Hard GCSE Algebraic fractions Questions

Challenging, exam-style GCSE Algebraic fractions questions with worked solutions. Stretch yourself on the hardest simplifying, difference of two squares, factorising quadratics, multiplying fractions problems.

simplifyingdifference of two squaresfactorising quadraticsmultiplying fractionscancellingdividing fractions
GCSE Higher34 questionsStep-by-step solutions
Question 1
6 markschallenging
Write 2x−1−1x+1+2x2−1\frac{2}{x-1} - \frac{1}{x+1} + \frac{2}{x^2-1} as a single fraction.
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Worked solution

  1. State the task

    2x−1−1x+1+2x2−1\frac{2}{x-1} - \frac{1}{x+1} + \frac{2}{x^2-1}

    Combine all three fractions.

  2. Factorise the third denominator

    x2−1=(x−1)(x+1)x^2 - 1 = (x-1)(x+1)

    A difference of two squares.

  3. Identify the common denominator

    (x−1)(x+1)(x-1)(x+1)

    It covers all three denominators.

  4. Convert the first fraction

    2x−1=2(x+1)(x−1)(x+1)\frac{2}{x-1} = \frac{2(x+1)}{(x-1)(x+1)}

    Multiply top and bottom by (x+11).

  5. Convert the second fraction

    1x+1=x−1(x−1)(x+1)\frac{1}{x+1} = \frac{x-1}{(x-1)(x+1)}

    Multiply top and bottom by (x-11).

  6. Keep the third fraction

    2(x−1)(x+1)\frac{2}{(x-1)(x+1)}

    It is already over the common denominator.

  7. Combine the numerators

    2(x+1)−(x−1)+22(x+1) - (x-1) + 2

    Subtract the middle term, keeping the bracket.

  8. Expand the first term

    2(x+1)=2x+22(x+1) = 2x + 2

    Multiply 22 through the bracket.

  9. Expand the second term

    −(x−1)=−x+1-(x-1) = -x + 1

    The minus sign flips both signs.

  10. Include the constant

    +2+ 2

    Add the third numerator.

  11. Collect like terms

    2x+2−x+1+2=x+52x + 2 - x + 1 + 2 = x + 5

    Combine to get x + 5.

  12. Write the single fraction

    x+5(x−1)(x+1)\frac{x+5}{(x-1)(x+1)}

    Place the numerator over the denominator.

  13. Check for cancelling

    x+5 shares no factor\text{x+5 shares no factor}

    Nothing cancels with the denominator.

  14. State the restrictions

    x≠1,  x≠−1x \neq 1,\; x \neq -1

    The denominators must be non-zero.

  15. State the answer

    x+5(x−1)(x+1)\frac{x+5}{(x-1)(x+1)}

    The single fraction is (x+55) over (x-11)(x+11).

Answer
x+5(x−1)(x+1)\frac{x+5}{(x-1)(x+1)}
Question 2
5 markschallenging
Which is the correct single fraction for 1x+1x+2\frac{1}{x} + \frac{1}{x+2}?
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Worked solution

  1. State the task

    1x+1x+2\frac{1}{x} + \frac{1}{x+2}

    Combine the two fractions.

  2. Note the denominators differ

    x≠x+2x \neq x+2

    A common denominator is needed.

  3. Choose the common denominator

    x(x+2)x(x+2)

    Multiply the two denominators together.

  4. Convert the first fraction

    1x=x+2x(x+2)\frac{1}{x} = \frac{x+2}{x(x+2)}

    Multiply top and bottom by (x+22).

  5. Convert the second fraction

    1x+2=xx(x+2)\frac{1}{x+2} = \frac{x}{x(x+2)}

    Multiply top and bottom by x.

  6. Add the numerators

    (x+2)+xx(x+2)\frac{(x+2) + x}{x(x+2)}

    Combine over the common denominator.

  7. Simplify the numerator

    2x+22x + 2

    x+2 plus x is 2x+2.

  8. Write the fraction

    2x+2x(x+2)\frac{2x+2}{x(x+2)}

    Place the numerator over the denominator.

  9. Factorise the numerator

    2x+2=2(x+1)2x + 2 = 2(x+1)

    Take out the common factor 22.

  10. Write in factorised form

    2(x+1)x(x+2)\frac{2(x+1)}{x(x+2)}

    Nothing cancels with the denominator.

  11. Check for cancelling

    no common factor\text{no common factor}

    (x+1) matches neither x nor (x+2).

  12. State the restrictions

    x≠0,  x≠−2x \neq 0,\; x \neq -2

    The denominators must be non-zero.

  13. Verify at a value

    12+14=34,  2(2)2⋅4=34\frac{1}{2} + \frac{1}{4} = \dfrac{3}{4},\; \frac{2(2)}{2 \cdot 4} = \dfrac{3}{4}

    At x=2x=2 both forms give three quarters.

  14. Reflect on a common error

    never add denominators\text{never add denominators}

    Adding tops and bottoms would give 2/(2x+22x+2), which is wrong.

  15. State the answer

    2(x+1)x(x+2)\frac{2(x+1)}{x(x+2)}

    The correct single fraction is 2(x+11) over x(x+22).

Answer
2(x+1)x(x+2)\frac{2(x+1)}{x(x+2)}
Question 3
6 markschallenging
Show that 1x−1−1x+1=2x2−1\frac{1}{x-1} - \frac{1}{x+1} = \frac{2}{x^2-1}. Which working is correct?
Show worked solution

Worked solution

  1. State the task

    1x−1−1x+1\frac{1}{x-1} - \frac{1}{x+1}

    Work on the left-hand side.

  2. Note the denominators differ

    x−1≠x+1x-1 \neq x+1

    A common denominator is needed.

  3. Choose the common denominator

    (x−1)(x+1)(x-1)(x+1)

    Multiply the two denominators together.

  4. Convert the first fraction

    1x−1=x+1(x−1)(x+1)\frac{1}{x-1} = \frac{x+1}{(x-1)(x+1)}

    Multiply top and bottom by (x+11).

  5. Convert the second fraction

    1x+1=x−1(x−1)(x+1)\frac{1}{x+1} = \frac{x-1}{(x-1)(x+1)}

    Multiply top and bottom by (x-11).

  6. Subtract the numerators

    (x+1)−(x−1)(x−1)(x+1)\frac{(x+1) - (x-1)}{(x-1)(x+1)}

    Combine over the common denominator.

  7. Expand carefully

    x+1−x+1x + 1 - x + 1

    The minus sign flips the second bracket.

  8. Simplify the numerator

    22

    The x terms cancel, leaving 2.

  9. Write the fraction

    2(x−1)(x+1)\frac{2}{(x-1)(x+1)}

    Place 22 over the factorised denominator.

  10. Recognise the denominator

    (x−1)(x+1)=x2−1(x-1)(x+1) = x^2 - 1

    A difference of two squares.

  11. Write the compact form

    2x2−1\frac{2}{x^2-1}

    This matches the right-hand side.

  12. State the restrictions

    x≠1,  x≠−1x \neq 1,\; x \neq -1

    The denominators must be non-zero.

  13. Reflect on a common error

    keep the bracket\text{keep the bracket}

    Forgetting the bracket would give x+1−x−1=0x+1-x-1 = 0, which is wrong.

  14. Verify at a value

    11−13=23,  24−1=23\frac{1}{1} - \frac{1}{3} = \dfrac{2}{3},\; \frac{2}{4-1} = \dfrac{2}{3}

    At x=2x=2 both sides give two thirds.

  15. Conclude

    1x−1−1x+1=2x2−1\frac{1}{x-1} - \frac{1}{x+1} = \frac{2}{x^2-1}

    The identity holds for the allowed values.

Answer
2x2−1\frac{2}{x^2-1}
Question 4
6 markschallenging
Simplify the compound fraction x−1x1+1x\dfrac{x - \frac{1}{x}}{1 + \frac{1}{x}}.
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Worked solution

  1. Recognise a complex fraction

    x−1x1+1x\frac{x - \frac{1}{x}}{1 + \frac{1}{x}}

    There are fractions inside the top and bottom.

  2. Focus on the top

    x−1xx - \frac{1}{x}

    Combine the top into a single fraction.

  3. Write x with denominator x

    x=x2xx = \frac{x^2}{x}

    This gives a common denominator on top.

  4. Simplify the top

    x2−1x\frac{x^2 - 1}{x}

    Subtract over the common denominator x.

  5. Focus on the bottom

    1+1x1 + \frac{1}{x}

    Combine the bottom into a single fraction.

  6. Write 1 with denominator x

    1=xx1 = \frac{x}{x}

    This gives a common denominator on the bottom.

  7. Simplify the bottom

    x+1x\frac{x + 1}{x}

    Add over the common denominator x.

  8. Rewrite the whole expression

    x2−1x÷x+1x\frac{x^2-1}{x} \div \frac{x+1}{x}

    The complex fraction is a division.

  9. Flip and multiply

    x2−1x×xx+1\frac{x^2-1}{x} \times \frac{x}{x+1}

    Multiply by the reciprocal of the bottom.

  10. Cancel the x

    x2−1x+1\frac{x^2-1}{x+1}

    The x on top and bottom cancels.

  11. Factorise x2−1x^2-1

    x2−1=(x−1)(x+1)x^2 - 1 = (x-1)(x+1)

    A difference of two squares.

  12. Cancel (x+11)

    (x−1)(x+1)x+1=x−1\frac{(x-1)(x+1)}{x+1} = x-1

    The factor (x+11) cancels.

  13. Write the simplified form

    x−1x - 1

    The compound fraction simplifies to x - 1.

  14. State the restrictions

    x≠0,  x≠−1x \neq 0,\; x \neq -1

    The denominators must be non-zero.

  15. Verify at a value

    2−121+12=3/23/2=1=2−1\frac{2 - \dfrac{1}{2}}{1 + \dfrac{1}{2}} = \frac{3/2}{3/2} = 1 = 2-1

    At x=2x=2 the value is 1, matching x - 1.

Answer
x−1x-1
Question 5
5 markschallenging
Write 3x2+x+2x+1\frac{3}{x^2+x} + \frac{2}{x+1} as a single fraction.
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Worked solution

  1. State the task

    3x2+x+2x+1\frac{3}{x^2+x} + \frac{2}{x+1}

    Combine the two fractions.

  2. Factorise the first denominator

    x2+x=x(x+1)x^2 + x = x(x+1)

    Take out the common factor x.

  3. Rewrite the first fraction

    3x(x+1)\frac{3}{x(x+1)}

    Use the factorised denominator.

  4. Identify the common denominator

    x(x+1)x(x+1)

    It already contains (x+11).

  5. Convert the second fraction

    2x+1=2xx(x+1)\frac{2}{x+1} = \frac{2x}{x(x+1)}

    Multiply top and bottom by x.

  6. Keep the first fraction

    3x(x+1)\frac{3}{x(x+1)}

    It is already over the common denominator.

  7. Add the numerators

    3+2xx(x+1)\frac{3 + 2x}{x(x+1)}

    Combine over the common denominator.

  8. Write the numerator neatly

    2x+3x(x+1)\frac{2x+3}{x(x+1)}

    Order the numerator as 2x+32x + 3.

  9. Check for cancelling

    2x+3 shares no factor\text{2x+3 shares no factor}

    Nothing cancels with x or (x+1).

  10. State the restrictions

    x≠0,  x≠−1x \neq 0,\; x \neq -1

    The denominators must be non-zero.

  11. Recall the expanded denominator

    x(x+1)=x2+xx(x+1) = x^2 + x

    This links back to the original.

  12. Verify at a value

    32+22=52,  51⋅2=52\frac{3}{2} + \frac{2}{2} = \dfrac{5}{2},\; \frac{5}{1 \cdot 2} = \dfrac{5}{2}

    At x=1x=1 both forms give five halves.

  13. Note to keep denominator factorised

    x(x+1)x(x+1)

    Leaving it factorised makes cancelling checks easier.

  14. Reflect on a common error

    never add denominators\text{never add denominators}

    Only numerators are added once denominators match.

  15. State the answer

    2x+3x(x+1)\frac{2x+3}{x(x+1)}

    The single fraction is (2x+32x+3) over x(x+11).

Answer
2x+3x(x+1)\frac{2x+3}{x(x+1)}

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