Álgebra 1Inequalities

Absolute Value Inequalities

Turn |x − a| < b into an "and" and |x − a| > b into an "or."

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Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.

You know that $|x - a|$ is the distance between $x$ and $a$, from absolute value equations, and how “and” and “or” work from compound inequalities. Put the two together and absolute value inequalities read like sentences about distance.

Por qué funciona

$|x - 3| < 5$ says “$x$ is less than $5$ units from $3$.” Those numbers are all within $5$ of $3$: between $-2$ and $8$. So it becomes the “and” statement $-5 < x - 3 < 5$. But $|x - 3| > 5$ says “$x$ is more than $5$ units from $3$,” which happens on either side: $x - 3 < -5$ or $x - 3 > 5$. Less than means one segment; greater than means two rays.
$|x - 3| < 5$: every number within $5$ units of $3$. Drag the test point past an end and the distance becomes too big.
$|x - 3| \ge 5$: every number at least $5$ units from $3$, on either side.

Idea clave

Isolate the absolute value first. Then, for $k > 0$: $|u| < k$ means $-k < u < k$ (one segment), and $|u| > k$ means $u < -k$ or $u > k$ (two rays). The same holds with $\le$ and $\ge$.

Worked examples

Ejemplo resuelto: Less than: one segment

Solve $|2x - 1| \le 7$.
  1. Distance at most $7$ from $0$: $-7 \le 2x - 1 \le 7$.
  2. Add $1$ to all three parts: $-6 \le 2x \le 8$.
  3. Divide by $2$: $-3 \le x \le 4$.

Respuesta: $-3 \le x \le 4$

Ejemplo resuelto: Greater than: two rays

Solve $|x + 4| - 3 > 2$.
  1. Isolate first: add $3$ to get $|x + 4| > 5$.
  2. More than $5$ from $0$: $x + 4 < -5$ or $x + 4 > 5$.
  3. Solve each: $x < -9$ or $x > 1$.

Respuesta: $x < -9$ or $x > 1$.

Ejemplo resuelto: When the number is negative

Solve $|x - 2| < -1$, then $|x - 2| \ge -1$.
  1. An absolute value is never negative, so it can never be less than $-1$: no solution.
  2. For the same reason it is always at least $-1$: every real number works.

Respuesta: No solution; all real numbers.

Error común

Writing $|x| > 3$ as $-3 > x > 3$. That chain says $-3 > 3$, which is false, so it describes no numbers at all. The two pieces point away from each other, so they must be joined with “or”: $x < -3$ or $x > 3$. Check with a number like $5$: $|5| > 3$ is true, so $5$ must be in your answer.

Práctica

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Fuentes y lecturas

  • OpenStax, Intermediate Algebra 2e, section 2.7 Solve Absolute Value Inequalities
  • Paul's Online Math Notes (Lamar University), Algebra: Absolute Value Inequalities