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Free holiday intensive · Week 1

Intensive algebra: choose the method, keep the restrictions

Start with three short checks, then work through algebraic fractions, indices, factorisation, equations and harder reasoning. Show your working before opening each solution. The 19 original questions carry 74 marks; allow about 90 minutes plus review.

Before you practise

1. Factor first, then decide what can cancel

A common factor can cancel from a numerator and denominator. Write the values excluded by the original denominator before cancelling.

Worked example: simplify (x²−4)(x²+3x+2).

(x−2)(x+2)[(x+1)(x+2)] = (x−2)(x+1). The original denominator is zero at x=−1 and x=−2, so both stay excluded.

Common mistake: cancelling the x+2 factor and then allowing x=−2.

2. Use a common base for index equations

Convert both sides to the same positive base. Equal powers of base 2 have equal exponents.

Worked example: solve 4x+1=8x.

22x+2=23x, so 2x+2=3x and x=2. Check: 4³=8²=64.

Common mistake: adding indices across a sum. Index laws for multiplication do not apply to addition.

3. Check equation answers in the original expression

Clearing a denominator is reversible only where that denominator is nonzero. Check every candidate against the original restrictions.

Worked example: solve 2(x−1)=x+1.

x≠1. Multiply by x−1: 2=(x+1)(x−1)=x²−1. Thus x=±√3; both are allowed because neither equals 1.

Warm-ups

Check the factorisation, index and denominator ideas you will need later.

Question 1

2 marks

Factorise x²−5x−14 completely. State its zeros.

See the worked solution
  1. Two numbers with product −14 and sum −5 are −7 and 2, so x²−5x−14=(x−7)(x+2).
  2. Zeros: x=7 or x=−2.

Question 2

2 marks

Evaluate 1634 without a calculator. Explain which operation the denominator 4 represents.

See the worked solution
  1. The denominator means fourth root: ⁴√16=2; cube that result to obtain 2³=8.

Question 3

2 marks

Simplify (x²−9)(x−3) and give its original restriction. Can the simplified expression give a value for the original fraction at x=3?

See the worked solution
  1. (x−3)(x+3)(x−3)=x+3 for x≠3.
  2. The original fraction is undefined at 3 even though x+3 has a value there, so substitution at 3 is invalid for the original expression.

Core applications

Choose a method, keep original restrictions and check the result.

Question 4

4 marks

Factorise 6x²−x−2. Hence solve 6x²−x−2=0 and check one root by substitution.

See the worked solution
  1. Split the middle term using product −12 and sum −1: 6x²+3x−4x−2=3x(2x+1)−2(2x+1)=(3x−2)(2x+1).
  2. Hence x=23 or x=−12.
  3. Check x=23: 6(49)−23−2=83−23−2=0.

Question 5

4 marks

Write 2(x−1)−3(x+2) as one simplified fraction. State every excluded value.

See the worked solution
  1. Use (x−1)(x+2) as the common denominator.
  2. Numerator 2(x+2)−3(x−1)=2x+4−3x+3=7−x.
  3. Thus (7−x)((x−1)(x+2)), with x≠1,−2.
  4. The minus sign applies to the whole second numerator.

Question 6

4 marks

Simplify [(x²−9)(x²+5x+6)] ÷ [(x−3)(x+2)]. Give all restrictions from the original expression, including the condition that the divisor is nonzero.

See the worked solution
  1. Factor: first fraction is (x−3)(x+3)[(x+2)(x+3)].
  2. Multiply by reciprocal (x+2)(x−3) to obtain 1.
  3. Original denominators exclude x=−2,−3; the divisor is zero at x=3, so x≠3 as well.
  4. Answer 1 for x≠−3,−2,3.

Question 7

4 marks

Solve 4x−1=82−x. Explain why comparing exponents is valid, then check your answer.

See the worked solution
  1. Rewrite as 22x−2=26−3x.
  2. Base 2 is positive and not 1, so equal powers have equal exponents: 2x−2=6−3x, giving x=85.
  3. Check exponents: both equal frac{6}{5}, so both original sides equal 265.

Question 8

4 marks

The quadratic x²+(k−3)x−3k has a root x=−2. Find k, factorise the quadratic and state its other root.

See the worked solution
  1. Substituting −2 gives 4−2(k−3)−3k=10−5k=0, so k=2.
  2. The quadratic becomes x²−x−6=(x−3)(x+2); its other root is x=3.
  3. Checking the original symbolic expression, (x−3)(x+k) expands to x²+(k−3)x−3k.

Challenging reasoning

Connect methods across longer problems and explain why your answer is valid.

Question 9

5 marks

Solve 2(x−1)+3(x+2)=1. State restrictions before clearing denominators and check that each candidate is allowed.

See the worked solution
  1. Require x≠1,−2.
  2. Multiply by (x−1)(x+2): 2(x+2)+3(x−1)=(x−1)(x+2).
  3. This gives 5x+1=x²+x−2, or x²−4x−3=0.
  4. By the quadratic formula, x=(4±√28)2=2±√7.
  5. Neither is 1 or −2, so both are solutions.
  6. Each satisfies the cleared equation, and denominator multiplication was reversible on the permitted domain.

Question 10

5 marks

A rectangle has width (x−1) cm and length (x+4) cm, where x>1. Its area is 36 cm². Find x, both dimensions and the perimeter. Explain why the other algebraic root is rejected.

See the worked solution
  1. (x−1)(x+4)=36 gives x²+3x−4=36, hence x²+3x−40=(x+8)(x−5)=0.
  2. Thus x=5 or −8.
  3. Since x>1, reject −8: it would give negative side lengths.
  4. Width 4 cm, length 9 cm, perimeter 2(4+9)=26 cm.
  5. Check area 4×9=36 cm².

Question 11

5 marks

Find the least positive integer n for which both 12n is a perfect square and 18n is a perfect cube. Prove your choice is least.

See the worked solution
  1. Write n=2a3b for the least candidate; introducing other prime factors can only increase n and their exponents would also have to satisfy both power conditions.
  2. Since 12=2²3, square condition requires a+2 even and b+1 even: a even, b odd.
  3. Since 18=2·3², cube condition requires a+1 and b+2 divisible by 3: a leaves remainder 2 when divided by 3, and b leaves remainder 1 when divided by 3.
  4. Least nonnegative choices are a=2, b=1, so n=2²·3=12.
  5. Check 12n=144=12²; 18n=216=6³.
  6. Any larger admissible exponents or extra primes make n larger.

Question 12

5 marks

Let G=(x²−4)(x²−x−6)+(x−2)(x+2). Simplify G to one fraction, state every original restriction, then find all x for which G=0.

See the worked solution
  1. Factor the first numerator and denominator: (x−2)(x+2)[(x−3)(x+2)].
  2. Original denominators require x≠−2,3.
  3. For allowed x, the first term is (x−2)(x−3).
  4. Use denominator (x−3)(x+2) for the sum: numerator (x−2)(x+2)+(x−2)(x−3)=(x−2)(2x−1).
  5. Thus G=(x−2)(2x−1)[(x−3)(x+2)], with x≠−2,3.
  6. A fraction is zero when its numerator is zero and denominator nonzero, so x=2 or x=12; both are permitted.

Question 13

5 marks

The equations x²−5x+6=0 and x²−(k+1)x+2k=0 have exactly one common real root. Find k and all roots of the second equation. Explain why the other root of the first equation cannot be common.

See the worked solution
  1. Factor the first equation as (x−2)(x−3)=0, giving candidates 2 and 3.
  2. Substituting x=2 into the second left side gives 4−2(k+1)+2k=2, so 2 is never a root.
  3. Substituting x=3 gives 9−3(k+1)+2k=6−k; hence k=6.
  4. The second equation is then x²−7x+12=(x−3)(x−4)=0, with roots 3 and 4.
  5. Its only shared root with {2,3} is 3.

Question 14

5 marks

Solve the simultaneous equations 2x+3y=17 and x2−y3=1. Show why your chosen elimination or substitution step is valid, then find x²−y² without expanding a new equation.

See the worked solution
  1. Multiply the second equation by 6 to clear its constant denominators: 3x−2y=6.
  2. This is reversible because 6 is nonzero.
  3. Multiply the first equation by 2 and the transformed second by 3: 4x+6y=34 and 9x−6y=18.
  4. Add to obtain 13x=52, so x=4.
  5. Substitute into 2x+3y=17: 8+3y=17, hence y=3.
  6. Check the second original equation: 42−33=1.
  7. Use difference of squares: x²−y²=(x−y)(x+y)=1·7=7.

Misconception analysis

Find the first invalid step and repair the reasoning.

Question 15

3 marks

A student claims 3(x−1)−2(x+1)=1(x²−1), saying they “subtracted the numerators and multiplied the denominators”. Identify the invalid step, find the correct single fraction and state its domain.

See the worked solution
  1. Numerators can be subtracted only after the fractions share a denominator.
  2. Use (x−1)(x+1)=x²−1 and form 3(x+1)−2(x−1)=3x+3−2x+2=x+5.
  3. The correct result is (x+5)(x²−1) for x≠−1,1.
  4. The student's 1(x²−1) loses the cross-multiplied terms.

Question 16

3 marks

A student solves (x²−9)(x−3)=6 by multiplying through, gets (x−3)²=0 and reports x=3. Is there a solution? Explain.

See the worked solution
  1. The original denominator requires x≠3.
  2. For allowed x, the left side simplifies to x+3; x+3=6 gives x=3, which is excluded.
  3. Thus there is no solution.
  4. Multiplying by x−3 without keeping its nonzero condition can introduce this invalid candidate.

Question 17

3 marks

A student writes 3x+3x=32x and therefore cannot solve 3x+3x=54. Repair the step and solve the equation.

See the worked solution
  1. Addition of equal terms doubles the term: 3x+3x=2·3x, not a product of powers.
  2. Thus 2·3x=54, so 3x=27=3³ and x=3.
  3. Check: 27+27=54.

Stretch

Work through parameter and exponent conditions within G3 Maths scope.

Question 18

4 marks

Suppose a>0 and 2x+1=a·4x. Find a so that x=−2 is a solution. Show that for this a the solution is unique, using only index laws.

See the worked solution
  1. At x=−2, 2−1=a·4−2, so 12=a16 and a=8.
  2. Then 2x+1=8·22x=22x+3.
  3. Equal powers of 2 give x+1=2x+3, hence x=−2.
  4. Since base 2 is injective, there is exactly one solution; no logarithms are needed.

Question 19

5 marks

For a real parameter k, solve (x+1)(x−2)−(x−1)(x+2)=k(x²−4) for x. Determine the two values of k for which the equation has no solution, and explain why.

See the worked solution
  1. Original restrictions: x≠2,−2.
  2. Multiply by x²−4=(x−2)(x+2), which is nonzero on the permitted domain.
  3. The left numerator is (x+1)(x+2)−(x−1)(x−2)=(x²+3x+2)−(x²−3x+2)=6x.
  4. Thus 6x=k, giving candidate x=k6.
  5. It is a solution only if k6≠2,−2, or k≠12,−12.
  6. For k=12 or −12, the sole candidate is an excluded denominator zero, so there is no solution.

Prefer paper?

The same 19 questions and worked solutions are available as optional PDFs.

Worksheet PDFWorked solutions PDF

What to practise next

Algebraic fractionsReview restrictions and cancellation IndicesRevisit powers and common bases Free diagnosticCheck your next learning priority