Solve distance conditions with absolute value

Try

Two locations on a straight path are exactly 4 km from the marker at coordinate 1 km. Find both coordinates. Draw the marker and move 4 km left and right; the coordinates are −3 and 5.

Predict

The condition is |x − 1| = 4, where the bars denote absolute value. Predict whether replacing 4 with 0 gives two distinct locations. Replacing it with −4 gives none: a distance cannot be negative.

Experiment

Draw a number line and shade all points whose distance from 1 is at most 4. The entire segment from −3 through 5 qualifies, including its endpoints. Now shade points farther than 4 away: two open rays remain outside the segment.

Condition	Geometric meaning
∣x − 1∣ = 4	Two boundary points
∣x − 1∣ ≤ 4	Closed segment
∣x − 1∣ > 4	Two outside rays

Understand

For a real quantity u, |u| is u when u is nonnegative and −u when u is negative. It is always nonnegative. Thus |u| = d with positive d means u = d or u = −d. If d = 0, both branches give the same solution; if d is negative, there is no solution.

For |3x + 2| = 5, solve 3x + 2 = 5 and 3x + 2 = −5. The candidates are x = 1 and x = −7/3. Substitution gives |5| and |−5|, both 5. The negative candidate is not automatically invalid: it is the absolute-value output, not x, that must be nonnegative.

For positive d, |u| ≤ d means −d ≤ u ≤ d. In contrast, |u| ≥ d means u ≤ −d or u ≥ d. These are intersection and union statements respectively. Always inspect a nonpositive d separately; mechanically copying a two-branch rule can obscure an empty set or an all-real set.

Apply

Use distance reasoning before solving the practice below. Exact fractions are accepted for rational roots. In the tolerance report, explain endpoint inclusion and use millimetres consistently; a passing numerical length without the condition does not document the specification.

Continue

Rational equations introduce excluded positions instead of distance branches. The next lesson records those exclusions before any operation can hide them.