Hard GCSE Algebraic vocabulary Questions

Challenging, exam-style GCSE Algebraic vocabulary questions with worked solutions. Stretch yourself on the hardest expression, equation, formula, identity problems.

expressionequationformulaidentityclassifyingterms
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
A taxi charges a fixed fee plus a rate per mile. The total cost is C=3+2mC = 3 + 2m, where mm is the number of miles. (a) Classify C=3+2mC = 3 + 2m. (b) Name the subject. (c) List the variables. (d) State the constant (fixed fee). (e) State the coefficient of mm and explain what it represents. (f) Classify 3+2m3 + 2m on its own.
Show worked solution

Worked solution

  1. Look at the whole statement

    C=3+2mC = 3 + 2m

    It has an equals sign and two letters.

  2. Check for a formula

    C←mC \leftarrow m

    It gives a rule for cost from miles, so it is a formula.

  3. Classify the whole statement

    formula\text{formula}

    So C=3+2mC = 3 + 2m is a formula.

  4. Identify the subject

    C=…C = \ldots

    The subject is the letter on its own: C.

  5. List the letters

    C,\; m

    The letters used are C and m.

  6. State the variables

    C,\; m

    The variables are C and m.

  7. Find the constant term

    33

    The fixed fee is the constant term 3.

  8. Interpret the constant

    fixed fee=3\text{fixed fee} = 3

    The 3 is a fixed charge that does not depend on distance.

  9. Find the term in m

    2m2m

    The term containing m is 2m.

  10. Read off the coefficient of m

    2m→22m \rightarrow 2

    The coefficient of m is 2.

  11. Interpret the coefficient

    2=cost per mile2 = \text{cost per mile}

    The 2 is the cost per mile travelled.

  12. Look at 3 + 2m alone

    3+2m  (no =)3 + 2m \;(\text{no } =)

    On its own there is no equals sign.

  13. Classify 3 + 2m alone

    expression\text{expression}

    So 3 + 2m by itself is an expression.

  14. Bring the parts together

    formula;  C;  C,m;  3;  2;  expression\text{formula};\; C;\; C,m;\; 3;\; 2;\; \text{expression}

    Collect all answers.

  15. State all the answers

    formula;  C;  C,m;  3;  2;  expression\text{formula};\; C;\; C,m;\; 3;\; 2;\; \text{expression}

    It is a formula; subject C; variables C and m; constant 3; coefficient of m is 2 (cost per mile); and 3 + 2m alone is an expression.

Answer
formula; subject C; variables C and m; constant 3; coefficient of m is 2 (cost per mile); 3 + 2m alone is an expression
Question 2
6 markschallenging
Classify each of the following: (a) 2p+3q2p + 3q (b) 2p+3q=122p + 3q = 12 (c) I=PRTI = PRT (d) 2(p+q)≡2p+2q2(p + q) \equiv 2p + 2q (e) 3q3q.
Show worked solution

Worked solution

  1. Recall what an expression is

    expression: terms, no =\text{expression: terms, no } =

    An expression is just a collection of terms with no equals sign.

  2. Recall what an equation is

    equation: = , some values\text{equation: } = \text{ , some values}

    An equation has an equals sign and is true only for particular value(s) of the letter.

  3. Recall what a formula is

    formula: = , linked letters\text{formula: } = \text{ , linked letters}

    A formula has an equals sign and links two or more different letters (variables).

  4. Recall what an identity is

    a≡b for all valuesa \equiv b \text{ for all values}

    An identity is true for every value of the letter and is written with the identity sign.

  5. Examine part (a)

    2p+3q2p + 3q

    There is no equals sign, so it is simply a collection of terms.

  6. Classify part (a)

    ⇒expression\Rightarrow \text{expression}

    No equals sign, so it is an expression.

  7. Examine part (b)

    2p+3q=122p + 3q = 12

    There is an equals sign and it is true only for a particular value of the letter.

  8. Classify part (b)

    ⇒equation\Rightarrow \text{equation}

    Has an equals sign; true for particular values of p and q.

  9. Examine part (c)

    I = PRT

    There is an equals sign linking two or more different letters.

  10. Classify part (c)

    ⇒formula\Rightarrow \text{formula}

    An equals sign links the different letters I, P, R and T.

  11. Examine part (d)

    2(p+q)≡2p+2q2(p + q) \equiv 2p + 2q

    Both sides are equal for every value of the letter, so the identity sign is used.

  12. Classify part (d)

    ⇒identity\Rightarrow \text{identity}

    Expanding gives 2p + 2q for all values.

  13. Examine part (e)

    3q3q

    There is no equals sign, so it is simply a collection of terms.

  14. Classify part (e)

    ⇒expression\Rightarrow \text{expression}

    A single term with no equals sign is an expression.

  15. State the classifications

    (a) expression, (b) equation, (c) formula, (d) identity, (e) expression\text{(a) expression, (b) equation, (c) formula, (d) identity, (e) expression}

    Each statement has now been correctly named.

Answer
(a) expression, (b) equation, (c) formula, (d) identity, (e) expression
Question 3
6 markschallenging
Decide whether each of the following is an identity or an equation: (a) 4(x+2)=4x+84(x + 2) = 4x + 8 (b) 3x−1=83x - 1 = 8 (c) x⋅x=x2x \cdot x = x^2 (d) 5x−2x=3x5x - 2x = 3x (e) 2x=x+x2x = x + x.
Show worked solution

Worked solution

  1. Recall equation versus identity

    = vs ≡= \text{ vs } \equiv

    An equation has an equals sign and is true only for particular value(s) of the letter. An identity is true for every value of the letter and is written with the identity sign.

  2. Expand part (a)

    4(x+2)=4x+84(x + 2) = 4x + 8

    Multiplying out matches the right side.

  3. Classify part (a)

    4(x+2)≡4x+84(x + 2) \equiv 4x + 8

    True for all x, so it is an identity.

  4. Solve part (b)

    3x−1=8⇒x=33x - 1 = 8 \Rightarrow x = 3

    True for one value only.

  5. Classify part (b)

    x=3 onlyx = 3 \text{ only}

    So it is an equation.

  6. Look at part (c)

    x⋅x=x2x \cdot x = x^2

    x times x is x2x^2 by definition.

  7. Classify part (c)

    x⋅x≡x2x \cdot x \equiv x^2

    True for all x, so it is an identity.

  8. Simplify part (d)

    5x−2x=3x5x - 2x = 3x

    Collecting like terms gives 3x.

  9. Classify part (d)

    5x−2x≡3x5x - 2x \equiv 3x

    True for all x, so it is an identity.

  10. Look at part (e)

    2x=x+x2x = x + x

    x + x is 2x.

  11. Classify part (e)

    2x≡x+x2x \equiv x + x

    True for all x, so it is an identity.

  12. Test the identities with a value

    x=2:  all holdx=2:\; \text{all hold}

    Checking x=2x = 2 confirms the identities.

  13. Count how many are identities

    4 identities4 \text{ identities}

    Parts a, c, d and e are identities.

  14. Note the single equation

    (b) is the equation\text{(b) is the equation}

    Only part b is a plain equation.

  15. State all the answers

    (a) id, (b) eq, (c) id, (d) id, (e) id(a)\,\text{id},\,(b)\,\text{eq},\,(c)\,\text{id},\,(d)\,\text{id},\,(e)\,\text{id}

    Four identities and one equation.

Answer
(a) identity, (b) equation, (c) identity, (d) identity, (e) identity
Question 4
6 markschallenging
Classify each of the following: (a) x2−1x^2 - 1 (b) x2−1=0x^2 - 1 = 0 (c) C=59(F−32)C = \frac{5}{9}(F - 32) (d) (x−1)(x+1)≡x2−1(x - 1)(x + 1) \equiv x^2 - 1 (e) x2+1=0x^2 + 1 = 0.
Show worked solution

Worked solution

  1. Recall what an expression is

    expression: terms, no =\text{expression: terms, no } =

    An expression is just a collection of terms with no equals sign.

  2. Recall what an equation is

    equation: = , some values\text{equation: } = \text{ , some values}

    An equation has an equals sign and is true only for particular value(s) of the letter.

  3. Recall what a formula is

    formula: = , linked letters\text{formula: } = \text{ , linked letters}

    A formula has an equals sign and links two or more different letters (variables).

  4. Recall what an identity is

    a≡b for all valuesa \equiv b \text{ for all values}

    An identity is true for every value of the letter and is written with the identity sign.

  5. Examine part (a)

    x2−1x^2 - 1

    There is no equals sign, so it is simply a collection of terms.

  6. Classify part (a)

    ⇒expression\Rightarrow \text{expression}

    No equals sign, so it is an expression.

  7. Examine part (b)

    x2−1=0x^2 - 1 = 0

    There is an equals sign and it is true only for a particular value of the letter.

  8. Classify part (b)

    ⇒equation\Rightarrow \text{equation}

    True only when x=1x = 1 or x=−1x = -1.

  9. Examine part (c)

    C=59(F−32)C = \tfrac{5}{9}(F - 32)

    There is an equals sign linking two or more different letters.

  10. Classify part (c)

    ⇒formula\Rightarrow \text{formula}

    An equals sign links the different letters C and F.

  11. Examine part (d)

    (x−1)(x+1)≡x2−1(x - 1)(x + 1) \equiv x^2 - 1

    Both sides are equal for every value of the letter, so the identity sign is used.

  12. Classify part (d)

    ⇒identity\Rightarrow \text{identity}

    Expanding the brackets gives x2−1x^2 - 1 for all x.

  13. Examine part (e)

    x2+1=0x^2 + 1 = 0

    There is an equals sign and it is true only for a particular value of the letter.

  14. Classify part (e)

    ⇒equation\Rightarrow \text{equation}

    An equation that is not satisfied by any real value of x.

  15. State the classifications

    (a) expression, (b) equation, (c) formula, (d) identity, (e) equation\text{(a) expression, (b) equation, (c) formula, (d) identity, (e) equation}

    Each statement has now been correctly named.

Answer
(a) expression, (b) equation, (c) formula, (d) identity, (e) equation
Question 5
6 markschallenging
The area of a trapezium is A=12(a+b)hA = \frac{1}{2}(a + b)h. (a) Name the subject. (b) List the variables. (c) Is 12\frac{1}{2} a variable or a constant? (d) How many terms are inside the bracket? (e) Classify A=12(a+b)hA = \frac{1}{2}(a + b)h. (f) Classify 12(a+b)h\frac{1}{2}(a + b)h on its own.
Show worked solution

Worked solution

  1. Look at the whole statement

    A=12(a+b)hA = \tfrac{1}{2}(a + b)h

    It has an equals sign and several letters.

  2. Identify the subject

    A=…A = \ldots

    The subject is the single letter on the left: A.

  3. List the letters

    A,\; a,\; b,\; h

    The letters used are A, a, b and h.

  4. State the variables

    A,\; a,\; b,\; h

    The variables are A, a, b and h.

  5. Look at the one half

    12\tfrac{1}{2}

    One half is a fixed number.

  6. Classify the one half

    12 is a constant\tfrac{1}{2} \text{ is a constant}

    It does not change, so it is a constant, not a variable.

  7. Look inside the bracket

    a+ba + b

    Count the terms in the bracket.

  8. Count the terms in the bracket

    a,  b→2a,\; b \rightarrow 2

    There are 2 terms inside the bracket.

  9. Check for an equals sign

    A=12(a+b)hA = \tfrac{1}{2}(a + b)h

    There is an equals sign.

  10. Check that letters are linked

    A←a,b,hA \leftarrow a, b, h

    It links several variables.

  11. Classify the whole statement

    formula\text{formula}

    A rule linking several variables is a formula.

  12. Look at the right-hand side alone

    12(a+b)h  (no =)\tfrac{1}{2}(a + b)h \;(\text{no } =)

    On its own there is no equals sign.

  13. Classify the right-hand side

    expression\text{expression}

    So it is an expression on its own.

  14. Bring the parts together

    A;  A,a,b,h;  constant;  2;  formula;  expressionA;\; A,a,b,h;\; \text{constant};\; 2;\; \text{formula};\; \text{expression}

    Collect all answers.

  15. State all the answers

    A;  A,a,b,h;  constant;  2;  formula;  expressionA;\; A,a,b,h;\; \text{constant};\; 2;\; \text{formula};\; \text{expression}

    Subject A; variables A, a, b, h; one half is a constant; 2 terms in the bracket; a formula; and the right side alone is an expression.

Answer
subject A; variables A, a, b, h; one half is a constant; 2 terms in the bracket; it is a formula; the right side alone is an expression

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