HW_Absolute Value Equations and Inequalities

Vocabulary flashcards covering absolute value equations, inequalities, domain constraints, word problems, and step-by-step solutions from the assignment.

Algebra 2 Honors: Absolute Value Equations & Inequalities

A study unit covering solution methods, graphing, domain constraints, and interval notation for linear inequalities, word problems, absolute value equations, and absolute value inequalities.

Problem 1 Inequality: $4(2 - x) - 3(1 + x) < 5(1 - x)$

Simplifies to $-2x < 0$, yielding the solution $x > 0$ or in interval notation $(0, \infty)$.

Problem 2 Compound Inequality: $-5 < 2(2 - x) + 1 \le 9$

Simplifies to $-10 < -2x \le 4$, which after dividing by $-2$ yields $-2 \le x < 5$ or in interval notation $[-2, 5)$.

Bingham Tunnel Toll Word Problem

A problem comparing a standard toll of $50\,\text{cents}$ per trip against a $\$5.50$ sticker with $35\,\text{cents}$ per trip; solved by setting $5.50 + 0.35x < 0.50x$, requiring at least $37\,\text{trips}$ for the sticker to cost less.

Problem 4 Equation: $3|10 - 4x| + 16 = 28$

Isolates to $|10 - 4x| = 4$, yielding two linear equations $10 - 4x = 4$ and $10 - 4x = -4$, with solution set $\left\{\frac{3}{2}, \frac{7}{2}\right\}$.

Problem 5 Inequality: $2|x + 6| + 8 \ge 24$

Isolates to $|x + 6| \ge 8$, leading to $x + 6 \ge 8$ or $x + 6 \le -8$, with interval notation solution $(-\infty, -14] \cup [2, \infty)$.

Problem 6 Inequality: $-2|4x + 1| \le -4$

Dividing by $-2$ flips the inequality to $|4x + 1| \ge 2$, yielding $x \ge \frac{1}{4}$ or $x \le -\frac{3}{4}$, written in interval notation as $\left(-\infty, -\frac{3}{4}\right] \cup \left[\frac{1}{4}, \infty\right)$.

Problem 7 Inequality: $3|x - 5| + 10 < 22$

Isolates to $|x - 5| < 4$, forming the compound inequality $-4 < x - 5 < 4$, with interval notation solution $(1, 9)$.

Problem 8 Inequality: $|x + 3| + 10 < 4$

Isolates to $|x + 3| < -6$; since an absolute value cannot be negative, there is no solution ($\emptyset$).

Problem 9 Inequality: $3|x - 9| + 12 > 12$

Isolates to $|x - 9| > 0$; true for all real numbers except where $x - 9 = 0$ ($x = 9$), yielding interval notation $(-\infty, 9) \cup (9, \infty)$.

Problem 10 Equation: $-2|x + 1| - 3 = 13$

Isolates to $|x + 1| = -8$; because an absolute value expression cannot equal a negative number, this equation has no solution ($\emptyset$).

Problem 11 Inequality: $7 + 5|c| \le 1 - 3|c|$

Isolates to $8|c| \le -6$ or $|c| \le -\frac{3}{4}$; since absolute value is non-negative, this inequality has no solution ($\emptyset$).

Problem 12 Equation: $|x - 4| = x + 1$

Solving $x - 4 = -(x + 1)$ gives $2x = 3 \implies x = \frac{3}{2}$, which checks as valid (while $x - 4 = x + 1$ yields no solution).

Problem 13 Double Inequality: $2 < 2|x - 3| - 4 < 12$

Isolates to $3 < |x - 3| < 8$, which splits into $-8 < x - 3 < -3$ and $3 < x - 3 < 8$, giving interval notation solution $(-5, 0) \cup (6, 11)$.

Problem 14 Domain Restricted Inequality: $|1 - x| - 7 < 3$

Simplifies to $-9 < x < 11$. For domain $\{N\}$ (Natural numbers), the solution is $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$; for domain \{\text{negative numbers}\}$, it is (-9, 0)$$.

Problem 15 Domain Restricted Equation: $\frac{2}{5}|2x - 3| = 14$

Isolates to $|2x - 3| = 35$, giving $x = 19$ or $x = -16$. For domain $\{W\}$ (Whole numbers), the solution is $x = 19$; for domain $\{Z\}$ (Integers), the solution set is $\{-16, 19\}$.

Problem 16 Domain Restricted Inequality: $4|2 - 3x| + 9 \ge 29$

Isolates to $|2 - 3x| \ge 5$, giving $x \le -1$ or $x \ge \frac{7}{3}$. Restricting to the domain $\{\text{positive numbers}\}$ yields $x \ge \frac{7}{3}$ or $\left[\frac{7}{3}, \infty\right)$.