🧲Absolute Value & Compound Inequalities

The two bars in |x| do one job: they measure how far x is from 0, ignoring which side of 0 it sits on. |7| = 7 and |-7| = 7, because both numbers are seven steps from 0.

That one idea explains everything else in this topic, including two facts that catch people out:

• ◆An absolute value is never negative. There is no such thing as a distance of -3.
• ◆Two different numbers share almost every distance. Only one number is 0 away from 0; every other distance is reached twice, once on each side.

The bars can hold more than a bare x. |x - 3| measures the distance from x to 3, not to 0. Try it: at x = 10 it gives |10 - 3| = 7, and at x = -4 it gives |-4 - 3| = 7. Both of those points are seven steps from 3 on the number line.

So read every absolute value as a distance and name what it is a distance from: |x - 3| is "how far x is from 3", and |x + 6|, which is really |x - (-6)|, is "how far x is from -6".

Because two numbers share each distance, an equation with bars around it usually splits into two ordinary equations.

|x - 3| = 7 says the value of x - 3 is 7 away from 0. The only numbers 7 away from 0 are 7 and -7, so

• ◆Case 1: x - 3 = 7, giving x = 10
• ◆Case 2: x - 3 = -7, giving x = -4

Both are genuine solutions, and the check confirms it: |10 - 3| = 7 and |-4 - 3| = 7.

Notice what became negative. It is the right-hand side, not the x. Case 2 is x - 3 = -7, never -x - 3 = 7.

Two special cases fall out of "a distance is never negative":

• ◆|x - 3| = -7 has no solution at all. Nothing is -7 away from anything.
• ◆|x - 3| = 0 has exactly one solution, x = 3, because the two cases collapse together.

If the bars are not alone, get them alone first. For 2|x - 3| + 1 = 9, subtract 1 and divide by 2 to reach |x - 3| = 4, and only then split. Splitting early gives the wrong two numbers.

A compound inequality joins two conditions with "and" or with "or", and the joining word decides what the answer looks like on a number line.

"And" needs both conditions at once. x > -2 and x <= 3 leaves only the numbers caught between them, so it is written as one sandwich with x in the middle:

-2 < x <= 3

On the number line that is a single segment with a circle at each end: open at -2 (left out) and filled at 3 (kept in).

"Or" needs only one of the two. x < 2 or x >= 4 gives two separate pieces heading in opposite directions — two rays, with the numbers from 2 to 4 left out. An "or" answer cannot be squeezed into a sandwich, so it is written as two statements: x < 2 or x >= 4.

Solving a sandwich is one calculation done three times over. For -3 < 2x + 1 <= 7, subtract 1 from all three parts to get -4 < 2x <= 6, then divide all three by 2: -2 < x <= 3. An "or" is solved one branch at a time, and the same rule as ever applies to each branch — multiplying or dividing by a negative reverses that branch's sign.

Once the bars mean distance, absolute value inequalities stop needing to be memorised.

|x - 3| < 5 asks which points are *less than* 5 away from 3. Being close to 3 traps you: you cannot wander further than 5 in either direction, so the answer is one stretch around 3.

-5 < x - 3 < 5, which gives -2 < x < 8

|x - 3| > 5 asks which points are *more than* 5 away from 3. Being far from 3 can be done in two ways — far to the left, or far to the right — so the answer is two rays.

x - 3 < -5 or x - 3 > 5, which gives x < -2 or x > 8

That is the whole rule, and the reason behind it: less than = and (one band), greater than = or (two rays). The endpoints follow the sign: < and > leave them out (open circles), while <= and >= keep them in (filled circles).

A coefficient changes nothing about which rule to use. |2x - 3| <= 7 becomes -7 <= 2x - 3 <= 7, then -4 <= 2x <= 10, then -2 <= x <= 5.

Outside a classroom, this topic is mostly about tolerance — how far from a target a real measurement is allowed to stray.

"A bolt must be within 0.5 mm of 20 mm" is exactly the sentence |x - 20| <= 0.5, and it unpacks into the acceptable range 19.5 <= x <= 20.5. The target sits inside the bars and the tolerance sits on the right; swapping them writes a different rule entirely.

The move runs backwards just as well. If a workshop accepts lengths from 19.5 mm to 20.5 mm, find the middle of the range, (19.5 + 20.5) / 2 = 20, and the distance from the middle to each end, 0.5. The rule behind the range is |x - 20| <= 0.5.

The same shape appears wherever measurement does: a temperature that must stay within 2 degrees of a set point, a scale accurate to within 5 grams, an error bound in science that says a measured value is within some margin of the true one.

Later on, the two-case split turns up again in piecewise functions — a function that follows one rule for x below a boundary and another above it — and the graph of y = |x - 3| is exactly that: two straight pieces meeting in a V at x = 3.