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Absolute Value Equations & Inequalities

Absolute value measures distance from zero, so an equation like 2x1=7|2x - 1| = 7 asks 'what makes the inside 7 units from zero?' The answer is two cases: the inside equals 77 or 7-7.

Inequalities split the same way but become compound statements. 'Less than' gives an 'and' (a band between two values); 'greater than' gives an 'or' (two outer pieces).

Solving the equation with two cases

For A=b|A| = b with b0b \geq 0, write A=bA = b or A=bA = -b and solve each. The two solutions are the two inputs that land bb units from zero.

If bb is negative there is no solution — absolute value can never be negative. Always check the right side before splitting.

Inequalities: and versus or

'Less than' (A<b|A| < b) means the inside is within bb of zero: b<A<b-b < A < b, a single band — an 'and.'

'Greater than' (A>b|A| > b) means the inside is farther than bb from zero: A<bA < -b or A>bA > b, two pieces — an 'or.' The memory aid is 'less-thand' and 'great-or.'

Worked examples

Example 1: an absolute value equation

Solve 2x1=7|2x - 1| = 7.

Split into two cases2x1=7;or;2x1=72x - 1 = 7 ;\text{or}; 2x - 1 = -7
Solve each2x=8;or;2x=62x = 8 ;\text{or}; 2x = -6
Divide by 2x=4;or;x=3x = 4 ;\text{or}; x = -3

Answer: x=4x = 4 or x=3x = -3

Example 2: a 'less than' inequality

Solve x<5|x| < 5.

Less than becomes a band5<x<5-5 < x < 5

Answer: 5<x<5-5 < x < 5

Example 3: a 'greater than' inequality

Solve x3>4|x - 3| > 4.

Greater than becomes two outer piecesx3<4;or;x3>4x - 3 < -4 ;\text{or}; x - 3 > 4
Add 3 to each sidex<1;or;x>7x < -1 ;\text{or}; x > 7

Answer: x<1x < -1 or x>7x > 7

Try one yourself

Common questions

When does an absolute value equation have no solution?

When it equals a negative number, like x=3|x| = -3. Absolute value is a distance and can never be negative.

How do I remember 'and' versus 'or'?

'Less-thand' — less than gives an AND (a band between two values). 'Great-or' — greater than gives an OR (two outer pieces).

Do I always split into two cases?

For equations and inequalities, yes — one case for the positive value inside, one for the negative. Just isolate the absolute value first.

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