} $$
The common denominator is $$(x+1)(x+2)$$.
$$ = \frac{3(x+2)}{(x+1)(x+2)} + \frac{2(x+1)}{(x+1)(x+2)} $$
$$ = \frac{3x+6+2x+2}{(x+1)(x+2)} $$
$$ = \frac{5x+8}{(x+1)(x+2)} $$
4.3 Introduction to Inequalities
An inequality is used when two quantities are not equal. We use these signs:
- > : Greater than
- < : Less than
- ≥ : Greater than or equal to
- ≤ : Less than or equal to
Example: To go on a ride, your height (h) must be at least 120 cm. We write this as $$h \ge 120$$.
We can show the solution to an inequality on a number line. For $$x > 2$$, the solution is all numbers bigger than 2. We show this with an open circle at 2 (because it's not equal to 2) and an arrow pointing to the right.
4.4 Solving Linear Inequalities
We solve these just like equations, with one VERY important new rule.
The DANGER RULE for Inequalities
If you multiply or divide both sides of an inequality by a NEGATIVE number, you MUST FLIP the inequality sign.
Example 1 (No flip needed): Solve $$2x + 5 < 11$$
$$2x < 11 - 5$$
$$2x < 6$$
$$x < 3$$ (We divided by positive 2, so the sign stays the same.)
Example 2 (Flip needed!): Solve $$-3x + 1 \ge 10$$
$$-3x \ge 10 - 1$$
$$-3x \ge 9$$
$$\frac{-3x}{-3} \le \frac{9}{-3}$$ (We are dividing by -3, so we FLIP the sign from ≥ to ≤!)
$$x \le -3$$
Key Takeaway for Part 4
We can use our equation-solving skills to work with formulas and algebraic fractions. Inequalities are solved in a similar way, but we must remember to flip the sign when multiplying or dividing by a negative number.