GCSE Algebraic fractions Practice Questions

Free GCSE Algebraic fractions practice questions with full step-by-step worked solutions. Covers simplifying, cancelling, index laws, factorising. Practise exam-style problems and check your method.

simplifyingcancellingindex lawsfactorisingmultiplying fractionsdividing fractions
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
Simplify 2x6\frac{2x}{6}.
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Worked solution

  1. Look for a common factor

    2x6\frac{2x}{6}

    The top 2x2x and the bottom 66 share a factor of 22.

  2. Divide top and bottom by 22

    2x6=x3\frac{2x}{6} = \frac{x}{3}

    Cancelling the 2 leaves x over 3.

  3. State the answer

    x3\frac{x}{3}

    The fraction is now in its simplest form.

Answer
x3\frac{x}{3}
Question 2
2 markseasy
Simplify 12x33x\frac{12x^3}{3x}.
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Worked solution

  1. Simplify the numbers

    123=4\frac{12}{3} = 4

    Twelve over 33 is 44.

  2. Simplify the powers of x

    x3x=x2\frac{x^3}{x} = x^2

    Subtract the indices: 33 minus 11 is 22.

  3. Combine the parts

    4x24x^2

    The simplified expression is 4x4x squared.

Answer
4x24x^2
Question 3
2 marksintermediate
Show that x2+6x+9x+3=x+3\frac{x^2+6x+9}{x+3} = x+3 for x≠−3x \neq -3. Which explanation is correct?
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Worked solution

  1. Factorise the numerator

    x2+6x+9=(x+3)2x^2 + 6x + 9 = (x+3)^2

    The numerator is a perfect square.

  2. Rewrite the fraction

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

    Write the numerator as (x+33)(x+33).

  3. Cancel one bracket

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

    One factor of (x+33) cancels.

  4. State the restriction

    x≠−3x \neq -3

    Cancelling needs the denominator to be non-zero.

  5. Check with a value

    164=4=1+3\frac{16}{4} = 4 = 1+3

    At x=1x=1 the original gives 44 and x+33 gives 44.

  6. Conclude

    x2+6x+9x+3=x+3\frac{x^2+6x+9}{x+3} = x+3

    The identity holds for all x not equal to -3.

Answer
x+3x+3
Question 4
3 markshard
Solve 2x+13=x+42\frac{2x+1}{3} = \frac{x+4}{2}.
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Worked solution

  1. Note the single fraction each side

    2x+13=x+42\frac{2x+1}{3} = \frac{x+4}{2}

    There is one fraction on each side, so cross-multiply.

  2. Cross-multiply

    2(2x+1)=3(x+4)2(2x+1) = 3(x+4)

    Multiply each numerator by the other denominator.

  3. Expand the left side

    4x+24x + 2

    Multiply out 2(2x+12x+1).

  4. Expand the right side

    3x+123x + 12

    Multiply out 3(x+44).

  5. Form the linear equation

    4x+2=3x+124x + 2 = 3x + 12

    Set the two expansions equal.

  6. Collect the x terms

    4x−3x=12−24x - 3x = 12 - 2

    Subtract 3x3x and 22 from both sides.

  7. Simplify

    x=10x = 10

    One x equals 10.

  8. Check the left side

    2(10)+13=213=7\frac{2(10)+1}{3} = \frac{21}{3} = 7

    Substituting x=10x=10 gives 77.

  9. Check the right side

    10+42=142=7\frac{10+4}{2} = \frac{14}{2} = 7

    Both sides equal 77, so the solution works.

  10. State the answer

    x=10x = 10

    The solution is x=10x = 10.

Answer
x=10x = 10
Question 5
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)}

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