Solve absolute-value equations and inequalities in one variable and graph the solution set, recognizing the two-case structure and no-solution cases (MA.912.AR.4.1, MA.912.AR.4.2).

What this topic is asking

MA.912.AR.4 asks you to solve absolute-value equations and inequalities and graph the solutions. Because absolute value measures distance from zero, every problem splits into two cases. The B.E.S.T. Algebra 1 EOC tests the two-case setup, the and versus or structure of inequalities, and the no-solution cases.

Why two cases

The absolute value $|a|$ is the distance of $a$ from zero, always nonnegative. So $|x| = 5$ has two answers, $x = 5$ and $x = -5$, both distance $5$ from zero. Every absolute-value equation inherits this two-case structure.

Solving equations

Isolate the bars, then split:

Solving inequalities: "and" versus "or"

The inequality direction sets the structure:

• $|x| < k$ (less than): a between solution, $-k < x < k$. Think "and," and on a number line it is a single shaded segment.
• $|x| > k$ (greater than): an outside solution, $x < -k$ or $x > k$. Think "or," and on a number line it is two rays pointing away.

A memory hook: "lessor" is wrong, less-than is and (the bounded middle); greater-than is or (the two ends).

No-solution cases

Because absolute value is never negative:

• $|expression| =$ a negative number: no solution.
• $|expression| <$ a negative number: no solution (nothing is that close to zero).
• $|expression| >$ a negative number: all real numbers (everything is at least that far).

How the B.E.S.T. EOC examines this topic

• Equation editor. Solve an absolute-value equation and type both values.
• GRID. Graph the compound solution on a number line (a segment or two rays).
• Multiple choice. Count solutions, or recognize a no-solution case.

A clarifying idea: the bars ask "how far from zero?", so solving means finding every input at the stated distance, which is why you get two values for an equation, a band for "less than," and two tails for "greater than."

Why "less than" is "and" but "greater than" is "or"

The and/or split comes straight from the distance meaning. "$|x| < 5$" asks for all numbers within distance 5 of zero, which is the connected stretch from $-5$ to $5$, and a number must be above $-5$ and below $5$ to lie in that stretch, hence "and." "$|x| > 5$" asks for all numbers farther than distance 5 from zero, which lies beyond $5$ on the right or beyond $-5$ on the left, two separate pieces, hence "or." Picturing the number line, a middle band for less-than, two tails for greater-than, lets you reconstruct the structure every time instead of memorizing a rule you might reverse. Always isolate the absolute value before splitting, or the cases come out wrong.

Graphing the solution set

The graph of an absolute-value solution depends on the type. An equation like $|x - 4| = 6$ gives two isolated points on the number line ($x = 10$ and $x = -2$). A less-than inequality like $|x - 4| < 6$ gives a single shaded segment between the two points ($-2 < x < 10$), drawn with open circles at the ends (or closed for $\le$). A greater-than inequality like $|x - 4| > 6$ gives two shaded rays pointing outward from $-2$ and $10$ in opposite directions. On a GRID item, the EOC may ask you to place the circles and shading, so matching the picture to the type, two dots, one band, or two tails, is the exact skill being scored. The center of the band or the gap between the tails always sits at the value that makes the inside zero, here $x = 4$, which is a quick way to position the graph.

Try this

Q1. Solve $|x - 4| = 6$. [2 points]

• Cue. $x - 4 = 6$ or $x - 4 = -6$, so $x = 10$ or $x = -2$.

Q2. Write $|x| < 3$ as a compound inequality. [1 point]

• Cue. $-3 < x < 3$ (an "and" between solution).