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Inequalities with algebraic fractions

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

F2 starts with inequalities you cannot solve by simply multiplying out. When the unknown is in a denominator, you have to think about its sign. Show your algebra: the scheme of work says that when a question asks for algebra, purely graphical answers with little working are heavily penalised.

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Why you cannot just cross-multiply

Multiplying an inequality by a negative number reverses it. A denominator such as (x + 2) is positive for some x and negative for others, so multiplying by it is unsafe.

A student who multiplies (4x - 1)/(x + 2) < 1 by (x + 2) gets 4x - 1 < x + 2, so x < 1, and misses the critical value x = -2 completely.

Method 1: one fraction and critical values

1. Move everything to one side so the other side is 0.

2. Combine into a single fraction and factorise the top and the bottom.

3. The critical values are where the top is 0 or the bottom is 0.

4. Test the sign of the fraction in each interval (a sign table or a quick sketch).

Values that make the denominator 0 are never included, even with ≤ or ≥.

Method 2: multiply by the square of the denominator

(x + 2)2 is never negative, so multiplying by it keeps the inequality the same way round:

(4x - 1)(x + 2) < (x + 2)2, so (x + 2)(4x - 1 - x - 2) < 0, so (x + 2)(3x - 3) < 0.

Do not cancel (x + 2): factorise it out. Then sketch the quadratic or cubic and read off where it is below or above the axis.

Writing the answer

Give every interval, for example x < -5 or x > 2, or in set notation {x : x < -5} ∪ {x : x > 2}.

Check one value from each interval in the original inequality. Exam tip: if the question says "use algebra", the critical values must come from algebra; a sketch alone only confirms which intervals to choose.

Worked example

Solve (4x - 1)/(x + 2) < 1.

  1. (4x - 1)/(x + 2) - 1 < 0, so (4x - 1 - x - 2)/(x + 2) < 0, so (3x - 3)/(x + 2) < 0.
  2. Critical values: 3x - 3 = 0 gives x = 1; x + 2 = 0 gives x = -2.
  3. Test x = 0: (-3)/2 < 0, true. Test x = 2: 3/4, false. Test x = -3: (-12)/(-1) = 12, false.

Answer: -2 < x < 1

Key idea: Never multiply by an expression whose sign you do not know. Bring everything to one side, find critical values from the top and bottom, then test each interval.

Check yourself

  1. Solve (3x + 1)/(x - 2) > 2.
    • -5 < x < 2
    • x > -5
    • x < 2 or x > 5
    • x < -5 or x > 2
    Show answer

    x < -5 or x > 2. (3x + 1)/(x - 2) - 2 = (3x + 1 - 2x + 4)/(x - 2) = (x + 5)/(x - 2) > 0. Critical values -5 and 2; the fraction is positive outside them.

  2. The solution of 2/(x - 3) < 1 is x < a or x > b. Find b.
    Show answer

    5. 2/(x - 3) - 1 = (2 - x + 3)/(x - 3) = (5 - x)/(x - 3) < 0. Critical values 3 and 5. The fraction is negative when x < 3 or x > 5.

  3. Solve x > 6/(x + 1).
    • x > 2
    • x < -3 or -1 < x < 2
    • -3 < x < 2
    • -3 < x < -1 or x > 2
    Show answer

    -3 < x < -1 or x > 2. x - 6/(x + 1) = (x2 + x - 6)/(x + 1) = (x + 3)(x - 2)/(x + 1) > 0. Critical values -3, -1 and 2. Testing: positive for -3 < x < -1 and for x > 2.

  4. Solve x/(x + 1) ≤ 2/(x - 2).
    • -1 < x ≤ 2 - √6 or 2 < x ≤ 2 + √6
    • -1 ≤ x ≤ 2 - √6 or 2 ≤ x ≤ 2 + √6
    • 2 - √6 ≤ x ≤ 2 + √6
    • x < -1 or 2 - √6 ≤ x < 2
    Show answer

    -1 < x ≤ 2 - √6 or 2 < x ≤ 2 + √6. One fraction: (x(x - 2) - 2(x + 1))/((x + 1)(x - 2)) = (x2 - 4x - 2)/((x + 1)(x - 2)) ≤ 0. Critical values -1, 2 - √6, 2, 2 + √6. The sign is negative on (-1, 2 - √6] and (2, 2 + √6]. x = -1 and x = 2 are excluded because the denominators are zero there.

  5. How many integers satisfy (x + 5)/(x - 2) < 0?
    Show answer

    6. Critical values -5 and 2; the fraction is negative for -5 < x < 2. The integers are -4, -3, -2, -1, 0 and 1.

  6. A student solves 5/(x - 1) > 2 by writing 5 > 2(x - 1), so x < 3.5. What is wrong?
    • Multiplying by x - 1 assumes it is positive, so x < 1 is wrongly included
    • Nothing: the answer x < 3.5 is correct
    • The inequality sign should have been reversed throughout
    • The student should have squared each side
    Show answer

    Multiplying by x - 1 assumes it is positive, so x < 1 is wrongly included. If x < 1 then x - 1 < 0 and the sign must flip. Doing it properly: (5 - 2x + 2)/(x - 1) = (7 - 2x)/(x - 1) > 0, so 1 < x < 3.5.

  7. To clear the fraction in (x + 4)/(x - 3) ≥ 2 safely, a student multiplies both sides by (x - 3)2. Which inequality results?
    • (x + 4) ≥ 2(x - 3)
    • (x + 4)(x - 3) ≤ 2(x - 3)2
    • (x + 4)(x - 3) ≥ 2(x - 3)2
    • (x + 4)2 ≥ 4(x - 3)2
    Show answer

    (x + 4)(x - 3) ≥ 2(x - 3)2. (x - 3)2 > 0 for x ≠ 3, so the sign is unchanged. Then (x - 3)(x + 4 - 2x + 6) ≥ 0, so (x - 3)(10 - x) ≥ 0, giving 3 < x ≤ 10 (x = 3 excluded).

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Last updated 8 November 2026. Spotted a mistake? Tell us.