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Inequalities: Linear, Compound & Absolute Value

Inequalities solve almost exactly like equations, with one rule you must respect: flip the sign when you multiply or divide by a negative. Then meet compound and absolute-value inequalities, and write answers in interval notation.

Solving a linear inequality

A linear inequality uses an inequality symbol (<, >, ≤, or ≥) instead of an equals sign, so its solution set is usually a whole range of numbers rather than a single value. You solve it with the very same balance steps as an equation, with one crucial exception: multiplying or dividing both sides by a negative number reverses the inequality symbol. Adding, subtracting, and multiplying or dividing by a positive all leave the symbol alone.

Solve   -2x + 1 < 9

Subtract 1 from both sides (symbol unchanged):
  -2x < 8
Divide both sides by -2  ->  FLIP the symbol:
  x > -4

Solution set: all x greater than -4.
Interval notation: (-4, infinity).
Quick test x = 0: -2(0) + 1 = 1 < 9  TRUE, and 0 > -4. Consistent.
Dividing by -2 flips < into >.

Compound inequalities: and / or

A compound inequality joins two conditions. An and statement like -1 ≤ 2x + 3 < 7 must hold on both counts at once; solve it by doing the same step to all three parts. An or statement, like x < -2 or x > 5, is satisfied when either piece is true, giving a solution in two separate chunks.

Solve   -1 <= 2x + 3 < 7   (an 'and')

Subtract 3 from all three parts:
  -4 <= 2x < 4
Divide all three parts by 2 (positive, no flip):
  -2 <= x < 2

Solution set: all x with -2 <= x < 2.
Interval notation: [-2, 2).
Apply each step to all three parts of an 'and' inequality.

Absolute-value inequalities

Since absolute value measures distance from zero, an absolute-value inequality becomes a compound inequality. Less-than means “close to zero,” an and: |x| < 5 turns into -5 < x < 5. Greater-than means “far from zero,” an or: |x| > 5 turns into x < -5 or x > 5. A handy mnemonic is “less-thand and great-or.”

Solve   |2x - 1| <= 7   (less-than -> 'and')

Rewrite as a double inequality:
  -7 <= 2x - 1 <= 7
Add 1 to all parts:
  -6 <= 2x <= 8
Divide all parts by 2:
  -3 <= x <= 4

Interval notation: [-3, 4].
A 'less-than' absolute value unfolds into a single 'and' chain.