Absolute Value Practice Quiz with a Step-by-Step Interactive Lesson
Use the quiz at the top of the page to practice absolute value: evaluating absolute value (like \(\lvert -7\rvert\)), simplifying absolute value expressions (including nested bars and negatives), using absolute value as distance on a number line (\(\lvert a-b\rvert\)), solving absolute value equations like \(\lvert ax+b\rvert=c\), solving absolute value inequalities like \(\lvert ax+b\rvert<c\) and \(\lvert ax+b\rvert\ge c\), writing solutions in interval notation, and understanding graphs of absolute value functions like \(y=\lvert x\rvert\) and \(y=\lvert x-h\rvert+k\). If you want a refresher, click Start lesson to open a step-by-step guide with examples and quick checks.
How this absolute value practice works
1. Take the quiz: answer the absolute value questions at the top of the page.
2. Open the lesson (optional): review the absolute value definition, distance meaning, and reliable solve steps for equations and inequalities.
3. Retry: return to the quiz and apply the absolute value rules immediately.
What you will learn in the absolute value lesson
Foundations & meaning
The definition of absolute value and why \(\lvert a\rvert \ge 0\)
Distance from zero and distance between two numbers: \(\lvert a-b\rvert\)
Piecewise form of \(\lvert x\rvert\) and when each case applies
Simplify absolute value expressions
Simplifying with nested absolute values and negatives
Order of operations with absolute value bars
Common mistakes (like confusing \(-\lvert a\rvert\) with \(\lvert -a\rvert\))
Graphing \(y=\lvert x\rvert\) and transformations \(y=\lvert x-h\rvert+k\)
Back to the quiz
When you are ready, return to the quiz at the top of the page and keep practicing absolute value.
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Absolute Value
Step-by-step guide
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Absolute Value Lesson
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Lesson Overview
Lesson overview
Purpose: Build a clear understanding of absolute value so you can simplify expressions, interpret distance, solve absolute value equations, solve absolute value inequalities, and connect everything to graphs and interval notation.
Success criteria
Evaluate absolute values like \(\lvert -7\rvert\) and explain why the result is nonnegative.
Use the piecewise definition of \(\lvert x\rvert\) and decide which case applies.
Simplify expressions with nested absolute values and negatives (for example, \(-\lvert \lvert -3\rvert\rvert\)).
Use \(\lvert a-b\rvert\) to find the distance between two numbers on a number line.
Solve equations of the form \(\lvert ax+b\rvert=c\) (and recognize when there is no solution).
Solve inequalities of the form \(\lvert ax+b\rvert<c\), \(\lvert ax+b\rvert\le c\), \(\lvert ax+b\rvert>c\), and \(\lvert ax+b\rvert\ge c\).
Write solution sets using interval notation and number-line reasoning.
Graph \(y=\lvert x\rvert\) and transformations like \(y=\lvert x-h\rvert+k\) (vertex shifts).
Key vocabulary
Absolute value: the distance of a number from \(0\) on the number line (always \(\ge 0\)).
Distance: between \(a\) and \(b\) is \(\lvert a-b\rvert\).
Piecewise definition: a definition that depends on whether a quantity is nonnegative or negative.
Solution set: all values that make an equation or inequality true.
Interval notation: a compact way to write solutions like \(( -3, 1 )\) or \((-\infty,-7)\cup(-3,\infty)\).
Quick pre-check
Pre-check 1: What is \(\lvert -8\rvert\)?
Hint: Absolute value is distance from \(0\), so it cannot be negative.
Pre-check 2: Solve \(\lvert x\rvert=5\). Which set is correct?
Hint: \(\lvert x\rvert=5\) means \(x\) is 5 units from 0, so there are two symmetric solutions.
Definition & Simplification
Definition of absolute value and simplifying expressions
Learning goal: Use the definition of \(\lvert x\rvert\) to simplify expressions correctly, including nested absolute values and negatives outside the bars.
Key idea
Absolute value can be defined piecewise: \[
\lvert x\rvert =
\begin{cases}
x, & x \ge 0 \\
-x, & x < 0
\end{cases}
\]
This definition explains why \(\lvert x\rvert\) is always nonnegative. A common trap is mixing up \(\lvert -x\rvert\) and \(-\lvert x\rvert\): \(\lvert -x\rvert=\lvert x\rvert\), but \(-\lvert x\rvert\) is the negative of a nonnegative number.
First simplify the inner absolute value: \(\lvert -1\rvert=1\). Now the expression becomes \( -\lvert -1 - 2 \rvert = -\lvert -3\rvert\). Since \(\lvert -3\rvert=3\), the final answer is: \[
-\lvert -\lvert -1 \rvert - 2 \rvert = -3.
\]
Try it
Try it 1: Simplify \(-\lvert \lvert -3 \rvert \rvert\).
Hint: Inner \(\lvert -3\rvert=3\). The outer absolute value keeps it \(3\), then the negative sign makes it \(-3\).
Try it 2: What is \(\lvert 3 - 8 \rvert\)?
Hint: \(\lvert 3-8\rvert=\lvert -5\rvert\).
Summary
\(\lvert x\rvert\) is always nonnegative and can be defined piecewise.
Be careful with a negative sign outside the bars: \(-\lvert x\rvert\) is negative (unless \(x=0\)).
Distance
Absolute value as distance on the number line
Learning goal: Use absolute value to measure distance from zero and distance between two numbers, and use that meaning to reason about inequalities.
Key idea
\(\lvert x\rvert\) is the distance from \(0\) to \(x\). More generally, the distance between two numbers \(a\) and \(b\) is: \[
\text{distance}(a,b)=\lvert a-b\rvert.
\]
This is why \(\lvert a-b\rvert=\lvert b-a\rvert\): distance does not depend on direction.
Worked example
Example: Find the distance between \(-1\) and \(4\).
Use \(\lvert a-b\rvert\): \[
\lvert -1-4\rvert=\lvert -5\rvert=5.
\]
So the numbers are 5 units apart on the number line.
Try it
Try it 1: Which is greater: \(\lvert 2 - 7 \rvert\) or \(\lvert -1 - 4 \rvert\)?
Hint: Both differences are \(-5\) inside the absolute value.
Try it 2: How many integers satisfy \(\lvert x + 1 \rvert < 2\)?
Hint: Rewrite as \(-2<x+1<2\) and list the integers that work.
Summary
\(\lvert x\rvert\) is distance from \(0\).
\(\lvert a-b\rvert\) is distance between \(a\) and \(b\).
Equations
Solving absolute value equations
Learning goal: Solve equations like \(\lvert ax+b\rvert=c\) by splitting into two linear cases.
Key idea
If \(c \ge 0\), then: \[
\lvert A\rvert=c \quad \Rightarrow \quad A=c \text{ or } A=-c.
\]
If \(c < 0\), there is no solution, because an absolute value cannot be negative.
Set the inside equal to \(7\) and \(-7\): \(3x+1=7 \Rightarrow 3x=6 \Rightarrow x=2\). \(3x+1=-7 \Rightarrow 3x=-8 \Rightarrow x=-\dfrac{8}{3}\). So the solution set is \(\left\{2,-\dfrac{8}{3}\right\}\).
Try it
Try it 1: Solve for \(x\): \(\lvert 2x - 5 \rvert = 3\).
Hint: Set \(\dfrac{x}{3}-2=1\) or \(\dfrac{x}{3}-2=-1\).
Summary
For \(\lvert A\rvert=c\) with \(c\ge 0\), solve \(A=c\) and \(A=-c\).
If \(c<0\), there are no solutions.
Inequalities Less Than
Absolute value inequalities: less than and at most
Learning goal: Solve \(\lvert A\rvert<c\) and \(\lvert A\rvert\le c\) by turning them into a single compound inequality.
Key idea
When \(c \ge 0\): \[
\lvert A\rvert < c \;\Rightarrow\; -c < A < c,
\qquad
\lvert A\rvert \le c \;\Rightarrow\; -c \le A \le c.
\]
This matches the distance meaning: \(\lvert A\rvert < c\) means "\(A\) is within \(c\) units of \(0\)".
Worked example
Example: How many integer values satisfy \(\lvert n + 4 \rvert < 3\)?
Rewrite as a compound inequality: \[
-3 < n+4 < 3.
\]
Subtract 4: \[
-7 < n < -1.
\]
The integers are \(-6,-5,-4,-3,-2\), so there are 5 integer solutions.
Try it
Try it 1: How many integer solutions satisfy \(\lvert 2k + 1 \rvert < 5\)?
Hint: Rewrite as \(-5<2k+1<5\), then solve for \(k\) and count integers.
Try it 2: How many integer values satisfy \(\lvert k + 2 \rvert \le 1\)?
Hint: \(\lvert k+2\rvert\le 1\) means \(-1\le k+2\le 1\).
Absolute value inequalities: greater than and at least
Learning goal: Solve \(\lvert A\rvert>c\) and \(\lvert A\rvert\ge c\) by splitting into two cases (a union of intervals).
Key idea
When \(c \ge 0\): \[
\lvert A\rvert > c \;\Rightarrow\; A > c \text{ or } A < -c,
\qquad
\lvert A\rvert \ge c \;\Rightarrow\; A \ge c \text{ or } A \le -c.
\]
This matches distance: \(\lvert A\rvert>c\) means "\(A\) is more than \(c\) units away from \(0\)", so you get two separated regions.
Worked example
Example: Solve \(\lvert x - 1\rvert > 2\).
Split into two cases: \[
x-1>2 \quad \text{or} \quad x-1<-2.
\]
Solve each: \[
x>3 \quad \text{or} \quad x<-1.
\]
In interval notation: \((-\infty,-1)\cup(3,\infty)\).
Try it
Try it 1: Solve \(\lvert x + 5 \rvert > 2\).
Hint: Write \(x+5>2\) or \(x+5<-2\).
Try it 2: Solve \(\lvert x - 2 \rvert \ge 5\).
Hint: Write \(x-2\ge 5\) or \(x-2\le -5\).
Summary
\(\lvert A\rvert>c\) becomes \(A>c\) or \(A<-c\) (a union of two intervals).
\(\lvert A\rvert\ge c\) becomes \(A\ge c\) or \(A\le -c\).
Graphs & Piecewise
Graphs of absolute value and the piecewise function view
Learning goal: Recognize the "V" shape of \(y=\lvert x\rvert\), use vertex shifts, and connect graphs to the piecewise definition.
Key idea
The graph of \(y=\lvert x\rvert\) is a "V" with vertex at \((0,0)\). You can view it as two lines: \[
y=\lvert x\rvert =
\begin{cases}
x, & x \ge 0 \\
-x, & x < 0
\end{cases}
\]
A useful transformation pattern is: \[
y=\lvert x-h\rvert + k,
\]
which shifts the vertex to \((h,k)\).
Worked example
Example: What is the vertex of \(y=\lvert x-2\rvert+1\)?
The vertex occurs where the inside equals zero: \(x-2=0 \Rightarrow x=2\). Then \(y=\lvert 0\rvert+1=1\). So the vertex is \((2,1)\).
Try it
Try it 1: What is the vertex of \(y=\lvert x+3\rvert-2\)?
Hint: The vertex is \((h,k)\) for \(y=\lvert x-h\rvert+k\). Rewrite \(\lvert x+3\rvert\) as \(\lvert x-(-3)\rvert\).
Try it 2: Which piecewise function equals \(\lvert x\rvert\)?
Hint: Absolute value keeps positives the same and flips negatives to become positive.
Summary
\(y=\lvert x\rvert\) is a V-shape with vertex at \((0,0)\).
\(y=\lvert x-h\rvert+k\) shifts the vertex to \((h,k)\).
The piecewise definition explains the shape: one line for \(x\ge 0\), another for \(x<0\).
Applications & Big Picture
Why absolute value matters
Learning goal: Connect absolute value to real-world tolerances and finish with a final check that reinforces the key rules.
Where absolute value shows up
Distance and geometry: how far apart values are on a line (\(\lvert a-b\rvert\)).
Error and tolerance: "within" a target value (\(\lvert x-\text{target}\rvert \le \text{tolerance}\)).
Algebra and graphs: V-shaped functions and transformations.
Inequalities: turning "within" or "at least this far" statements into intervals.
Worked example: tolerance inequality
Example: A measurement \(T\) is within \(0.5\) units of \(20\). Write the inequality and solve it.
"Within \(0.5\) of \(20\)" means: \[
\lvert T-20\rvert \le 0.5.
\]
Rewrite as a compound inequality: \[
-0.5 \le T-20 \le 0.5 \quad \Rightarrow \quad 19.5 \le T \le 20.5.
\]
Try it
Try it 1: Solve \(\lvert x-10\rvert \le 0.5\) in interval notation.
Hint: "\(\le\)" becomes a closed interval between the two endpoints.
Try it 2: How many solutions does \(\lvert 3x+1\rvert = -2\) have?
Hint: Absolute value is always \(\ge 0\), so it cannot equal a negative number.
Final recap
\(\lvert x\rvert\) is distance and is always nonnegative.
Simplify carefully: \(\lvert -x\rvert=\lvert x\rvert\), but \(-\lvert x\rvert\) is negative (unless \(x=0\)).
Equation: \(\lvert A\rvert=c \Rightarrow A=c\) or \(A=-c\) (for \(c\ge 0\)).
Inequality (greater than): \(\lvert A\rvert>c \Rightarrow A>c\) or \(A<-c\).
Graph: \(y=\lvert x-h\rvert+k\) has vertex \((h,k)\).
Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the absolute value skill you need.
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