Absolute Value Equations and Inequalities: Additional Practice (Set 2)

Both sides are absolute values, so set the insides equal, then opposite.

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

Both are genuine, and their sum is $-4 + \frac{2}{3} = -\frac{10}{3}$.

Split into cases and check each against $2x - 7 \ge 0$.

$x - 2 = 2x - 7 \Rightarrow x = 5, \qquad x - 2 = -(2x - 7) \Rightarrow 3x = 9 \Rightarrow x = 3$

At $x = 5$ the right side is $3 \ge 0$ and $\lvert 5 - 2 \rvert = 3$ checks. At $x = 3$ the right side is $-1 < 0$, extraneous, so the solution set is $\{5\}$.

The anchors $5$ and $-1$ are $6$ apart and the target $8$ exceeds it, so solve on the two outer intervals.

$x < -1:\ (5 - x) + (-x - 1) = 8 \Rightarrow x = -2; \qquad x > 5:\ (x - 5) + (x + 1) = 8 \Rightarrow x = 6$

Each root lies inside its interval, and the middle contributes nothing, so the solution set is $\{-2, 6\}$.

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

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

Both candidates are below $4$, so the right side is negative at each and both are extraneous. There is no solution.

The insides share the same coefficient of $x$, so the "equal" case dies and only the "opposite" case survives.

$2x - 1 = -(2x + 5) \quad\Longrightarrow\quad 4x = -4 \quad\Longrightarrow\quad x = -1$

The single solution is $x = -1$.

Split into cases and check each against $2x + 1 \ge 0$.

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

At $x = 2$ the right side is $5 \ge 0$ and $\lvert 5 \rvert = 5$ checks. At $x = -\frac{4}{3}$ the right side is $-\frac{5}{3} < 0$, extraneous, so the solution set is $\{2\}$.

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

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

At $x = 3$ the right side is $3 \ge 0$ and $\lvert 3 \rvert = 3$; at $x = 1$ it is $1 \ge 0$ and $\lvert -1 \rvert = 1$. Both pass, so the solution set is $\{1, 3\}$.

The square root of a square is the nonnegative size of the base, which is its absolute value.

$\sqrt{(3x - 7)^2} = \lvert 3x - 7 \rvert$

Dropping the bars would give a negative result whenever $3x - 7 < 0$, which a root can never be.

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

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

At $x = 1$ the right side is $2 \ge 0$ and $\lvert 3 - 1 \rvert = 2$ checks. At $x = -3$ the right side is $-6 < 0$, extraneous, so the solution set is $\{1\}$.

An absolute value equals the negative of its inside exactly when that inside is nonpositive.

$\lvert x + 2 \rvert = -(x + 2) \quad\Longleftrightarrow\quad x + 2 \le 0$

That gives $x \le -2$, the whole ray from $-2$ leftward.

The anchors $-4$ and $2$ are $6$ apart, exactly the target on the right.

$g = \lvert 2 - (-4) \rvert = 6 = \text{target}$

Between the anchors the two distances always sum to $6$, so every point of the segment $[-4, 2]$ is a solution.

An absolute value is positive everywhere except where its inside is zero.

$4x - 8 = 0 \quad\Longrightarrow\quad x = 2$

At $x = 2$ the value is $0$, not greater than $0$, so the solution set is every real number except $x = 2$.