Absolute Value Equations and Graphs: Practice

Isolate the bars before splitting into cases.

$3|x - 2| = 12 \;\Rightarrow\; |x - 2| = 4$

Then $x - 2 = 4$ or $x - 2 = -4$, giving $x = 6$ or $x = -2$.

An absolute value is never negative, so it cannot equal $-7$.

$|x + 4| \ge 0 > -7$

The negative right side means there is no solution.

A 'less than or equal to' absolute value traps the inside in a band.

$-5 \le x - 3 \le 5$

Adding $3$ to every part gives $-2 \le x \le 8$.

Isolate the bars first.

$-|x| = -3 \;\Rightarrow\; |x| = 3$

Then $x = 3$ or $x = -3$.

Split into two cases, then check against the requirement $2x \ge 0$.

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

Testing, $x = 3$ gives $|6| = 6 = 2(3)$, but $x = -1$ gives $|2| = 2 \ne -2$. Only $x = 3$ works.

Isolate the bars.

$2|x - 1| = 0 \;\Rightarrow\; |x - 1| = 0 \;\Rightarrow\; x - 1 = 0$

Only zero has absolute value zero, so $x = 1$ is the single solution.

A 'greater than or equal to' absolute value gives two rays.

$x \ge 3 \quad \text{or} \quad x \le -3$

These are the numbers at least three units from zero.

The inside $x - 5$ gives $h = 5$, the constant gives $k = 2$, and $a = 1 > 0$ opens upward.

$(h, k) = (5, 2), \qquad y \ge 2$

Only zero has absolute value zero.

$2x + 6 = 0 \;\Rightarrow\; x = -3$

The right side $0$ gives exactly one solution.

Split into two cases.

$5x - 2 = 13 \;\Rightarrow\; x = 3, \qquad 5x - 2 = -13 \;\Rightarrow\; 5x = -11 \;\Rightarrow\; x = -\tfrac{11}{5}$

The distance between two numbers is the absolute value of their difference.

$\text{distance} = |x - 4|$

The bars make the result positive no matter which number is larger.

Opening downward needs a negative coefficient, and a vertex at height $4$ needs $+4$ outside.

$y = -|x| + 4$