Absolute Value Equations and Inequalities: Practice

Split into two cases, keeping each as a candidate until the right side is checked for being nonnegative.

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

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

The right side $x - 1$ must be nonnegative, so only $x \ge 1$ is eligible. Split and check.

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

Both candidates are below $1$, so the right side is negative at each and both are extraneous. The solution set is $\varnothing$.

Two absolute values equal means the insides are equal or opposite, and no check is needed.

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

Both are genuine solutions, so the solution set is $\left\{3, -\frac{1}{2}\right\}$.

Set the insides equal, then equal to opposites.

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

Both cases give genuine solutions, so the solution set is $\left\{-6, \frac{4}{3}\right\}$.

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

$5x - 2 = 0 \quad\Longrightarrow\quad x = \tfrac{2}{5}$

At $x = \frac{2}{5}$ the value is $0$, not greater than $0$, so the solution set is every real number except $x = \frac{2}{5}$.

A strict "less than" absolute value gives an open band around the center.

$-5 < x - 3 < 5 \quad\Longrightarrow\quad -2 < x < 8$

That is the open interval $(-2, 8)$, endpoints excluded because the inequality is strict.

"Within $4$ of $-1$" means $\lvert x - (-1) \rvert \le 4$, a closed band around $-1$.

$-4 \le x + 1 \le 4 \quad\Longrightarrow\quad -5 \le x \le 3$

That is the closed interval $[-5, 3]$, centered at $-1$ with radius $4$.

Between the anchors, the two distances split the gap between them, so their sum is that gap.

$g = \lvert 7 - 1 \rvert = 6$

For every $x$ in $[1, 7]$ the sum is exactly $6$, constant across the whole middle segment.

The smallest the left side can be is the gap between the anchors, reached across the middle segment.

$g = \lvert 7 - 1 \rvert = 6 > 4$

Since the target $4$ is below the minimum $6$, the sum never reaches it and there is no solution.

The target equals the gap between the anchors, and the sum equals that gap everywhere between them.

$g = \lvert 7 - 1 \rvert = 6 = \text{target}$

Every point of the segment $[1, 7]$ satisfies the equation, so there are infinitely many solutions.

The gap is $6$ and the target $10$ exceeds it, so there are two solutions, one outside each anchor. The excess is split evenly, $(10 - 6)/2 = 2$ beyond each end.

$x = 1 - 2 = -1, \qquad x = 7 + 2 = 9$

Checking, $\lvert -1 - 1 \rvert + \lvert -1 - 7 \rvert = 2 + 8 = 10$ and $\lvert 9 - 1 \rvert + \lvert 9 - 7 \rvert = 8 + 2 = 10$, so the solutions are $-1$ and $9$.

When $x < 2$ the inside $x - 2$ is negative, so the absolute value negates it.

$\lvert x - 2 \rvert = -(x - 2) = 2 - x \quad (x < 2)$

The result $2 - x$ is positive there, as a distance must be.