MISSION · Number sense
Absolute Value
How far from zero, regardless of direction
LEVEL 1 · EXPLORE
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LEVEL 2 · LEARN
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The idea in one sentence
Absolute value measures distance from zero, and distance has no direction — so the answer is never negative.
\[|7| = 7 \qquad |-7| = 7 \qquad |0| = 0\]Distance, not sign-stripping
It’s tempting to learn \(|x|\) as “delete the minus sign”. That works on plain numbers and then fails on everything else.
The distance reading keeps working. \(|-7| = 7\) because \(-7\) sits seven units from zero. And \(|x - 3|\) isn’t “\(x - 3\) with the sign removed” — it’s the distance between \(x\) and 3. That one reinterpretation makes most absolute-value problems solvable by thinking rather than by procedure.
A temperature of \(-5°\) and one of \(5°\) are equally far from freezing. An error of \(-0.3\) mm and one of \(+0.3\) mm are equally wrong. Absolute value is what you use when size matters and direction doesn’t.
Two solutions, always
| $$ | x | = 5\(has two answers, 5 and\)-5$$, because two points sit five units from zero. |
That’s the pattern for any equation of this shape: strip the bars in two ways, once positive and once negative.
\[|x - 3| = 4 \quad \Rightarrow \quad x - 3 = 4 \;\text{ or }\; x - 3 = -4\]giving \(x = 7\) or \(x = -1\). Read as distance it’s immediate — which numbers are four units from 3? One on each side.
And \(|x| = -2\) has no solutions at all. No distance is negative, so nothing satisfies it. Recognising that saves you from solving an equation with no answer.
The graph is a V
\(y = |x|\) falls at a slope of \(-1\) until it hits zero, then rises at a slope of \(+1\). The sharp corner at the origin is characteristic: absolute value is continuous everywhere but has no single slope at the corner, which is the standard example of a function that’s continuous but not differentiable.
Inside \(|x - 3|\), the corner moves to \(x = 3\) — the point the distance is measured from.
Inequalities split two ways
| $$ | x | < 3\(means "within 3 units of zero", a segment:\)-3 < x < 3$$. |
| $$ | x | > 3\(means "more than 3 units away", two separate rays:\)x < -3\(or\)x > 3$$. |
Less-than gives you one piece in the middle; greater-than gives two pieces on the outside. The distance picture tells you which without memorising a rule.
Common mix-ups
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Giving one solution to $$ x = k$$ when there are two. - Making the inside negative and stopping. \(|x| = -4\) has no solution; the answer is “none”, not \(-4\).
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**Treating $$ a + b \(as\) a + b \(.**\) 3 + (-5) = 2$$, not 8. - Splitting a less-than inequality into two rays. That’s the greater-than case.
Check yourself
| Solve $$ | x + 2 | = 6\(, and describe the solutions of\) | x | \le 4$$. |
\(x + 2 = 6\) or \(x + 2 = -6\), so \(x = 4\) or \(x = -8\) — the two points six units from \(-2\). And \(|x| \le 4\) is everything from \(-4\) to 4 inclusive.
LEVEL 3 · REFLECT
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