Expressions and Simplifying

Algebra is the language we use to describe relationships between quantities when one or more of those quantities is unknown or can vary. Instead of a fixed number we write a letter, and the rules of arithmetic carry straight over. This lesson lays the foundations for the whole Algebra strand of OCR GCSE Mathematics (J560): writing and reading expressions, collecting like terms, multiplying and dividing terms, applying the index laws in an algebraic setting, and substituting numbers into expressions and formulae. The main assessment objective here is AO1 — using and applying standard techniques fluently and accurately — but the way you set out your working also feeds AO2 (reasoning and communicating clearly), which examiners reward throughout the paper.

Key Vocabulary

Term	Meaning
Variable	A letter that stands for an unknown or changing number, e.g. $n$n, $t$t, $x$x
Term	A single number, a single variable, or a product of numbers and variables, e.g. $5x^{2}$5x2, $-3ab$−3ab, $7$7
Coefficient	The number multiplying the variable part of a term; in $-4y$−4y the coefficient is $-4$−4
Constant	A term that is just a number, with no variable, e.g. $9$9
Expression	A collection of terms joined by $+$+ or $-$−, with no equals sign, e.g. $3a + 2b - 5$3a+2b−5
Like terms	Terms with exactly the same variable part, e.g. $4x$4x and $-7x$−7x, or $2x^{2}y$2x2y and $9x^{2}y$9x2y
Simplify	Rewrite an expression in its shortest equivalent form by collecting like terms and using index laws
Substitute	Replace each variable with a given number, then evaluate

Why does algebra matter so much? Because almost every later topic in the course — equations, graphs, sequences, even the formulae you meet in geometry and statistics — is written in this language. If you are fluent at handling expressions, the rest of GCSE mathematics becomes far more accessible; if you are shaky, small slips multiply through every question. The good news is that algebra obeys exactly the same arithmetic rules you already know for numbers. The only new idea is that a letter stands in for a number we either do not yet know or are allowing to change. Throughout this lesson, notice how often the "rule" is really just an arithmetic fact you have used since primary school, now written with letters.

Reading and Writing Expressions

Before you can manipulate algebra, you have to translate between ordinary English and symbols. Reading algebra is a skill in its own right: a single short expression can pack in several operations, and the order in which they are written matters. Look for the operation hidden inside the wording — words such as "sum", "total" and "more than" signal addition; "difference" and "less than" signal subtraction; "product", "times" and "of" signal multiplication; and "shared", "split" and "per" signal division. Be especially careful with subtraction wording: "$4$4 less than $k$k" is $k - 4$k−4, not $4 - k$4−k, because we start from $k$k and take $4$4 away.

Words	Expression
$7$7 more than $m$m	$m + 7$m+7
$4$4 less than $k$k	$k - 4$k−4
double $p$p	$2p$2p
the product of $a$a and $b$b	$ab$ab
$w$w shared equally between $5$5	$\dfrac{w}{5}$5w​
the square of $t$t	$t^{2}$t2
$6$6 times the sum of $x$x and $3$3	$6(x + 3)$6(x+3)

Notice that "the square of $t$t" is $t^{2}$t2, but "twice $t$t" is $2t$2t — multiplying by $2$2 and raising to the power $2$2 are completely different.

Worked Example 1

A bag of crisps costs $c$c pence and a chocolate bar costs $b$b pence. Write an expression for the total cost, in pence, of $5$5 bags of crisps and $2$2 chocolate bars.

Five bags cost $5c$5c and two bars cost $2b$2b, so the total is $5c + 2b$5c+2b pence.

Answer: $5c + 2b$5c+2b

Worked Example 2

Priya is $n$n years old. Her cousin is $4$4 years younger, and her uncle is three times Priya's age.

(a) Cousin's age: $n - 4$n−4

(b) Uncle's age: $3n$3n

(c) The sum of all three ages: $n + (n - 4) + 3n = 5n - 4$n+(n−4)+3n=5n−4

Answer: $5n - 4$5n−4

Collecting Like Terms

To simplify by collecting like terms, you add or subtract the coefficients of terms that have an identical variable part. Terms with different variable parts must be left separate. Think of the variable part as a "label": you can only combine terms that carry exactly the same label, in the same way that you can add £$3$3 to £$5$5 but you cannot add £$3$3 to $5$5 apples. The number in front (the coefficient) tells you how many of that labelled object you have, so collecting like terms is simply counting.

• $6x$6x and $x$x are like terms (both are multiples of $x$x); note $x$x means $1x$1x.
• $6x$6x and $6x^{2}$6x2 are not like terms — the powers differ, so the labels $x$x and $x^{2}$x2 are different.
• $3pq$3pq and $5qp$5qp are like terms, because $pq = qp$pq=qp (multiplication can be done in any order).
• $7$7 and $-2$−2 are like terms (both are constants, with no variable label at all).

A neat habit, especially with longer expressions, is to underline or highlight each type of term in a different way before you combine them. This stops you accidentally pairing an $x$x-term with an $x^{2}$x2-term, which is one of the most common slips at this level.

Worked Example 3

Simplify $8a + 5b - 3a + 2b - 6$8a+5b−3a+2b−6.

Group each type of term:

• $a$a-terms: $8a - 3a = 5a$8a−3a=5a
• $b$b-terms: $5b + 2b = 7b$5b+2b=7b
• constants: $-6$−6

Answer: $5a + 7b - 6$5a+7b−6

Worked Example 4

Simplify $5x^{2} + 4x - 2x^{2} - x + 3$5x2+4x−2x2−x+3.

• $x^{2}$x2-terms: $5x^{2} - 2x^{2} = 3x^{2}$5x2−2x2=3x2
• $x$x-terms: $4x - x = 3x$4x−x=3x
• constants: $+3$+3

Answer: $3x^{2} + 3x + 3$3x2+3x+3

Common error: treating $4x - x$4x−x as $4$4 (forgetting that $x$x means $1x$1x, so $4x - x = 3x$4x−x=3x, not $4$4).

Worked Example 5

Simplify $7mn - 2nm + 4m^{2}n$7mn−2nm+4m2n.

Because $mn = nm$mn=nm, the first two terms combine: $7mn - 2mn = 5mn$7mn−2mn=5mn. The last term, $4m^{2}n$4m2n, has a different variable part and stays as it is.

Answer: $5mn + 4m^{2}n$5mn+4m2n

Multiplying and Dividing Terms

When you multiply terms, multiply the coefficients and combine the letters using the index laws. When you divide, divide the coefficients and subtract the powers of matching letters.

Worked Example 6

Simplify $4a \times 5ab$4a×5ab.

Multiply the numbers: $4 \times 5 = 20$4×5=20. Combine the letters: $a \times a = a^{2}$a×a=a2 and there is one $b$b.

Answer: $20a^{2}b$20a2b

Worked Example 7

Simplify $\dfrac{18x^{4}y^{3}}{6xy}$6xy18x4y3​.

Divide the coefficients: $18 \div 6 = 3$18÷6=3. Subtract powers: $x^{4} \div x = x^{3}$x4÷x=x3 and $y^{3} \div y = y^{2}$y3÷y=y2.

Answer: $3x^{3}y^{2}$3x3y2

Common error: writing $x^{4} \div x = x^{4}$x4÷x=x4 by forgetting that the single $x$x on the bottom is $x^{1}$x1, so the power drops by $1$1.

Worked Example 7b

Simplify $2x \times 3x \times 4x$2x×3x×4x.

Multiply the coefficients together: $2 \times 3 \times 4 = 24$2×3×4=24. Multiply the variables: $x \times x \times x = x^{3}$x×x×x=x3.

Answer: $24x^{3}$24x3

It is worth pausing on the difference between $2x \times 3x$2x×3x and $2x + 3x$2x+3x. The first is a multiplication, giving $6x^{2}$6x2 (multiply numbers, add the single powers of $x$x), whereas the second is collecting like terms, giving $5x$5x. Students who blur these two operations lose marks throughout the paper, so always check whether the symbol between the terms is a $\times$× or a $+$+.

Index Laws in Algebra

The laws of indices work for letters exactly as they do for numbers.

Law	Rule	Example
Multiplying	$a^{m} \times a^{n} = a^{m+n}$am×an=am+n	$x^{4} \times x^{3} = x^{7}$x4×x3=x7
Dividing	$a^{m} \div a^{n} = a^{m-n}$am÷an=am−n	$y^{8} \div y^{5} = y^{3}$y8÷y5=y3
Power of a power	$(a^{m})^{n} = a^{mn}$(am)n=amn	$(x^{2})^{5} = x^{10}$(x2)5=x10
Power of a product	$(ab)^{n} = a^{n}b^{n}$(ab)n=anbn	$(2x)^{3} = 8x^{3}$(2x)3=8x3
Zero index	$a^{0} = 1$a0=1	$7^{0} = 1$70=1

Worked Example 8

Simplify $(3x^{4})^{2}$(3x4)2.

Square both factors inside the bracket: $3^{2} = 9$32=9 and $(x^{4})^{2} = x^{8}$(x4)2=x8.

Answer: $9x^{8}$9x8

Common error: writing $(3x^{4})^{2} = 3x^{8}$(3x4)2=3x8 — the coefficient must be squared too, giving $9x^{8}$9x8.

Worked Example 8b

Simplify $\dfrac{x^{6} \times x^{2}}{x^{3}}$x3x6×x2​.

Deal with the top first using the multiplying rule: $x^{6} \times x^{2} = x^{8}$x6×x2=x8. Then divide: $x^{8} \div x^{3} = x^{5}$x8÷x3=x5.

Answer: $x^{5}$x5

When several index laws appear together, work through them one at a time in the natural order — simplify any products or powers first, then carry out the division. Trying to do everything in a single leap is where errors creep in.

Substitution

To substitute, replace each letter with its value and evaluate, following the order of operations (often remembered as BIDMAS: Brackets, Indices, Division and Multiplication, Addition and Subtraction). Always write negative numbers inside brackets so the signs behave correctly. Substitution appears on every paper and underpins every formula you will use across the whole subject, so it is worth practising until it is second nature.

Worked Example 9

Given $a = 4$a=4, $b = -3$b=−3 and $c = 2$c=2, work out $3a - 2b + c$3a−2b+c.

$3(4) - 2(-3) + 2 = 12 + 6 + 2 = 20$3(4)−2(−3)+2=12+6+2=20.

Answer: $20$20

Worked Example 10

Using the same values, work out $a^{2} - bc$a2−bc.

$(4)^{2} - (-3)(2) = 16 - (-6) = 16 + 6 = 22$(4)2−(−3)(2)=16−(−6)=16+6=22.

Answer: $22$22

Common error: writing $-(-3)(2) = -6$−(−3)(2)=−6 by mishandling the double negative. Because $bc = (-3)(2) = -6$bc=(−3)(2)=−6, the calculation is $16 - (-6) = 16 + 6 = 22$16−(−6)=16+6=22.

Extended Worked Examples

Worked Example 11 — Power of a product with a negative coefficient

Simplify $(-2p^{3})^{2} \times 4p$(−2p3)2×4p.

Square the bracket first: $(-2)^{2} = 4$(−2)2=4 and $(p^{3})^{2} = p^{6}$(p3)2=p6, giving $4p^{6}$4p6. Then $4p^{6} \times 4p = 16p^{7}$4p6×4p=16p7.

Answer: $16p^{7}$16p7

The sign is positive because squaring removes the minus. A very common slip is to leave the result negative.

Worked Example 12 — Simplify, then substitute

Simplify $\dfrac{12x^{3}y}{4xy}$4xy12x3y​ and then find its value when $x = 3$x=3 and $y = 5$y=5.

Simplifying first: $\dfrac{12x^{3}y}{4xy} = 3x^{2}$4xy12x3y​=3x2. Substituting: $3(3)^{2} = 3 \times 9 = 27$3(3)2=3×9=27.

Answer: $27$27

Simplifying before substituting is faster and far less error-prone on the non-calculator paper.

Worked Example 13 — Substituting into a formula

The surface area of a closed cuboid is $S = 2lw + 2lh + 2wh$S=2lw+2lh+2wh. Work out $S$S when $l = 5$l=5, $w = 3$w=3 and $h = 4$h=4 (lengths in cm).

$S = 2(5)(3) + 2(5)(4) + 2(3)(4) = 30 + 40 + 24 = 94$S=2(5)(3)+2(5)(4)+2(3)(4)=30+40+24=94.

Answer: $94\ \text{cm}^{2}$94 cm2

Worked Example 14 — Why brackets matter

Work out $5 - t^{2}$5−t2 when $t = -3$t=−3.

Write the substitution with a bracket: $5 - (-3)^{2} = 5 - 9 = -4$5−(−3)2=5−9=−4. Because the squaring applies to the whole of $-3$−3, we get $+9$+9 inside, then subtract.

Answer: $-4$−4

Worked Example 15 — A two-variable formula in context

The cost in pounds of hiring a hall is given by $C = 40 + 15h$C=40+15h, where $h$h is the number of hours. Work out the cost of a $4$4-hour booking, and the cost if the booking is extended to $6$6 hours.

For $4$4 hours: $C = 40 + 15(4) = 40 + 60 = 100$C=40+15(4)=40+60=100, so the cost is £$100$100. For $6$6 hours: $C = 40 + 15(6) = 40 + 90 = 130$C=40+15(6)=40+90=130, so the cost is £$130$130.

Answer: £$100$100 for $4$4 hours and £$130$130 for $6$6 hours.

Notice the structure of this formula: the $40$40 is a fixed charge and the $15h$15h grows with the hours. Recognising the fixed part and the variable part of a formula is an idea you will meet again when you study straight-line graphs.

Worked Example 16 — Simplifying a longer expression

Simplify $4a + 3a^{2} - a + 5 - a^{2} + 2$4a+3a2−a+5−a2+2.

Sort the terms by type. The $a^{2}$a2-terms: $3a^{2} - a^{2} = 2a^{2}$3a2−a2=2a2. The $a$a-terms: $4a - a = 3a$4a−a=3a. The constants: $5 + 2 = 7$5+2=7.

Answer: $2a^{2} + 3a + 7$2a2+3a+7

Writing the answer with the highest power first ($a^{2}$a2, then $a$a, then the constant) is the conventional order and makes your work easy for a marker to read.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: "Simplify fully $6a^{2}b \times 3ab^{4}$6a2b×3ab4. Then substitute $a = 2$a=2 and $b = -1$b=−1 to work out the value of your simplified expression."

Grades 3–4 response: Recognises a multiplication and multiplies the numbers to get $18$18, but often stops at $18a^{2}b \times ab^{4}$18a2b×ab4 without combining the letters, or writes $18a^{3}b^{4}$18a3b4 but loses one index. On substitution may forget to bracket $-1$−1 and mishandle the sign.

Grades 5–6 response: Applies the index laws correctly: $6 \times 3 = 18$6×3=18, $a^{2} \times a = a^{3}$a2×a=a3, $b \times b^{4} = b^{5}$b×b4=b5, giving $18a^{3}b^{5}$18a3b5. Substitutes with brackets: $18(2)^{3}(-1)^{5} = 18 \times 8 \times (-1) = -144$18(2)3(−1)5=18×8×(−1)=−144.

Grades 7–9 response: Does all of the above and names the rule ($a^{m} \times a^{n} = a^{m+n}$am×an=am+n), then checks the sign is sensible — because $b^{5}$b5 has an odd power, a negative $b$b keeps the answer negative. Writes the final line clearly as $18a^{3}b^{5} = -144$18a3b5=−144 when $a = 2$a=2, $b = -1$b=−1.

Examiner-style commentary: the method mark is for the index laws; the accuracy mark needs the correct sign, which is exactly where weaker candidates slip.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: "Given that $p = -2$p=−2 and $q = 5$q=5, work out the value of $2p^{2} - pq$2p2−pq."

Grades 3–4 response: Substitutes but may write $2(-2)^{2}$2(−2)2 as $2 \times -4 = -8$2×−4=−8, squaring incorrectly, or treat $-pq$−pq as $-(-2)(5) = 10$−(−2)(5)=10 with uncertain reasoning, arriving at a muddled total.

Grades 5–6 response: $2(-2)^{2} = 2 \times 4 = 8$2(−2)2=2×4=8 and $pq = (-2)(5) = -10$pq=(−2)(5)=−10, so $2p^{2} - pq = 8 - (-10) = 18$2p2−pq=8−(−10)=18. Working shown line by line.

Grades 7–9 response: All of the above, presented compactly with each substitution bracketed, and a final check that $-(-10) = +10$−(−10)=+10 is handled correctly to give $18$18.

Examiner-style commentary: OCR rewards bracketed substitution; the most common lost mark is squaring a negative as if it were negative.

Worked Example 17 — Dividing terms with the same letter on top and bottom

Simplify $\dfrac{15a^{5}b^{2}}{20a^{2}b^{2}}$20a2b215a5b2​.

Deal with the number and each letter separately. The numbers simplify like an ordinary fraction: $\dfrac{15}{20} = \dfrac{3}{4}$2015​=43​. For the $a$a-terms, subtract the powers: $a^{5} \div a^{2} = a^{3}$a5÷a2=a3. For the $b$b-terms, $b^{2} \div b^{2} = b^{0} = 1$b2÷b2=b0=1, so the $b$b disappears entirely.

Answer: $\dfrac{3a^{3}}{4}$43a3​

Common error: writing $\dfrac{15}{20} = \dfrac{3}{4}$2015​=43​ but then leaving a stray $b$b in the answer, forgetting that $b^{2} \div b^{2} = 1$b2÷b2=1, not $b$b. When the powers of a letter are equal on top and bottom, that letter cancels to $1$1 and is simply not written.

Common Mistakes and Misconceptions

As a teacher I see the same handful of slips again and again. Watch for adding unlike terms — $3x + 2y$3x+2y is not $5xy$5xy; it is already simplified. Be careful that $2x^{2}$2x2 means $2 \times x \times x$2×x×x, whereas $(2x)^{2} = 4x^{2}$(2x)2=4x2 — the index in $2x^{2}$2x2 applies only to the $x$x. Remember that $x + x = 2x$x+x=2x but $x \times x = x^{2}$x×x=x2; addition and multiplication are not interchangeable. When using index laws, indices are added when multiplying, not multiplied: $a^{3} \times a^{2} = a^{5}$a3×a2=a5, not $a^{6}$a6. Finally, always bracket negatives in substitution so that, for example, $(-3)^{2} = 9$(−3)2=9 rather than an accidental $-9$−9.

Exam Tips

• AO1 (use and apply): "Simplify" on an OCR paper means collect like terms and use index laws to reach the shortest form. Show each stage — a $2$2- or $3$3-mark "simplify" expects more than one step.
• AO2 (reason and communicate): when a question says "Show that", the answer is given to you, so every line of working must be visible and logically ordered; a correct final value with no working scores little.
• AO3 (solve problems): "Write down an expression for…" in a worded context means stay in algebra — do not try to find a number. Read the units and the command word ("Work out", "Find", "Write down") carefully.
• Paper 1 of J560 is non-calculator, so practise substitution with negatives and fractions by hand until it is automatic.

Summary

• An expression is a collection of terms with no equals sign.
• Like terms share an identical variable part and can be collected by adding or subtracting coefficients.
• Multiply terms by multiplying coefficients and adding indices; divide by dividing coefficients and subtracting indices.
• The index laws ($a^{m} \times a^{n} = a^{m+n}$am×an=am+n, $a^{m} \div a^{n} = a^{m-n}$am÷an=am−n, $(a^{m})^{n} = a^{mn}$(am)n=amn) work for letters just as for numbers.
• Substitute by replacing letters with values, always bracketing negatives.
• Clear, step-by-step working earns method marks even when the final answer slips.

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This content is aligned with the OCR GCSE Mathematics (J560) specification.