Algebra I - Absolute Value

• Mathematics •

---

The **absolute value** of a number $n$, denoted as $\vert{}n\vert{}$, is its distance from $0$. Because it represents distance, $\vert{}n\vert{}$ is always nonnegative. * **Basic Examples:** $\vert{}3\vert{} = 3$, $\vert{}-5\vert{} = 5$, and $\vert{}0\vert{} = 0$. * **Formal Definition:** Formally, absolute value is defined as: $$\vert{}x\vert{} = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$$ * **Distance Between Two Numbers:** The distance between $a$ and $b$ is the positive value $\vert{}a - b\vert{} = \vert{}b - a\vert{}$. For example, $\vert{}5 - 8\vert{} = 3$. * **Order of Operations:** Absolute value bars act like parentheses when determining the order of operations. --- ### Properties of Absolute Value * $\vert{}a\vert{} = b$ means $a = b$ or $a = -b$. * $\vert{}a\vert{} = 0$ means $a = 0$. * **When $b \ge 0$:** * $\vert{}a\vert{} = b$ means $a = b$ or $a = -b$. * $\vert{}a\vert{} < b$ means $-b < a < b$. * $\vert{}a\vert{} > b$ means $a < -b$ or $a > b$. * **When $b < 0$:** * $\vert{}a\vert{} \le b$ is impossible. * $\vert{}a\vert{} > b$ means $a$ could be anything. --- ### Solving Equations with Absolute Value To solve equations containing absolute value expressions, follow these steps: 1. **Isolate the absolute value expression:** Change the equation until the absolute value expression is alone on one side. * You can factor out positive constants: for example, $\vert{}2x - 4\vert{} = 6$ is equivalent to $\vert{}x - 2\vert{} = 3$. * To factor out negative constants, use the rule $\vert{}a\vert{} = \vert{}-a\vert{}$. For example, $\vert{}-x - 1\vert{} = 4$ is equivalent to $\vert{}x + 1\vert{} = +4$, which has the solutions $\{3, -5\}$ (whereas the equation $\vert{}x + 1\vert{} = -4$ has no solutions). 2. **Unravel the absolute value:** Use the Properties of Absolute Value to split the equation into two equalities (unless using the property that $\vert{}a\vert{} = 0$ implies $a = 0$). 3. **Solve each equality separately:** There may be no solutions, $1$ or $2$ solutions, or all real numbers may be solutions. 4. **Check your solutions:** Plug specific solutions back into the original equation. If there are infinitely many solutions or no solutions, test two numbers of large magnitude, one positive and one negative. 5. **Watch out for variables outside the bars:** Be especially careful if the equation contains variables both inside and outside the absolute value bars, and keep track of which equalities and inequalities hold true in each case. **Worked Examples for Equations:** * **Example 1 ($\vert{}3x - 5\vert{} = 2x$):** If $2x \ge 0$, then $3x - 5 = 2x$ or $3x - 5 = -2x$. The first equation gives $x = 5$, and the second gives $x = 1$; both solutions work. * **Example 2 ($\vert{}3x + 5\vert{} = 3x$):** If $3x \ge 0$, we can rewrite this as $3x + 5 = 3x$ or $3x + 5 = -3x$. The first equation ($3x + 5 = 3x$) gives no solutions. The second equation seems to give the solution $x = -\frac{5}{6}$. However, the equation $3x + 5 = -3x$ only holds if $3x \ge 0$ (meaning $x \ge 0$), so $-\frac{5}{6}$ does not work, and there is **no solution**. --- ### Solving Inequalities with Absolute Value **Method 1: Unraveling Tricks** * $\vert{}a\vert{} < b$ is true when $b \ge 0$ and $-b < a < b$. * $\vert{}a\vert{} > b$ is true when $b < 0$ **OR** $\{b \ge 0 \text{ and } a > b\}$ **OR** $\{b \ge 0 \text{ and } a < -b\}$. * **Example ($\vert{}10x + 1\vert{} < 7x + 3$):** This is equivalent to $7x + 3 \ge 0$ and $-7x - 3 < 10x + 1 < 7x + 3$. Thus, the three inequalities $7x + 3 \ge 0$, $-7x - 3 < 10x + 1$, and $10x + 1 < 7x + 3$ must all hold simultaneously. Solving them gives $x \ge -\frac{3}{7}$, $x > -\frac{4}{17}$, and $x < \frac{2}{3}$. Since the second condition ($x > -\frac{4}{17}$) forces the first ($x \ge -\frac{3}{7}$), the final solutions are all $x$ with $-\frac{4}{17} < x < \frac{2}{3}$. **Method 2: Least-Thinking Method (Trial and Error)** * Replace every $\vert{} \text{expression} \vert{}$ with $\pm(\text{expression})$ and find all solutions to the associated equalities. Also find all solutions obtained by replacing the absolute value with $0$. * All of these values act as potential boundary points on the real number line. Determine the solution intervals by testing a point in every interval (such as $x = 0$) along with every boundary point. * **Example ($\vert{}10x + 1\vert{} < 7x + 3$):** Solve the three equations $10x + 1 = 7x + 3$, $-10x - 1 = 7x + 3$, and $7x + 3 = 0$ to find the potential boundary points: $\frac{2}{3}$, $-\frac{4}{17}$, and $-\frac{3}{7}$. Testing the three boundary points and a point from each of the four resulting intervals yields the solution $-\frac{4}{17} < x < \frac{2}{3}$. --- ### Graphing Absolute Value Statements Because absolute value represents distance, $\vert{}x - a\vert{} = b$ means the distance between $a$ and $x$ is $b$ (where $b$ must be non-negative): * **$\vert{}x - a\vert{} = b$:** The distance from $a$ to $x$ is exactly $b$. Plot the two points $x = a + b$ and $x = a - b$. * **$\vert{}x - a\vert{} < b$ (or $\vert{}x - a\vert{} \le b$):** The distance from $a$ to $x$ is less than (or no more than) $b$, meaning $x$ is closer than $b$ to $a$. Plot the open or closed interval $a - b < x < a + b$ (or $a - b \le x \le a + b$). * **$\vert{}x - a\vert{} > b$ (or $\vert{}x - a\vert{} \ge b$):** The distance from $a$ to $x$ is more than (or no less than) $b$, meaning $x$ is further than $b$ away from $a$. Plot the open or closed double rays $x < a - b$ and $x > a + b$ (or $x \le a - b$ and $x \ge a + b$).