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

Split the positive right side into two cases.

$3x - 5 = 7 \;\Rightarrow\; x = 4, \qquad 3x - 5 = -7 \;\Rightarrow\; 3x = -2 \;\Rightarrow\; x = -\tfrac{2}{3}$

Isolate the bars.

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

A right side of $0$ gives exactly one solution.

Split into two cases.

$7 - 2x = 3 \;\Rightarrow\; x = 2, \qquad 7 - 2x = -3 \;\Rightarrow\; -2x = -10 \;\Rightarrow\; x = 5$

The vertex is at $(2, -4)$, and $a = 3 > 0$ opens the V upward.

$y \ge k = -4$

A 'greater than' absolute value gives two rays.

$x > 6 \quad \text{or} \quad x < -6$

These are the numbers more than six units from zero.

An absolute value is never negative, so it cannot equal $-5$.

$|2x + 3| \ge 0 > -5$

The negative right side means there is no solution.

Set $y = 0$ and solve the absolute value equation.

$|x + 2| - 3 = 0 \;\Rightarrow\; |x + 2| = 3 \;\Rightarrow\; x + 2 = 3 \text{ or } x + 2 = -3$

This gives $x = 1$ or $x = -5$.

Split into two cases.

$6 - x = 2 \;\Rightarrow\; x = 4, \qquad 6 - x = -2 \;\Rightarrow\; x = 8$

Only zero has absolute value zero.

$x - 9 = 0 \;\Rightarrow\; x = 9$

A right side of $0$ gives exactly one solution.

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

$2x - 1 \ge 7 \quad \text{or} \quad 2x - 1 \le -7$

The first gives $x \ge 4$; the second gives $2x \le -6$, so $x \le -3$.

An absolute value equation has no solution exactly when the right side is negative.

$c < 0 \;\Rightarrow\; \text{no solution}$

Among the choices, only $c = -3$ is negative; $c = 0$ gives one solution and the positive values give two.

Split into two cases, then check that the right side $2x - 1$ is not negative.

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

Testing, $x = 5$ gives $|9| = 9 = 2(5) - 1$, but $x = -1$ makes the right side negative, so it fails. Only $x = 5$ works.