Expressions and formulae
Section: Algebra | Syllabus: Cambridge Lower Secondary Checkpoint Mathematics (0862)
What is an Algebraic Expression?
An algebraic expression is a collection of terms - numbers, letters (variables) and operations - combined together, with no equals sign.
- Term: a single number, variable, or product of numbers and variables, e.g. 5x, -3y², 7.
- Variable: a letter that represents an unknown or changing value, e.g. x, n.
- Coefficient: the number multiplying a variable, e.g. in 5x, the coefficient is 5.
- Constant: a term with no variable, e.g. in 3x + 7, the constant is 7.
- Like terms: terms with exactly the same variable(s) and power(s), e.g. 4x and 9x, or 2x² and 5x².
Example: in the expression 4x² + 3x - 7, there are three terms: 4x² (coefficient 4), 3x (coefficient 3), and the constant -7.
Order of Operations and Simplifying Expressions
The same order of operations used with numbers (brackets, indices, multiplication/division, addition/subtraction) applies to algebraic terms and expressions. To simplify an expression, combine like terms only - add or subtract their coefficients and keep the variable part unchanged.
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Question: Simplify 7a - 3b - 2a + 5b.
- Step 1: Group like terms: (7a - 2a) + (-3b + 5b)
- Step 2: 7a - 2a = 5a, and -3b + 5b = 2b
- Answer: 5a + 2b
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Question: Simplify 5 + 2x × 3, using the order of operations.
- Step 1: Multiplication happens before addition: 2x × 3 = 6x
- Step 2: 5 + 6x
- Answer: 6x + 5
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Question: Simplify 2x² + 3x² - x².
- Step 1: All three terms are like terms (x²), so combine coefficients: 2 + 3 - 1 = 4
- Answer: 4x²
Substituting Negative Numbers into Expressions
Substituting a negative number is exactly the same process as substituting a positive one - but the sign rules for multiplying need extra care, especially when a power is involved.
- An even power of a negative number is always positive, e.g. (-2)² = 4.
- An odd power of a negative number stays negative, e.g. (-2)³ = -8.
- Subtracting a negative number is the same as adding, e.g. 16 - (-6) = 16 + 6.
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Question: Work out the value of 4y² - 3y when y = -2.
- Step 1: Square first: y² = (-2)² = 4, so 4y² = 4 × 4 = 16
- Step 2: -3y = -3 × (-2) = 6
- Step 3: Combine: 16 + 6
- Answer: 22
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Question: Work out the value of 2(m + 4)² - m³ when m = -3.
- Step 1: Work out the bracket first: m + 4 = -3 + 4 = 1
- Step 2: Square it, then double: 2 × 1² = 2 × 1 = 2
- Step 3: Cube m: m³ = (-3)³ = -27 (an odd power, so it stays negative)
- Step 4: Combine: 2 - (-27) = 2 + 27
- Answer: 29
Expanding Single Brackets
To expand a bracket, multiply everything inside it by the term outside - this is the distributive law: a(b + c) = ab + ac.
The area model for a(b + c) = ab + ac: one rectangle split into two smaller areas
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Question: Expand 3(x + 4).
- Step 1: Multiply 3 by x: 3 × x = 3x
- Step 2: Multiply 3 by 4: 3 × 4 = 12
- Answer: 3x + 12
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Question: Expand -2(x - 5).
- Step 1: -2 × x = -2x
- Step 2: -2 × -5 = +10
- Answer: -2x + 10
Expanding Two Brackets
To expand two brackets like (x + 3)(x + 5), multiply every term in the first bracket by every term in the second bracket, then simplify by combining like terms.
The area model for (x + 3)(x + 5): four smaller areas that add up to the expansion
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Question: Expand and simplify (x + 3)(x + 5).
- Step 1: x × x = x². x × 5 = 5x. 3 × x = 3x. 3 × 5 = 15
- Step 2: Combine: x² + 5x + 3x + 15
- Answer: x² + 8x + 15
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Question: Expand and simplify (x - 2)(x + 7).
- Step 1: x × x = x². x × 7 = 7x. -2 × x = -2x. -2 × 7 = -14
- Step 2: Combine: x² + 7x - 2x - 14
- Answer: x² + 5x - 14
Laws of Indices in Algebra
The same index laws used with numbers apply to algebraic terms - just remember to deal with coefficients and indices separately.
- Multiplying: add the indices - am × an = am+n
- Dividing: subtract the indices - am ÷ an = am-n
- Power of a power: multiply the indices - (am)n = amn
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Question: Simplify 3x² × 4x³.
- Step 1: Multiply the coefficients: 3 × 4 = 12
- Step 2: Add the indices: x² × x³ = x⁵
- Answer: 12x⁵
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Question: Simplify 12x⁵ ÷ 3x².
- Step 1: Divide the coefficients: 12 ÷ 3 = 4
- Step 2: Subtract the indices: x⁵ ÷ x² = x³
- Answer: 4x³
Simplifying Algebraic Fractions
An algebraic fraction can be simplified by dividing the numerator and denominator by their highest common factor - if the numerator has more than one term, factorise it first.
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Question: Simplify 6x²/3x.
- Step 1: Divide the coefficients: 6 ÷ 3 = 2
- Step 2: Divide the powers of x: x² ÷ x = x
- Answer: 2x
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Question: Simplify (x² + 3x)/x.
- Step 1: Factorise the numerator: x² + 3x = x(x + 3)
- Step 2: Cancel the common factor of x: x(x + 3)/x
- Answer: x + 3
Adding and Subtracting Algebraic Fractions
Two separate algebraic fractions combine the same way as numerical fractions: rewrite both over a common denominator, then add or subtract the numerators.
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