Absolute Value Equations and Inequalities: Additional Practice (Set 1)

The anchors $-6$ and $2$ are $8$ apart, exactly the target on the right.

$g = \lvert 2 - (-6) \rvert = 8 = \text{target}$

Between the anchors the two distances always sum to $8$, so every point of the segment $[-6, 2]$ is a solution.

The anchors $4$ and $-2$ are $6$ apart and the target $10$ exceeds that gap, so split the excess $(10 - 6)/2 = 2$ beyond each end.

$x = -2 - 2 = -4, \qquad x = 4 + 2 = 6$

Checking, $\lvert -4 - 4 \rvert + \lvert -4 + 2 \rvert = 8 + 2 = 10$ and $\lvert 6 - 4 \rvert + \lvert 6 + 2 \rvert = 2 + 8 = 10$, so the solutions are $-4$ and $6$.

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

$5 - 2x = x - 4 \Rightarrow x = 3, \qquad 5 - 2x = -(x - 4) \Rightarrow x = 1$

Both candidates are below $4$, so the right side is negative at each and both are extraneous. The solution set is empty.

Set the insides equal, then opposite.

$x - 1 = 4 - x \Rightarrow 2x = 5 \Rightarrow x = \tfrac{5}{2}, \qquad x - 1 = -(4 - x) \Rightarrow -1 = -4 \ \text{(impossible)}$

The second case is a contradiction, so the only solution is $x = \frac{5}{2}$, the midpoint of $1$ and $4$.

The anchors $3$ and $-5$ are $8$ apart, exactly the target on the right.

$g = \lvert 3 - (-5) \rvert = 8 = \text{target}$

Between the anchors the two distances always sum to $8$, so every point of the segment $[-5, 3]$ is a solution.

The anchors $3$ and $-5$ are $8$ apart, and the sum can never fall below that gap.

$g = \lvert 3 - (-5) \rvert = 8 > 4$

Since the target $4$ is below the minimum $8$, no $x$ works and the solution set is $\varnothing$.

An absolute value is never negative, so it can never be less than $-1$.

$\lvert 7x + 3 \rvert \ge 0 > -1$

No value of $x$ works, so the solution set is $\varnothing$.

The insides share the coefficient $4x$, so the "equal" case is impossible and only the "opposite" case survives.

$4x - 1 = -(4x + 9) \quad\Longrightarrow\quad 8x = -8 \quad\Longrightarrow\quad x = -1$

That leaves exactly one solution.

Since $x - 6 = -(6 - x)$, the equation reads $\lvert 6 - x \rvert = -(6 - x)$, which holds when the inside is nonpositive.

$\lvert 6 - x \rvert = -(6 - x) \quad\Longleftrightarrow\quad 6 - x \le 0$

That gives $x \ge 6$, the whole ray from $6$ rightward.

An absolute value is always at least $0$, so it can be less than or equal to $0$ only when it is exactly $0$.

$\lvert x - 8 \rvert \le 0 \quad\Longleftrightarrow\quad x - 8 = 0$

That forces $x = 8$, so the solution set is the single point $\{8\}$.

The anchors $6$ and $-2$ are $8$ apart and the target $12$ exceeds that gap, so split the excess $(12 - 8)/2 = 2$ beyond each end.

$x = -2 - 2 = -4, \qquad x = 6 + 2 = 8$

Checking, $\lvert -4 - 6 \rvert + \lvert -4 + 2 \rvert = 10 + 2 = 12$ and $\lvert 8 - 6 \rvert + \lvert 8 + 2 \rvert = 2 + 10 = 12$, so the solutions are $-4$ and $8$.

Split into cases and check each against $3x + 5 \ge 0$.

$x - 1 = 3x + 5 \Rightarrow x = -3, \qquad x - 1 = -(3x + 5) \Rightarrow 4x = -4 \Rightarrow x = -1$

At $x = -3$ the right side is $-4 < 0$, extraneous. At $x = -1$ it is $2 \ge 0$ and $\lvert -1 - 1 \rvert = 2$ checks, so the solution set is $\{-1\}$.