Solve absolute-value equations and inequalities in one variable, splitting into two cases and representing solution sets symbolically and on a number line (A.EI.3).

What this topic is asking

A.EI.3 asks you to solve absolute-value equations and inequalities in one variable. On the Virginia Algebra I SOL these are Equations and Inequalities items: solve and type the two solutions, write the solution set, or recognize a no-solution case. They appear as fill-in-the-blank, multiple choice, and number-line hot-spot items.

Why two cases

The absolute value $|a|$ is the distance of $a$ from zero, and two different numbers can be the same distance from zero (for instance $5$ and $-5$ are both distance $5$). So when $|{\text{something}}| = k$, the inside can be either $k$ or $-k$. This is the source of the two cases, and it is why you must solve both.

Solving absolute-value equations

Isolate the absolute value first, then split into two equations.

If, after isolating, the absolute value equals a negative number, stop: there is no solution, because distance cannot be negative.

Absolute-value inequalities: the and/or distinction

The structure of the solution depends on the direction of the inequality.

• Less than ($|x| < k$, with $k > 0$). The inside is within distance $k$ of zero, a single band: $-k < x < k$. This is an "and" compound (both conditions hold at once).
• Greater than ($|x| > k$, with $k > 0$). The inside is farther than distance $k$, two outward rays: $x < -k$ or $x > k$. This is an "or" compound.

So $|x - 3| < 5$ becomes $-5 < x - 3 < 5$, giving $-2 < x < 8$. And $|x + 1| \ge 4$ becomes $x + 1 \le -4$ or $x + 1 \ge 4$, giving $x \le -5$ or $x \ge 3$.

Graphing the solution

A less-than ("and") solution graphs as a segment between two endpoints. A greater-than ("or") solution graphs as two rays pointing away from each other. Use closed circles for $\le$ or $\ge$ and open circles for $<$ or $>$, the same endpoint rule as ordinary inequalities.

Why these patterns are really about distance

Every absolute-value statement is a sentence about distance, and translating it that way makes the and/or rule obvious rather than memorized. "$|x - 3| < 5$" says "the distance from $x$ to $3$ is less than $5$," which describes all points within $5$ units of $3$, a single interval centered at $3$. "$|x - 3| > 5$" says "the distance from $x$ to $3$ is more than $5$," which describes points far from $3$ in either direction, two intervals. Read the bars as "distance," the center as the number you subtract, and the value as the radius, and you can write the solution directly from the meaning. This distance view also explains the no-solution case: a distance can never be less than zero, and it can never equal a negative number.

How the SOL examines this topic

• Fill-in-the-blank. Solve an absolute-value equation and type both solutions.
• Multiple choice. Pick the solution set of an inequality, or identify a no-solution case.
• Hot spot / number line. Graph the band or the two rays with the correct circles.

Try this

Q1. Solve $|x - 5| = 8$. [2 points]

• Cue. $x - 5 = 8$ or $x - 5 = -8$, so $x = 13$ or $x = -3$.

Q2. Write the solution of $|x| \le 4$ as a compound inequality. [1 point]

• Cue. Within distance 4: $-4 \le x \le 4$ (an "and").