Absolute Value Equations and Inequalities: Practice

The right side $4 - x$ must be nonnegative, so only $x \le 4$ is eligible. Split and check.

$x - 6 = 4 - x \Rightarrow x = 5, \qquad x - 6 = -(4 - x) \Rightarrow -6 = -4 \ \text{(impossible)}$

The first candidate $x = 5$ fails $x \le 4$, and the second case is a contradiction, so there is no solution.

The anchors $3$ and $-2$ are $5$ apart, and the target $9$ exceeds that gap, so use the three breakpoint intervals.

$x < -2:\ (3 - x) + (-x - 2) = 9 \Rightarrow x = -4; \qquad x > 3:\ (x - 3) + (x + 2) = 9 \Rightarrow x = 5$

Both roots land inside their intervals, and the middle interval gives $5 = 9$ (no solution), so the answer is $\{-4, 5\}$.

The anchors $-1$ and $4$ are $5$ apart, exactly the target on the right.

$g = \lvert 4 - (-1) \rvert = 5 = \text{target}$

Between the anchors the two distances always sum to $5$, so every point of the segment $[-1, 4]$ is a solution.

Equal coefficients of $x$ make the "insides equal" case impossible, leaving only the "opposite" case.

$3x - 2 = -(3x + 4) \quad\Longrightarrow\quad 6x = -2 \quad\Longrightarrow\quad x = -\tfrac{1}{3}$

That is the one point equidistant from the two shifted lines, so there is exactly one solution.

Split into cases and check each against $2x - 5 \ge 0$.

$x - 7 = 2x - 5 \Rightarrow x = -2, \qquad x - 7 = -(2x - 5) \Rightarrow 3x = 12 \Rightarrow x = 4$

At $x = -2$ the right side is $-9 < 0$, extraneous. At $x = 4$ it is $3 \ge 0$ and $\lvert 4 - 7 \rvert = 3$ checks, so the solution set is $\{4\}$.

Split into cases and check each against $3x + 1 \ge 0$.

$x + 5 = 3x + 1 \Rightarrow x = 2, \qquad x + 5 = -(3x + 1) \Rightarrow 4x = -6 \Rightarrow x = -\tfrac{3}{2}$

At $x = 2$ the right side is $7 \ge 0$ and $\lvert 7 \rvert = 7$ checks. At $x = -\frac{3}{2}$ the right side is $-\frac{7}{2} < 0$, extraneous, so the solution set is $\{2\}$.

Because both sides are nonnegative, $\lvert X \rvert = \lvert Y \rvert$ is equivalent to $X^2 = Y^2$.

$(x - 4)^2 = (3x)^2 = 9x^2$

The square of $3x$ is $9x^2$, so the squared equation is $(x - 4)^2 = 9x^2$.

Set the insides equal, then opposite.

$x - 4 = 3x \Rightarrow x = -2, \qquad x - 4 = -3x \Rightarrow 4x = 4 \Rightarrow x = 1$

Both are genuine solutions, so the solution set is $\{-2, 1\}$.

An absolute value is always at least zero, so the inequality holds with nothing excluded.

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

The solution set is all of $\mathbb{R}$.

An absolute value equals its own inside exactly when that inside is nonnegative.

$\lvert x + 2 \rvert = x + 2 \quad\Longleftrightarrow\quad x + 2 \ge 0$

That gives $x \ge -2$, the whole ray from $-2$ rightward.

Since $5 - x = -(x - 5)$, the equation reads $\lvert x - 5 \rvert = -(x - 5)$, which holds when the inside is nonpositive.

$\lvert x - 5 \rvert = -(x - 5) \quad\Longleftrightarrow\quad x - 5 \le 0$

That gives $x \le 5$, the whole ray from $5$ leftward.

The two identical terms combine, so the equation is a single absolute value.

$2\lvert x - 1 \rvert = 6 \quad\Longrightarrow\quad \lvert x - 1 \rvert = 3$

The inside is $3$ or $-3$, giving $x = 4$ or $x = -2$, so there are exactly two solutions.