Absolute Value Equations and Inequalities: Practice

The absolute value branches on the sign of its input, returning the input when it is nonnegative and the negated input when it is negative.

$\lvert x \rvert = x \ (x \ge 0), \qquad \lvert x \rvert = -x \ (x < 0)$

The negated form $-x$ is positive precisely because $x$ is negative there, so the output is a nonnegative distance either way.

Here $x = -6$ is negative, so the second branch applies and returns the input negated.

$\lvert -6 \rvert = -(-6) = 6$

Negating a negative number gives a positive one, matching the distance of $6$ from zero.

An absolute value is never negative, so it can never equal $-5$.

$\lvert x + 2 \rvert \ge 0 > -5$

No value of $x$ works, so the solution set is empty. Splitting into two cases here would manufacture false answers.

A distance cannot be less than a negative number, since distances start at $0$.

$\lvert 3x - 1 \rvert \ge 0 > -2$

Nothing satisfies the inequality, so the solution set is $\varnothing$. The band template $-c < X < c$ needs $c > 0$ and does not apply.

An absolute value is always at least $0$, which already exceeds $-1$.

$\lvert 5x + 4 \rvert \ge 0 > -1 \quad\text{for every } x$

Every real number satisfies the inequality, so the solution set is all of $\mathbb{R}$.

The distance from $x$ to zero is positive everywhere except at the one point where it is zero.

$\lvert x \rvert > 0 \quad\Longleftrightarrow\quad x \neq 0$

Only $x = 0$ gives $\lvert x \rvert = 0$, so it is the single value excluded.

An absolute value is a distance, so it is always zero or greater, with no exceptions.

$\lvert x - 7 \rvert \ge 0 \quad\text{for every } x$

The inequality holds for every real number, so its solution set is all of $\mathbb{R}$.

The left side $\lvert X \rvert$ is never negative, so the equation secretly requires $Y \ge 0$, a condition the bare split $X = Y$ or $X = -Y$ forgets.

$\lvert X \rvert = Y \quad\Longleftrightarrow\quad Y \ge 0 \ \text{ and } \ (X = Y \ \text{ or } \ X = -Y)$

Dropping $Y \ge 0$ is a non-reversible step that can add roots, so every candidate must be confirmed.

The two cases are $x = x - 8$, which is impossible, and $x = -(x - 8)$.

$x = -(x - 8) \quad\Longrightarrow\quad 2x = 8 \quad\Longrightarrow\quad x = 4$

The surviving solution is $x = 4$, the midpoint of $0$ and $8$, as the distance reading predicts.

A breakpoint is where an inside expression changes sign, that is where it equals zero.

$x - 2 = 0 \Rightarrow x = 2, \qquad x + 3 = 0 \Rightarrow x = -3$

So the breakpoints are $x = 2$ and $x = -3$, and they cut the line into three intervals.

An absolute value cannot equal a negative number, so test the sign of the right side.

$3(1) - 5 = -2 < 0$

The right side is negative while the left side $\lvert 1 + 4 \rvert = 5$ cannot be, so the equation fails and $x = 1$ is extraneous.

Substitute $x = 0$ into each absolute value and add.

$\lvert 0 - 3 \rvert + \lvert 0 + 1 \rvert = 3 + 1 = 4$

The value $4$ is the distance from $0$ to $3$ plus the distance from $0$ to $-1$.