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IMAT Mathematics · Equations and Inequalities

Linear, Quadratic, Fractional and Absolute Value Inequalities

An inequality is solved like an equation, except that multiplying by a negative flips it; quadratic and fractional inequalities are read from a sign chart.

Why this matters

Inequalities are the most asked topic in this chapter since 2023: quadratic inequalities in 2023 and 2026, a pair of linear inequalities in 2023, and a fraction containing an absolute value in 2024. Older papers (2012, 2013) combined a quadratic and a linear inequality.

Concept 1 of 4: Linear inequalities and systems of inequalities

Adding or subtracting the same number keeps an inequality true, and so does multiplying by a positive number. Multiplying by a negative number reverses the order: 2 < 3, but −2>−3-2 > -3. That single rule is the whole difference from solving an equation.

Definition

  • Solve as for a linear equation, but reverse the sign whenever you multiply or divide both sides by a negative number.
  • A system (both conditions must hold) is solved one inequality at a time, then the solution sets are intersected on a number line.
  • Conditions joined by or give the union of the solution sets.
  • An empty intersection means no solution (∅\varnothing); an inequality that reduces to a true statement such as 0<40 < 4 holds for every real xx (R\mathbb{R}).

Multiplying by a negative number

a<b  and  c<0  ⇒  ac>bca < b \;\text{and}\; c < 0 \;\Rightarrow\; ac > bc

Worked example

(a) Solve x+42−2x−13>1\dfrac{x + 4}{2} - \dfrac{2x - 1}{3} > 1. (b) Solve the system 3−2x≤73 - 2x \le 7 and 5x−4<2x+115x - 4 < 2x + 11.
Practice this conceptself-check · 4 quick reps

Dividing by a negative reverses the sign

From −3x<6-3x < 6 the answer is x>−2x > -2, not x<−2x < -2. The unreversed answer is always among the options.

Concept 2 of 4: Quadratic inequalities with a sign chart

The graph of ax2+bx+cax^2 + bx + c with a>0a > 0 is a parabola opening upwards. It is below the axis only between its roots and above it outside them. So find the roots, and the shape tells you the answer. With no roots the parabola never crosses the axis, so its sign never changes.

Definition

  • Move everything to one side so the other side is 0. If a<0a < 0, multiply by −1-1 and reverse the sign.
  • With a>0a > 0 and roots x1<x2x_1 < x_2: the expression is negative between the roots and positive outside them.
  • Δ<0\Delta < 0: always positive. So >0> 0 holds for all of R\mathbb{R}, and <0< 0 has no solution.
  • Δ=0\Delta = 0: a(x−r)2a(x - r)^2 is positive except at x=rx = r, where it is 0.
  • Include the roots for ≤\le and ≥\ge; exclude them for << and >>.

Sign of a quadratic with a > 0 and roots x₁ < x₂

a(x−x1)(x−x2)<0  for  x1<x<x2,>0  for  x<x1  or  x>x2a(x - x_1)(x - x_2) < 0 \;\text{for}\; x_1 < x < x_2, \qquad > 0 \;\text{for}\; x < x_1 \;\text{or}\; x > x_2

Worked example

Solve (a) x2−x−12>0x^2 - x - 12 > 0, (b) −x2+4x−5<0-x^2 + 4x - 5 < 0.
Practice this conceptself-check · 4 quick reps

x² < 16 is not x < 4

Taking the square root of both sides loses the negative half: x=−10x = -10 satisfies x<4x < 4 but not x2<16x^2 < 16. Write (x−4)(x+4)<0(x - 4)(x + 4) < 0 and read the answer −4<x<4-4 < x < 4 from the sign chart.

Concept 3 of 4: Fractional inequalities: signs of numerator and denominator

You cannot multiply both sides by a denominator such as x−1x - 1 without knowing its sign, because a negative one would reverse the inequality. Instead, keep the fraction and ask when it is positive: when the top and bottom have the same sign.

Definition

  • Move everything to one side and combine into a single fraction N(x)D(x)\frac{N(x)}{D(x)} compared with 0.
  • Find where N=0N = 0 and where D=0D = 0; mark them on a number line and record the sign of each factor in each region.
  • The fraction is positive where NN and DD have the same sign, negative where they differ.
  • Zeros of DD are always excluded. Zeros of NN are included only for ≤\le or ≥\ge.
  • A factor that is always positive (such as x2+1x^2 + 1) never changes the sign: ignore it.

Worked example

Solve 2x+1x−1<1\dfrac{2x + 1}{x - 1} < 1.
Practice this conceptself-check · 4 quick reps

Do not multiply by a denominator of unknown sign

Multiplying 3x−1<1\frac{3}{x - 1} < 1 by x−1x - 1 assumes x−1>0x - 1 > 0 and loses every solution with x<1x < 1. Bring everything to one side and use a sign chart.

Concept 4 of 4: Inequalities with absolute value

∣A∣<k|A| < k says A is less than kk away from 0, so A lies in one piece between −k-k and kk. ∣A∣>k|A| > k says A is more than kk away, so A lies in one of two outer pieces. (Absolute value is defined in Numbers, Powers and Logarithms.)

Definition

For k>0k > 0:

  • ∣A∣<k|A| < k means −k<A<k-k < A < k: one interval.
  • ∣A∣>k|A| > k means A<−kA < -k or A>kA > k: two intervals.
  • For k<0k < 0: ∣A∣<k|A| < k has no solution and ∣A∣>k|A| > k holds wherever AA is defined.
  • ∣A∣≥0|A| \ge 0 always, and ∣A∣=0|A| = 0 only where A=0A = 0. So in a fraction, a numerator like ∣A∣+1|A| + 1 is always positive and only the denominator decides the sign.

Absolute value inequalities (k > 0)

∣A∣<k  ⇔  −k<A<k∣A∣>k  ⇔  A<−k  or  A>k|A| < k \;\Leftrightarrow\; -k < A < k \qquad |A| > k \;\Leftrightarrow\; A < -k \;\text{or}\; A > k

Worked example

Solve (a) ∣2x−1∣≤5|2x - 1| \le 5, (b) ∣x+2∣>3|x + 2| > 3.
Practice this conceptself-check · 4 quick reps

Greater than gives two pieces, not one

∣x+2∣>3|x + 2| > 3 is x<−5x < -5 or x>1x > 1. Writing it as −5>x>1-5 > x > 1 describes no number at all. Only the less-than form gives a single interval.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Linear inequalities and systems of inequalities

    Multiplying by a negative number

    a<b  and  c<0  ⇒  ac>bca < b \;\text{and}\; c < 0 \;\Rightarrow\; ac > bc
  • Quadratic inequalities with a sign chart

    Sign of a quadratic with a > 0 and roots x₁ < x₂

    a(x−x1)(x−x2)<0  for  x1<x<x2,>0  for  x<x1  or  x>x2a(x - x_1)(x - x_2) < 0 \;\text{for}\; x_1 < x < x_2, \qquad > 0 \;\text{for}\; x < x_1 \;\text{or}\; x > x_2
  • Inequalities with absolute value

    Absolute value inequalities (k > 0)

    ∣A∣<k  ⇔  −k<A<k∣A∣>k  ⇔  A<−k  or  A>k|A| < k \;\Leftrightarrow\; -k < A < k \qquad |A| > k \;\Leftrightarrow\; A < -k \;\text{or}\; A > k

Watch out for (4)

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