Math I · A-REI.3.1

Solving and Graphing One-Variable Absolute-Value Equations and Inequalities in Context

Absolute value models distance from a target, making it essential for tolerances, error, deviation, safety margins, and acceptable ranges.

Concept Algebra
Domain Reasoning with Equations and Inequalities
Read time 10 minutes

What this learning objective is really asking you to learn

This objective is about understanding absolute value as distance and then using that distance idea to solve equations and inequalities. Absolute value is often introduced as “make the number positive,” but that shortcut is not enough for algebra. The deeper meaning is distance from zero. The expression \(|x|\) means the distance between \(x\) and 0 on the number line. The expression \(|x - 5|\) means the distance between \(x\) and 5. The expression \(|T - 72|\) means how far the temperature \(T\) is from 72 degrees.

An absolute-value equation often asks for values at an exact distance from a target. For example, \(|x - 5| = 3\) asks: what numbers are exactly 3 units away from 5? The answers are 8 and 2. One is 3 units to the right; the other is 3 units to the left. This is why absolute-value equations often split into two cases.

An absolute-value inequality asks for values within a distance or outside a distance. For example, \(|x - 5| \le 3\) asks for numbers whose distance from 5 is at most 3. The answer is the interval from 2 to 8, written \(2 \le x \le 8\). By contrast, \(|x - 5| \ge 3\) asks for numbers whose distance from 5 is at least 3. The answer is \(x \le 2\) or \(x \ge 8\). The first is an inside interval; the second is two outside rays.

This objective also asks students to graph solutions. For one-variable absolute-value problems, the graph is usually on a number line. An equation like \(|x - 5| = 3\) has two points: 2 and 8. An inequality like \(|x - 5| \le 3\) has a closed interval from 2 to 8. An inequality like \(|x - 5| > 3\) has two open rays going left of 2 and right of 8.

The context piece is essential. Absolute value appears whenever the important issue is not direction but distance from a standard, target, center, or expected value. A thermostat target, manufacturing tolerance, measurement error, test-score deviation, delivery-time window, lane position, budget difference, or prediction error can all use absolute value. The student must be able to translate phrases like “within,” “no more than,” “at least,” “more than,” “tolerance,” and “deviation” into the correct absolute-value statement and then interpret the solution.

Why students should learn this math

Absolute value is the mathematics of closeness, error, tolerance, and deviation. That is a huge part of real life. Many systems do not require a quantity to be exactly perfect; they require it to be close enough. A machine part does not need to be exactly 10.000 centimeters long in a philosophical sense. It needs to be within an acceptable tolerance, perhaps within 0.005 centimeters. A refrigerator does not need to stay at exactly 37 degrees every second. It needs to stay within a safe range. A delivery may not need to arrive at exactly 2:00 p.m. It may need to arrive within 15 minutes of the scheduled time.

Students should learn this math because the real world is full of acceptable ranges. In school, students may be trained to look for exact answers. But engineering, medicine, manufacturing, statistics, navigation, electronics, construction, and computer science often care about how far a value is from a target. Absolute value is the first algebraic language students learn for that idea.

Think about quality control. A company manufactures bolts that should be 50 millimeters long, with an allowed error of 0.2 millimeters. The condition is \(|L - 50| \le 0.2\). That single inequality captures a full inspection rule. It means \(49.8 \le L \le 50.2\). Any bolt in that interval passes. Any bolt outside fails. The absolute-value notation is compact, precise, and directly tied to measurement.

Think about sports. A coach may track how far a runner's lap times deviate from a target pace. Being 2 seconds faster and 2 seconds slower are different strategically, but both are 2 seconds from the target. Absolute value measures the size of the deviation. Think about finance. A budget forecast might be off by $500. Whether the actual cost is $500 above or $500 below the prediction, the absolute error is $500. Think about science. Experimental measurements are compared to accepted values by calculating error. Absolute value prevents positive and negative errors from canceling when the size of error is what matters.

This objective also helps students develop number-line reasoning. Many algebra mistakes happen because students manipulate symbols without a mental picture. Absolute value forces students to think spatially: where is the center, how far is the radius, which values are inside, and which values are outside? That number-line picture supports later work with intervals, domains, inequalities, piecewise functions, distance formulas, and statistics.

The “why” is especially important here because absolute value can feel artificial if taught only as a symbol. But it is not artificial. It is one of the most practical mathematical ideas students meet. It tells us how wrong, how far, how close, how tolerant, and how safe.

The historical machinery: distance, magnitude, and measurement

The idea behind absolute value is older than the notation. Humans have always needed to measure magnitude without caring about direction. A debt and a credit have opposite signs financially, but their sizes can be compared. A movement east and a movement west have opposite directions, but their distances can be equal. A temperature 5 degrees above a target and 5 degrees below a target are different states, but both are 5 degrees away.

Negative numbers took a long time to become fully accepted in mathematics. In some ancient and medieval traditions, negative quantities were treated with suspicion because they did not always fit concrete counting interpretations. Over time, commerce, algebra, and coordinate geometry made negative numbers increasingly natural. Once numbers could be placed on a line extending in both directions, distance from zero became a clear geometric idea. The number -7 lies 7 units from zero, just as 7 does.

The absolute-value notation using vertical bars became standard much later, but the concept of magnitude is fundamental. In more advanced mathematics, absolute value is the simplest example of a norm, a function that measures size or distance. For real numbers, \(|x|\) measures distance from zero. For points in the plane, the distance formula measures distance from one point to another. For vectors, norms measure length. In statistics, absolute deviations measure distance from a center. In machine learning, loss functions measure how far predictions are from actual values.

This means that an Integrated Math I absolute-value inequality is not a small side topic. It is the beginning of a major mathematical idea: measuring difference without letting direction erase size. If one error is \(+5\) and another is -5, their average signed error is zero, but that does not mean there was no error. Absolute value keeps the magnitude visible.

Absolute value also connects to the history of precision. As science and industry developed, measurement tolerance became increasingly important. Building interchangeable parts, designing machines, calibrating instruments, and controlling industrial processes all required language for acceptable deviation. Absolute-value inequalities are a clean algebraic way to express that language.

Where this fits in the big map of mathematics

This objective sits at the intersection of algebra, geometry, measurement, and modeling. Algebra supplies the equations and inequalities. Geometry supplies the number-line distance model. Measurement supplies the idea of error and tolerance. Modeling supplies the real-world situations where distance from a target matters.

It builds on one-variable inequalities. Students already need to understand symbols like \(<\), \(\le\), \(>\), and \(\ge\), as well as graphing solution sets on a number line. Absolute value adds a distance layer. Instead of simply asking whether \(x\) is greater than 5, it asks whether \(x\) is within 3 units of 5 or more than 3 units away from 5.

It prepares students for piecewise functions. The absolute-value function \(f(x) = |x|\) can be written as two pieces: \(f(x) = x\) when \(x \ge 0\), and \(f(x) = -x\) when \(x < 0\). That split explains the V-shape of its graph and previews the idea that a function can behave differently on different parts of its domain.

It prepares students for transformations of functions. The graph \(y = |x|\) is centered at zero. The graph \(y = |x - 5|\) shifts right 5 units. The equation \(|x - 5| \le 3\) describes the x-values whose transformed absolute-value output is at most 3. This connects algebraic inequalities to graph features.

It prepares students for distance formulas. On a number line, the distance between \(x\) and \(a\) is \(|x - a|\). In the coordinate plane, the distance between \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) becomes \(\sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}\). The structure is more complex, but the idea is the same: measure separation between positions.

It prepares students for statistics. Mean absolute deviation uses absolute value to measure how far data values are from a mean. Residuals in regression can be positive or negative, but their sizes matter. Error analysis often uses absolute value or squared error to prevent cancellation. The simple expression \(|actual - predicted|\) is one of the most important practical formulas in data work.

In the big map, absolute value is the first formal tool for treating distance as a quantity in itself.

How to execute the skill technically

The first technical rule is to isolate the absolute-value expression before splitting into cases. If \(2|x - 4| + 3 = 13\), subtract 3 first to get \(2|x - 4| = 10\), then divide by 2 to get \(|x - 4| = 5\). Only then split into \(x - 4 = 5\) or \(x - 4 = -5\), giving \(x = 9\) or \(x = -1\).

For equations of the form \(|expression| = number\), there are three possibilities. If the number is positive, there are usually two cases. If the number is zero, there is one case because the expression inside must equal zero. If the number is negative, there is no solution because absolute value cannot be negative. For example, \(|x + 2| = -3\) has no solution.

For inequalities, the distance interpretation is the safest guide. If \(|x - a| < b\), with \(b > 0\), then x is within b units of a: \(a - b < x < a + b\). If \(|x - a| \le b\), then \(a - b \le x \le a + b\). These are inside intervals. If \(|x - a| > b\), then x is more than b units away from a: \(x < a - b\) or \(x > a + b\). If \(|x - a| \ge b\), then \(x \le a - b\) or \(x \ge a + b\). These are outside rays.

The compound-inequality form is also useful. \(|x - 5| \le 3\) becomes \(-3 \le x - 5 \le 3\). Add 5 to all three parts and get \(2 \le x \le 8\). For a greater-than inequality, do not write a single compound inequality with “and.” Use two cases with “or”: \(x - 5 \ge 3\) or \(x - 5 \le -3\), giving \(x \ge 8\) or \(x \le 2\).

Graphing requires attention to open and closed endpoints. Use closed circles for \(\le\) and \(\ge\), because endpoints are included. Use open circles for \(<\) and \(>\), because endpoints are excluded. For \(2 \le x \le 8\), use closed circles at 2 and 8 and shade between. For \(x < 2 or x > 8\), use open circles at 2 and 8 and shade outward.

Checking is especially useful for absolute value because sign errors are common. If your solution to \(|x - 5| \le 3\) is \(2 \le x \le 8\), test an inside point such as 5: \(|5 - 5| = 0\), which is less than or equal to 3. Test an outside point such as 10: \(|10 - 5| = 5\), which is not less than or equal to 3. The solution makes sense.

A worked example: manufacturing tolerance

A metal rod is supposed to be 12 inches long. It passes inspection if its length is within 0.03 inches of the target. What lengths pass?

Let \(L\) be the rod's length in inches. “Within 0.03 inches of 12” means the distance between \(L\) and 12 is no more than 0.03:

\[|L - 12| \le 0.03\].

Rewrite as a compound inequality:

\[-0.03 \le L - 12 \le 0.03\].

Add 12 to all three parts:

\[11.97 \le L \le 12.03\].

The rod passes if its length is at least 11.97 inches and at most 12.03 inches. On a number line, this is a closed interval because the endpoints are allowed.

This example shows the practical meaning of absolute value. The rod can be slightly short or slightly long. What matters is the size of the deviation from the target.

A worked example: being too far from a safe range center

A sensor is calibrated to a center value of 100. A warning should trigger if the reading is more than 8 units away from 100. What readings trigger the warning?

Let \(r\) be the reading. “More than 8 units away from 100” means

\[|r - 100| > 8\].

This is an outside-rays situation:

\(r - 100 < -8\) or \(r - 100 > 8\).

So

\(r < 92\) or \(r > 108\).

The warning triggers below 92 or above 108. On a number line, use open circles at 92 and 108 and shade outward because the condition says more than 8, not at least 8.

Common mistakes and how to avoid them

One common mistake is forgetting that absolute-value equations usually have two solutions. \(|x| = 7\) means \(x = 7\) or \(x = -7\). The negative solution is not a negative distance; it is a number located 7 units from zero in the negative direction.

Another common mistake is splitting before isolating the absolute value. In \(3|x + 1| - 4 = 8\), do not split \(3(x + 1) - 4 = 8\) and \(3(x + 1) - 4 = -8\). First isolate: \(3|x + 1| = 12\), then \(|x + 1| = 4\), then split.

A third mistake is mixing up “and” and “or.” Less-than absolute-value inequalities create an interval and use “and”: \(|x - 5| < 3\) means \(2 < x < 8\). Greater-than absolute-value inequalities create two separate regions and use “or”: \(|x - 5| > 3\) means \(x < 2 or x > 8\).

Students also sometimes ignore the meaning of strict versus inclusive inequalities. “Within 3 units” usually means \(\le 3\), including the endpoints. “Less than 3 units away” means \(< 3\), excluding the endpoints. “At least 3 units away” means \(\ge 3\), including the boundary values. The words matter.

The deepest mistake is thinking of absolute value as a symbol trick instead of distance. Once students ask, “Distance from what target, and how far is allowed?” the topic becomes much easier.

What students should be able to say

A student who has mastered this objective should be able to say: “Absolute value measures distance on a number line. An equation like \(|x - a| = b\) asks for values exactly b units from a. A less-than absolute-value inequality gives values inside a distance range, and a greater-than absolute-value inequality gives values outside the range. I can graph the solutions with correct endpoints, and I can interpret them in real contexts such as tolerance, error, safe ranges, and deviation.”

That is more than a procedural skill. It is a way to measure closeness, and closeness is one of the most common ideas in applied mathematics.

Problem Library

Problems in the App From This Objective

144 problems across 12 archetypes in the app.

split into two cases.
12 problems Warmup Practice Mixed Review Assessment
Problem 1

Solve the absolute-value equation |x - 3| = 5.

Problem 2

Solve the absolute-value equation |x + 4| = 7.

Problem 3

Solve the absolute-value equation |x - 10| = 2.

Problem 4

Solve the absolute-value equation |x| = 6.

Problem 5

Solve the absolute-value equation |x - 7| = 3.

Problem 6

Solve the absolute-value equation |x + 2| = 9.

Problem 7

Solve the absolute-value equation |x - 15| = 10.

Open in simulator
Problem 8

Solve the absolute-value equation |x + 12| = 4.

Problem 9

Solve the absolute-value equation |x - 5| = 1.

Problem 10

Solve the absolute-value equation |x + 8| = 2.

Problem 11

Solve the absolute-value equation |x - 6| = 5.

Problem 12

Solve the absolute-value equation |x + 1| = 3.

isolate absolute value before splitting.
12 problems Warmup Practice Mixed Review Assessment
Problem 13

Solve the absolute-value equation 2|x - 1| = 10 after isolating the absolute-value expression.

Problem 14

Solve the absolute-value equation 3|x + 2| - 4 = 8 after isolating the absolute-value expression.

Problem 15

Solve the absolute-value equation 5|x - 7| + 1 = 16 after isolating the absolute-value expression.

Problem 16

Solve the absolute-value equation 4|x + 3| = 20 after isolating the absolute-value expression.

Open in simulator
Problem 17

Solve the absolute-value equation 2|x - 5| - 3 = 7 after isolating the absolute-value expression.

Problem 18

Solve the absolute-value equation 6|x + 1| + 2 = 14 after isolating the absolute-value expression.

Problem 19

Solve the absolute-value equation 7|x| - 1 = 20 after isolating the absolute-value expression.

Problem 20

Solve the absolute-value equation 4|x - 2| + 6 = 18 after isolating the absolute-value expression.

Problem 21

Solve the absolute-value equation 2|x + 4| - 1 = 9 after isolating the absolute-value expression.

Problem 22

Solve the absolute-value equation 3|x - 8| + 5 = 14 after isolating the absolute-value expression.

Problem 23

Solve the absolute-value equation 5|x + 6| - 2 = 13 after isolating the absolute-value expression.

Problem 24

Solve the absolute-value equation 8|x - 10| = 24 after isolating the absolute-value expression.

produce bounded interval.
12 problems Warmup Practice Mixed Review Assessment
Problem 25

Solve the absolute-value inequality |x - 4| <= 3 and write the interval solution.

Problem 26

Solve the absolute-value inequality |x + 2| < 5 and write the interval solution.

Problem 27

Solve the absolute-value inequality |x - 10| <= 1 and write the interval solution.

Problem 28

Solve the absolute-value inequality |x| < 6 and write the interval solution.

Problem 29

Solve the absolute-value inequality |x - 5| < 2 and write the interval solution.

Problem 30

Solve the absolute-value inequality |x + 3| <= 4 and write the interval solution.

Open in simulator
Problem 31

Solve the absolute-value inequality |x - 8| < 0.5 and write the interval solution.

Problem 32

Solve the absolute-value inequality |x + 1| <= 10 and write the interval solution.

Problem 33

Solve the absolute-value inequality |x - 20| < 5 and write the interval solution.

Problem 34

Solve the absolute-value inequality |x + 15| < 3 and write the interval solution.

Problem 35

Solve the absolute-value inequality |x| <= 1.5 and write the interval solution.

Problem 36

Solve the absolute-value inequality |x - 7| <= 7 and write the interval solution.

produce two-ray solution set.
12 problems Warmup Practice Mixed Review Assessment
Problem 37

Solve the absolute-value inequality |x - 2| >= 5 and write the union solution.

Open in simulator
Problem 38

Solve the absolute-value inequality |x + 1| > 4 and write the union solution.

Problem 39

Solve the absolute-value inequality |x - 6| >= 2 and write the union solution.

Problem 40

Solve the absolute-value inequality |x| >= 3 and write the union solution.

Problem 41

Solve the absolute-value inequality |x| > 7 and write the union solution.

Problem 42

Solve the absolute-value inequality |x - 5| >= 1 and write the union solution.

Problem 43

Solve the absolute-value inequality |x + 3| > 2 and write the union solution.

Problem 44

Solve the absolute-value inequality |x - 10| >= 4 and write the union solution.

Problem 45

Solve the absolute-value inequality |x + 8| > 5 and write the union solution.

Problem 46

Solve the absolute-value inequality |x - 1| >= 0.5 and write the union solution.

Problem 47

Solve the absolute-value inequality |x - 4| > 0.1 and write the union solution.

Problem 48

Solve the absolute-value inequality |x + 2| >= 6 and write the union solution.

open/closed endpoints and interval/ray notation.
12 problems Warmup Practice Mixed Review Assessment
Problem 49

Graph the solution set -2 <= x < 5 on a number line and write interval or union notation.

Problem 50

Graph the solution set x < 1 or x >= 4 on a number line and write interval or union notation.

Problem 51

Graph the solution set 3 < x < 9 on a number line and write interval or union notation.

Problem 52

Graph the solution set 0 <= x <= 7 on a number line and write interval or union notation.

Problem 53

Graph the solution set -5 <= x < -1 on a number line and write interval or union notation.

Problem 54

Graph the solution set -3 < x <= 2 on a number line and write interval or union notation.

Open in simulator
Problem 55

Graph the solution set x < -4 or x > 0 on a number line and write interval or union notation.

Problem 56

Graph the solution set x <= -6 or x >= -2 on a number line and write interval or union notation.

Problem 57

Graph the solution set x <= 2 or x > 5 on a number line and write interval or union notation.

Problem 58

Graph the solution set -10 < x < -5 on a number line and write interval or union notation.

Problem 59

Graph the solution set x <= 3 on a number line and write interval or union notation.

Problem 60

Graph the solution set x > -2 on a number line and write interval or union notation.

connect algebraic cases to context.
12 problems Warmup Practice Mixed Review Assessment
Problem 61

Interpret the absolute-value statement |T - 72| <= 3 in the context room temperature T in degrees.

Open in simulator
Problem 62

Interpret the absolute-value statement |x - 10| > 0.5 in the context part length x in centimeters.

Problem 63

Interpret the absolute-value statement |s - 80| = 5 in the context test score s.

Problem 64

Interpret the absolute-value statement |p - 15| <= 2 in the context pressure p in psi.

Problem 65

Interpret the absolute-value statement |w - 50| > 10 in the context weight w in kilograms.

Problem 66

Interpret the absolute-value statement |h - 60| = 3 in the context height h in inches.

Problem 67

Interpret the absolute-value statement |c - 20| < 1 in the context concentration c in mg/L.

Problem 68

Interpret the absolute-value statement |v - 120| >= 5 in the context voltage v in volts.

Problem 69

Interpret the absolute-value statement |t + 5| <= 2 in the context temperature t in degrees Celsius.

Problem 70

Interpret the absolute-value statement |d - 100| > 10 in the context distance d in miles.

Problem 71

Interpret the absolute-value statement |a| = 7 in the context altitude a in feet.

Problem 72

Interpret the absolute-value statement |r - 250| < 5 in the context resistance r in ohms.

reason about nonnegative distance.
12 problems Warmup Practice Mixed Review Assessment
Problem 73

Determine whether |x - 4| = -3 is possible. Explain using the nonnegative meaning of absolute value.

Problem 74

Determine whether |x + 2| < -1 is possible. Explain using the nonnegative meaning of absolute value.

Problem 75

Determine whether |x - 5| >= -2 is possible. Explain using the nonnegative meaning of absolute value.

Open in simulator
Problem 76

Determine whether |x + 7| = -5 is possible. Explain using the nonnegative meaning of absolute value.

Problem 77

Determine whether |2x - 1| < -4 is possible. Explain using the nonnegative meaning of absolute value.

Problem 78

Determine whether |x/3 + 2| <= -6 is possible. Explain using the nonnegative meaning of absolute value.

Problem 79

Determine whether |x - 10| > -1 is possible. Explain using the nonnegative meaning of absolute value.

Problem 80

Determine whether |3x + 5| >= -8 is possible. Explain using the nonnegative meaning of absolute value.

Problem 81

Determine whether |x + 9| = -12 is possible. Explain using the nonnegative meaning of absolute value.

Problem 82

Determine whether |x - 8| < -20 is possible. Explain using the nonnegative meaning of absolute value.

Problem 83

Determine whether |5x - 2| >= -15 is possible. Explain using the nonnegative meaning of absolute value.

Problem 84

Determine whether |x/2 - 3| > -7 is possible. Explain using the nonnegative meaning of absolute value.

determine center, radius/tolerance, and inequality direction.
12 problems Warmup Practice Mixed Review Assessment
Problem 85

Choose the absolute-value model for A measurement x must be within 0.2 cm of 5 cm.

Problem 86

Choose the absolute-value model for A score s must be more than 10 points away from 70.

Problem 87

Choose the absolute-value model for A temperature T is exactly 4 degrees from 68.

Problem 88

Choose the absolute-value model for A value y must be strictly within 3 units of 10.

Problem 89

Choose the absolute-value model for A price p must be at least $5 away from $25.

Problem 90

Choose the absolute-value model for A height h must be no more than 1 inch from 100 inches.

Problem 91

Choose the absolute-value model for A weight w is exactly 10 pounds from 150 pounds.

Problem 92

Choose the absolute-value model for A velocity v must be more than 7 m/s away from 0 m/s.

Open in simulator
Problem 93

Choose the absolute-value model for A distance d must not be more than 0.5 miles from 12 miles.

Problem 94

Choose the absolute-value model for A time t must be strictly outside 2 hours of 32 hours.

Problem 95

Choose the absolute-value model for A resistance r must be at most 1.5 ohms from 50 ohms.

Problem 96

Choose the absolute-value model for A temperature k is exactly 8 degrees from -5 degrees.

isolate, split cases, and check for extraneous solutions.
12 problems Warmup Practice Mixed Review Assessment
Problem 97

Solve the absolute-value equation |x - 1| = x + 3 and check for extraneous solutions.

Problem 98

Solve the absolute-value equation |2x - 4| = x + 1 and check for extraneous solutions.

Problem 99

Solve the absolute-value equation |x + 2| = 3x - 6 and check for extraneous solutions.

Problem 100

Solve the absolute-value equation |x + 5| = 2x + 1 and check for extraneous solutions.

Problem 101

Solve the absolute-value equation |3x - 2| = x + 4 and check for extraneous solutions.

Problem 102

Solve the absolute-value equation |x + 1| = x - 2 and check for extraneous solutions.

Problem 103

Solve the absolute-value equation |x - 5| = 2x - 1 and check for extraneous solutions.

Open in simulator
Problem 104

Solve the absolute-value equation |5x + 3| = x + 9 and check for extraneous solutions.

Problem 105

Solve the absolute-value equation |-x + 7| = x + 1 and check for extraneous solutions.

Problem 106

Solve the absolute-value equation |2x + 1| = -x + 7 and check for extraneous solutions.

Problem 107

Solve the absolute-value equation |3x - 1| = 3x - 5 and check for extraneous solutions.

Problem 108

Solve the absolute-value equation |(1/2)x + 3| = x + 1 and check for extraneous solutions.

undo outside operations before interval/ray split.
12 problems Warmup Practice Mixed Review Assessment
Problem 109

Solve the absolute-value inequality 2|x - 3| + 1 <= 9 after isolating the absolute-value expression.

Problem 110

Solve the absolute-value inequality 3|x + 1| - 2 > 10 after isolating the absolute-value expression.

Problem 111

Solve the absolute-value inequality 5|x - 2| + 6 >= 16 after isolating the absolute-value expression.

Problem 112

Solve the absolute-value inequality 4|x + 5| - 3 <= 13 after isolating the absolute-value expression.

Problem 113

Solve the absolute-value inequality 2|x - 4| + 5 > 15 after isolating the absolute-value expression.

Problem 114

Solve the absolute-value inequality 6|x + 2| - 1 >= 11 after isolating the absolute-value expression.

Problem 115

Solve the absolute-value inequality 7|x - 1| + 2 < 23 after isolating the absolute-value expression.

Open in simulator
Problem 116

Solve the absolute-value inequality -2|x + 3| + 10 <= 4 after isolating the absolute-value expression.

Problem 117

Solve the absolute-value inequality -3|x - 5| + 1 < -8 after isolating the absolute-value expression.

Problem 118

Solve the absolute-value inequality (1/2)|x + 4| - 1 <= 1 after isolating the absolute-value expression.

Problem 119

Solve the absolute-value inequality 12 <= 3|x - 6| + 3 after isolating the absolute-value expression.

Problem 120

Solve the absolute-value inequality 5|2x - 1| - 7 < 8 after isolating the absolute-value expression.

connect strict/non-strict symbols to open/closed endpoints.
12 problems Warmup Practice Mixed Review Assessment
Problem 121

Explain whether endpoints are included in the graph of |x - 4| <= 3.

Problem 122

Explain whether endpoints are included in the graph of |x + 2| < 5.

Problem 123

Explain whether endpoints are included in the graph of |x - 1| >= 6.

Problem 124

Explain whether endpoints are included in the graph of |x + 3| > 4.

Problem 125

Explain whether endpoints are included in the graph of |x - 5| <= 2.

Problem 126

Explain whether endpoints are included in the graph of |x + 1| < 8.

Problem 127

Explain whether endpoints are included in the graph of |x - 2| >= 1.

Problem 128

Explain whether endpoints are included in the graph of |x| > 6.

Open in simulator
Problem 129

Explain whether endpoints are included in the graph of |x| <= 9.

Problem 130

Explain whether endpoints are included in the graph of |2x - 6| < 4.

Problem 131

Explain whether endpoints are included in the graph of |3x + 9| >= 3.

Problem 132

Explain whether endpoints are included in the graph of |x + 6| < 10.

translate between graph and symbolic distance statement.
12 problems Warmup Practice Mixed Review Assessment
Problem 133

Choose which absolute-value statement matches the number-line graph: closed interval from 2 to 8.

Problem 134

Choose which absolute-value statement matches the number-line graph: open rays left of -4 and right of 6.

Problem 135

Choose which absolute-value statement matches the number-line graph: closed rays x <= -2 or x >= 10.

Open in simulator
Problem 136

Choose which absolute-value statement matches the number-line graph: open interval between -3 and 7.

Problem 137

Choose which absolute-value statement matches the number-line graph: closed interval from -1 to 5.

Problem 138

Choose which absolute-value statement matches the number-line graph: open rays left of -7 and right of -1.

Problem 139

Choose which absolute-value statement matches the number-line graph: closed rays x <= 0 or x >= 10.

Problem 140

Choose which absolute-value statement matches the number-line graph: open interval between -6 and 2.

Problem 141

Choose which absolute-value statement matches the number-line graph: closed interval from -9 to -3.

Problem 142

Choose which absolute-value statement matches the number-line graph: open rays left of -2 and right of 8.

Problem 143

Choose which absolute-value statement matches the number-line graph: closed rays x <= -8 or x >= -4.

Problem 144

Choose which absolute-value statement matches the number-line graph: open interval between 1 and 9.