Algebraic Expressions and Simplifying

In algebra we use letters to represent unknown values or variables. This lesson covers the foundations of algebra for Edexcel GCSE Mathematics (1MA1): writing algebraic expressions, collecting like terms, substitution, and using the laws of indices in algebraic contexts.

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Key Vocabulary

Term	Meaning	Example
Variable	A letter representing an unknown or changing value	x, y, n
Expression	A collection of terms (no equals sign)	3x + 2y − 7
Term	A single number, variable, or their product	$5x^{2}$5x2, −3y, 7
Coefficient	The number in front of a variable	In $5x^{2}$5x2, the coefficient is 5
Constant	A term with no variable — just a number	7, −3
Like terms	Terms with exactly the same variable parts	3x and −5x; $2x^{2}y$2x2y and $7x^{2}y$7x2y
Equation	A statement that two expressions are equal	3x + 1 = 10
Formula	An equation showing the relationship between variables	$A = \pi r^{2}$A=πr2

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Writing Algebraic Expressions

When translating words into algebra, look for the mathematical operation hidden in the language.

English phrase	Algebraic expression
5 more than x	x + 5
3 less than y	y − 3
twice n	2n
the product of a and b	ab
x divided by 4	x/4
the square of p	$p^{2}$p2
3 times the sum of x and 2	3(x + 2)

Worked Example 1

A pen costs p pence and a ruler costs r pence. Write an expression for the total cost of 4 pens and 3 rulers.

Solution: 4p + 3r

Worked Example 2

Tara is t years old. Her brother is 5 years older. Their mother is three times Tara's age.

(a) Write an expression for her brother's age: t + 5

(b) Write an expression for their mother's age: 3t

(c) Write an expression for the sum of all three ages: t + (t + 5) + 3t = 5t + 5

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Collecting Like Terms

To simplify an expression, combine terms that have identical variable parts.

Rules

• Only like terms can be added or subtracted.
• 3x and 5x are like terms (both have x).
• 3x and $3x^{2}$3x2 are not like terms (different powers).
• 2xy and 5xy are like terms; 2xy and 2x are not.

Worked Example 3

Simplify: 7a + 3b − 2a + 5b − 4

Group the like terms:

• a-terms: 7a − 2a = 5a
• b-terms: 3b + 5b = 8b
• Constants: −4

Answer: 5a + 8b − 4

Worked Example 4

Simplify: $4x^{2} + 3x - 2x^{2} + 7 - x + 1$4x2+3x−2x2+7−x+1

• $x^{2}-$x2−terms: $4x^{2} - 2x^{2} = 2x^{2}$4x2−2x2=2x2
• x-terms: 3x − x = 2x
• Constants: 7 + 1 = 8

Answer: $2x^{2} + 2x + 8$2x2+2x+8

Worked Example 5

Simplify: $5ab - 3ba + 2a^{2}b$5ab−3ba+2a2b

Since ab = ba (multiplication is commutative): $5ab - 3ab + 2a^{2}b =$5ab−3ab+2a2b= $2ab + 2a^{2}b$2ab+2a2b

Note: 2ab and $2a^{2}b$2a2b are NOT like terms — they have different variable parts.

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Multiplying and Dividing Algebraic Terms

When multiplying:

• Multiply the coefficients.
• Multiply the variables using the index laws.

When dividing:

• Divide the coefficients.
• Divide the variables using the index laws.

Worked Example 6

Simplify: $3a \times 4ab$3a×4ab

$3 \times 4 = 12$3×4=12; $a \times a = a^{2}$a×a=a2; $\times b = b$×b=b

Answer: $12a^{2}b$12a2b

Worked Example 7

Simplify: $12x^{3}y^{2} \div 4xy$12x3y2÷4xy

$12 \div 4 = 3$12÷4=3; $x^{3} \div x = x^{2}$x3÷x=x2; $y^{2} \div y = y$y2÷y=y

Answer: $3x^{2}y$3x2y

Simplifying an Algebraic Expression — Decision Flow

flowchart TD
    EXPR[Algebraic expression] --> OP{Operation present?}
    OP -->|Addition / subtraction| LIKE{Are there like terms?}
    OP -->|Multiplication| MULT["Multiply coefficients<br/>add indices on same letter"]
    OP -->|Division| DIV["Divide coefficients<br/>subtract indices on same letter"]
    OP -->|Power of a product| POW["Raise each factor to that power<br/>multiply indices"]
    LIKE -->|Yes: same variable parts| COLLECT["Add or subtract<br/>the coefficients"]
    LIKE -->|No| LEAVE[Leave as separate terms]
    COLLECT --> SIMP[Simplified expression]
    LEAVE --> SIMP
    MULT --> SIMP
    DIV --> SIMP
    POW --> SIMP
    SIMP --> NEG{"Any negative<br/>or fractional indices?"}
    NEG -->|Yes| INDEX["Apply index laws:<br/>a to the minus n = 1 over a to the n;<br/>a to the half = root a"]
    NEG -->|No| FINAL[Final answer]
    INDEX --> FINAL

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Index Laws in Algebra

The laws of indices apply to algebraic expressions in exactly the same way as for numbers.

Law	Rule	Algebraic Example
Multiply	$a^{m} \times a^{n} = a^{m+n}$am×an=am+n	$x^{3} \times x^{5} = x^{8}$x3×x5=x8
Divide	$a^{m} \div a^{n} = a^{m-n}$am÷an=am−n	$y^{7} \div y^{2} = y^{5}$y7÷y2=y5
Power of a power	$(a^{m})^{n} = a^{mn}$(am)n=amn	$(x^{4})^{3} = x^{12}$(x4)3=x12
Zero index	$a^{0} = 1$a0=1	$x^{0} = 1$x0=1
Negative index	$a^{-n} = 1/a^{n}$a−n=1/an	$x^{-3} = 1/x^{3}$x−3=1/x3

Worked Example 8

Simplify: $(2x^{3})^{4}$(2x3)4

$= 2^{4} \times (x^{3})^{4} = 16 \times x^{12} =$=24×(x3)4=16×x12= $16x^{12}$16x12

Worked Example 9

Simplify: $(3a^{2}b)^{3} \div 9a^{4}b$(3a2b)3÷9a4b

Step 1: $(3a^{2}b)^{3} = 27a^{6}b^{3}$(3a2b)3=27a6b3

Step 2: $27a^{6}b^{3} \div 9a^{4}b = 3a^{2}b^{2}$27a6b3÷9a4b=3a2b2

Answer: $3a^{2}b^{2}$3a2b2

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Substitution

Replace each variable with its given value, then evaluate.

Key rule: Always use brackets when substituting, especially for negative numbers.

Worked Example 10

Given a = 3, b = −2, c = 5, evaluate:

(a) 4a + 2b = 4(3) + 2(−2) = 12 − 4 = 8

$(b) a^{2} - bc = (3)^{2} - (-2)(5) = 9 - (-10) = 9 + 10 =$(b)a2−bc=(3)2−(−2)(5)=9−(−10)=9+10= 19

$(c) 2(a + c)^{2} = 2(3 + 5)^{2} = 2(8)^{2} = 2 \times 64 =$(c)2(a+c)2=2(3+5)2=2(8)2=2×64= 128

Worked Example 11

The formula for the area of a trapezium is A = ½(a + b)h.

Find A when a = 6, b = 10, h = 4.

A = ½(6 + 10)(4) = ½ $\times 16 \times 4 =$×16×4= 32

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Common Mistakes and Misconceptions

Mistake	Why it's wrong	Correct version
3x + 2y = 5xy	You cannot add unlike terms	3x + 2y (it's already simplified)
$2x^{2}$2x2 means $(2x)^{2}$(2x)2	The index applies only to x, not to 2	$2x^{2} = 2 \times x \times x$2x2=2×x×x; $(2x)^{2} = 4x^{2}$(2x)2=4x2
Forgetting to multiply the sign	$-2 \times -3$−2×−3 is positive, not negative	(−2)(−3) = +6
$x + x = x^{2}$x+x=x2	Adding is not the same as multiplying	x + x = 2x; $x \times x = x^{2}$x×x=x2
$a^{3} \times a^{2} = a^{6}$a3×a2=a6	Indices are added, not multiplied	$a^{3} \times a^{2} = a^{5}$a3×a2=a5

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Edexcel Exam Tips

• Paper 1 is non-calculator — practise substitution by hand, especially with negative numbers and fractions.
• On Edexcel papers, "Simplify" means collect like terms and/or use index laws to write in simplest form. Show each step of working.
• "Write an expression for…" means leave it in algebraic form — do NOT try to solve it.
• When a question says "Show that…", you must demonstrate every step of working clearly — the answer is given to you.
• Edexcel GCSE has 3 papers, each 1 hour 30 minutes. Algebra appears heavily across all three papers.
• Look for marks allocated: a 1-mark "simplify" needs one step; a 3-mark "simplify" means there are multiple stages.

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Practice Problems

1. Simplify: 6x + 3y − 2x + 4y − 1
2. Simplify: $5a^{2}b \times 3ab^{2}$5a2b×3ab2
3. Simplify: $20x^{5}y^{3} \div 5x^{2}y$20x5y3÷5x2y
4. Simplify: $(4m^{2})^{3}$(4m2)3
5. If p = −3 and q = 4, evaluate: $2p^{2} - 3pq + q$2p2−3pq+q
6. Write an expression for: "Five less than twice the square of n."
7. Simplify: $3x + 7 - x^{2} + 2x^{2} - 5x + 1$3x+7−x2+2x2−5x+1
8. Simplify: $(2a^{3}b^{2})^{2} \times 3a^{2}b$(2a3b2)2×3a2b

Answers

1. 4x + 7y − 1
2. $15a^{3}b^{3}$15a3b3
3. $4x^{3}y^{2}$4x3y2
4. $64m^{6}$64m6
5. 2(9) − 3(−3)(4) + 4 = 18 + 36 + 4 = 58
6. $2n^{2} - 5$2n2−5
7. $x^{2} - 2x + 8$x2−2x+8
8. $4a^{6}b^{4} \times 3a^{2}b = 12a^{8}b^{5}$4a6b4×3a2b=12a8b5

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Summary

• An expression contains terms but no equals sign.
• Like terms have identical variable parts and can be collected.
• The index laws (multiply $\to add$→add indices; divide $\to$→ subtract indices; power of a power $\to$→ multiply indices) work the same way for algebra as for numbers.
• Substitution means replacing letters with values — always use brackets for negative numbers.
• On Edexcel GCSE, simplifying expressions and substitution appear on every paper series — practise until they are automatic.

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Extended Worked Examples

Worked Example 12 — Mixed Powers with Negative Bases

Simplify: $(-2a^{3})^{2} \times 5a^{4}$(−2a3)2×5a4

Apply the power of a product rule. Everything inside the bracket is squared, so $(-2)^{2} = 4$(−2)2=4 and $(a^{3})^{2} = a^{6}$(a3)2=a6.

$(-2a^{3})^{2} = 4a^{6}$(−2a3)2=4a6.

Then $4a^{6} \times 5a^{4} = 20a^{10}$4a6×5a4=20a10.

Answer: $20a^{10}$20a10.

Note that the overall sign is positive because $(-2)^{2} = 4$(−2)2=4 — a very common slip is to leave the sign as negative.

Worked Example 13 — Fractional Index Law in an Algebraic Context

Simplify: $x^{\frac{1}{2}} \times x^{\frac{3}{2}}$x21​×x23​.

Add the indices: $\frac{1}{2} + \frac{3}{2} = 2$21​+23​=2.

Answer: $x^{2}$x2.

Then try: $\frac{x^{\frac{5}{2}}}{x^{\frac{1}{2}}} = x^{\frac{5}{2} - \frac{1}{2}} = x^{2}$x21​x25​​=x25​−21​=x2. The same answer, obtained by the subtract-indices rule.

Worked Example 14 — Simplifying Before Substituting

Let $p = 4$p=4 and $q = -3$q=−3. Evaluate $\frac{3p^{2}q - pq^{2}}{pq}$pq3p2q−pq2​ after simplifying first.

Simplify: Divide every term on the numerator by $pq$pq.

$\frac{3p^{2}q}{pq} = 3p$pq3p2q​=3p, and $\frac{pq^{2}}{pq} = q$pqpq2​=q.

So the expression becomes $3p - q$3p−q.

Substitute: $3(4) - (-3) = 12 + 3 = 15$3(4)−(−3)=12+3=15.

Answer: 15.

Contrast with substituting first: $\frac{3(16)(-3) - (4)(9)}{(4)(-3)} = \frac{-144 - 36}{-12} = \frac{-180}{-12} = 15$(4)(−3)3(16)(−3)−(4)(9)​=−12−144−36​=−12−180​=15. Same answer — but simplifying first is faster and less error-prone, especially on Paper 1 (non-calculator).

Worked Example 15 — Working with Algebraic Fractions

Simplify: $\frac{6x^{2}y}{9xy^{3}}$9xy36x2y​.

Cancel common factors: the HCF of 6 and 9 is 3; $x^{2} \div x = x$x2÷x=x; $y \div y^{3} = \frac{1}{y^{2}}$y÷y3=y21​.

Answer: $\frac{2x}{3y^{2}}$3y22x​.

Worked Example 16 — Using Formulae by Substitution

The kinetic energy of a moving object is given by $E = \frac{1}{2}mv^{2}$E=21​mv2, where m is mass in kg and v is speed in m/s.

Find E when m = 8 kg and v = 5 m/s.

$E = \frac{1}{2}(8)(5^{2}) = \frac{1}{2}(8)(25) = 4 \times 25 = 100$E=21​(8)(52)=21​(8)(25)=4×25=100 J.

Answer: 100 J (joules).

Why Bracketing Matters in Substitution

Consider $f = a - b^{2}$f=a−b2 with a = 7, b = −3.

• Correct: $f = 7 - (-3)^{2} = 7 - 9 = -2$f=7−(−3)2=7−9=−2.
• Incorrect: $f = 7 - -3^{2} = 7 - -9 = 16$f=7−−32=7−−9=16 (treating $-3^{2}$−32 as $-(3^{2})$−(32) but then getting signs muddled).

Always bracket negative substitutions: write $(-3)^{2}$(−3)2 so that the square applies to the whole of −3, giving +9 rather than accidentally −9.

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Answering at different grade levels

Exam-style question: "Simplify fully $5a^{2}b \times 2ab^{3}$5a2b×2ab3, and then substitute $a = 2$a=2, $b = -1$b=−1 to find the value of the simplified expression."

Grade 3–4 response: A student at this level recognises that this is a multiplication and multiplies the numbers 5 × 2 = 10, but often stops there or forgets to use the index laws on the variables. A typical answer is $10a^{2}b \cdot ab^{3}$10a2b⋅ab3 without combining, or $10a^{3}b^{3}$10a3b3 (missing one of the indices). On substitution they may not bracket $(-1)$(−1) and lose the sign.

Grade 5–6 response: Applies the index laws correctly: $5 \times 2 = 10$5×2=10, $a^{2} \times a = a^{3}$a2×a=a3, $b \times b^{3} = b^{4}$b×b3=b4, giving $10a^{3}b^{4}$10a3b4. When substituting: $10(2)^{3}(-1)^{4} = 10 \times 8 \times 1 = 80$10(2)3(−1)4=10×8×1=80. Working is shown line-by-line, brackets used correctly.

Grade 7–9 response: Does all of the above, but also states the rule being applied ("using $a^{m} \times a^{n} = a^{m+n}$am×an=am+n"), verifies by checking whether the final sign is sensible (since $b^{4}$b4 is always non-negative), and writes the final answer as $10a^{3}b^{4} = 80$10a3b4=80 when $a = 2$a=2, $b = -1$b=−1. A top-band candidate would add a brief note: "Because the index 4 is even, any negative sign on b disappears — so the answer is always positive for non-zero b."

The distinction: Grade 3–4 candidates see one operation at a time; Grade 5–6 candidates execute the expand / factorise / index-law machinery correctly; Grade 7–9 candidates also anticipate and justify why their answer has the form it does.

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Edexcel alignment: This content is aligned with Edexcel GCSE Mathematics (1MA1) specification — specifically Topic A [A1 Notation, vocabulary and manipulation; A2 Indices and index laws; A3 Expand and factorise; A4 Algebraic formulae and substitution]. Assessed on Papers 1, 2, 3.