This deck teaches absolute value from a single picture - distance from zero on the number line - and then uses it to solve every standard problem type: isolate-then-split equations, the no-solution and single-solution cases, absolute value on both sides, less-than inequalities as one interval, greater-than inequalities as a union of two rays, the always-true and never-true cases, and tolerance and error-bound applications. It targets the four errors that sink students here: splitting before isolating, swapping "and" with "or", negating only the constant on the other side, and forgetting to check candidates against a variable right-hand side.
Subject: College Algebra · 135 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
College Algebra - Deck 05
One picture runs this entire topic: distance from zero.
Objectives
By the end of this deck you can:
Warm-up
Discussion prompt
Before we open Absolute Value Equations and Inequalities: without looking back, what was the main idea of Linear Equations and Inequalities, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck solves linear equations end to end. It starts with the balance-scale model, then covers clearing fractions and decimals, telling identities from contradictions, literal equations, applied linear models, and inequalities on the number line written in interval notation. It targets the forgotten inequality flip, multiplying only part of the equation by the LCD, calling a true collapsed statement "no solution", and writing an "or" union as an "and" intersection.
Section
Section 1
Picture it
Figure (svg): Number line with zero in the middle, dots at negative five and five, each labeled five units from zero
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
The bars around a number ask exactly one question: how far is this number from zero?
Concept
The bars around a number ask exactly one question: how far is this number from zero?
\[ \left| -5 \right| = 5 \qquad \left| 5 \right| = 5 \]
Figure (svg): Number line with zero in the middle, dots at negative five and five, each labeled five units from zero
Counterexample
Discussion prompt
The bars around a number ask exactly one question: how far is this number from zero?
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Picture it
Animation
Shows: The V, and what shifts it — a rendered Manim animation.
Rendered with Manim.
Takeaway: The vertex sits wherever the inside is zero.
Intuition
Walking five blocks east and five blocks west are different trips. But both are five blocks. Absolute value keeps the size of the trip and throws away the direction.
That is why an absolute value is never negative. A distance of negative three does not exist.
absolute value — The distance between a number and zero on the number line. Because it is a distance, the output is always zero or positive.
Analogy
Discussion prompt
Explain Distance forgets direction by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Walking five blocks east and five blocks west are different trips. But both are five blocks. Absolute value keeps the size of the trip and throws away the direction.
Concept
The bars behave like parentheses. Finish everything inside first, and only then measure the distance.
\[ \left| 3 - 7 \right| = \left| -4 \right| = 4 \]
They are not a switch that flips signs term by term.
Explain it
Discussion prompt
Explain The bars are a grouping symbol to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The bars behave like parentheses. Finish everything inside first, and only then measure the distance.
Ranking
Put in order
Put the moves of Worked example: evaluate an expression with bars into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The bars group what is inside them, exactly like parentheses.
Worked example
Evaluate this expression.
\[ 5 - 2\left| 3 - 7 \right| \]
Work inside the bars first
Why: The bars group what is inside them, exactly like parentheses. Inside, 3 minus 7 is negative 4.
\[ 5 - 2\left| 3 - 7 \right| = 5 - 2\left| -4 \right| \]
Measure the distance from zero
Why: Negative 4 sits 4 units away from zero, so the absolute value is 4. Notice the answer came out positive.
\[ 5 - 2\left| -4 \right| = 5 - 2(4) \]
Multiply, then subtract
Why: Order of operations puts multiplication before subtraction. Two times 4 is 8, and 5 minus 8 is negative 3.
\[ 5 - 8 = -3 \]
Verify by recomputing the whole expression in one clean pass
Why: Inside gives negative 4, the bars give 4, twice that is 8, and 5 minus 8 is negative 3. The second pass agrees with the first.
\[ 5 - 2\left| 3 - 7 \right| = -3 \]
Picture it
Animation
Shows: Each line of the worked example "evaluate an expression with bars", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Inside gives negative 4, the bars give 4, twice that is 8, and 5 minus 8 is negative 3. The second pass agrees with the first.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Treating the bars as something you can distribute across the terms inside.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Three minus 7 is negative 4, so this claims a distance of negative 4.
Finish the inside, then measure the distance.
Why: Three minus 7 is negative 4, so this claims a distance of negative 4. No such distance exists, which is the tell that the move was illegal.
Trap
Treating the bars as something you can distribute across the terms inside.
\[ \left| 3 - 7 \right| \stackrel{?}{=} \left| 3 \right| - \left| 7 \right| \]
This produces a negative distance
Why: Three minus 7 is negative 4, so this claims a distance of negative 4. No such distance exists, which is the tell that the move was illegal.
\[ \left| 3 \right| - \left| 7 \right| = 3 - 7 = -4 \]
Finish the inside, then measure the distance.
\[ \left| 3 - 7 \right| = \left| -4 \right| \]
The answer is a real distance
Why: Negative 4 lies 4 units from zero. An absolute value can never come out negative, so any method that gives a negative answer is wrong.
\[ \left| -4 \right| = 4 \]
Notation
Annotate
From Trap: splitting the bars across a subtraction — read this one piece at a time. What is each part doing?
On: \( \left| 3 - 7 \right| \stackrel{?}{=} \left| 3 \right| - \left| 7 \right| \)
Concept
Here is the distance idea written as a rule you can use in algebra.
\[ \left| x \right| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases} \]
If the inside is already positive or zero, leave it alone. If the inside is negative, take its opposite.
Intuition
That second line looks like it hands back a negative answer. It does not.
\[ x = -6 \;\Longrightarrow\; -x = -(-6) = 6 \]
The opposite of a negative number is positive. That line is the one that strips the minus sign off.
Step zero
Discussion prompt
Worked example: rewrite an absolute value without bars — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Ask when the inside is zero or positive
Answer:
Worked example
Write this expression as a piecewise rule with no bars in it.
\[ \left| x - 3 \right| \]
Ask when the inside is zero or positive
Why: The definition splits on the sign of what is inside the bars, not on the sign of x by itself. This is the step students skip.
\[ x - 3 \ge 0 \iff x \ge 3 \]
On that branch the bars simply disappear
Why: When the inside is already zero or positive, the absolute value returns it unchanged.
\[ \left| x - 3 \right| = x - 3 \quad \text{when } x \ge 3 \]
On the other branch take the opposite of the entire inside
Why: When x is less than 3 the inside is negative, so the absolute value returns its opposite. The opposite of x minus 3 is 3 minus x - note that both terms changed sign.
\[ \left| x - 3 \right| = -(x - 3) = 3 - x \quad \text{when } x < 3 \]
Verify with one number from each branch
Why: At x equal to 5 the first branch gives 2, and the original expression is the absolute value of 2, which is 2. At x equal to 1 the second branch gives 2, and the original is the absolute value of negative 2, which is 2. Both branches match the original.
\[ \left| 5 - 3 \right| = 2 = 5 - 3 \qquad \left| 1 - 3 \right| = 2 = 3 - 1 \]
Picture it
Animation
Shows: Each line of the worked example "rewrite an absolute value without bars", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 5 the first branch gives 2, and the original expression is the absolute value of 2, which is 2. At x equal to 1 the second branch gives 2, and the original is the absolute value of negative 2, which is 2. Both branches match the original.
Concept
Subtract, then take the absolute value, and you have measured the gap between two numbers.
\[ d(a, b) = \left| a - b \right| = \left| b - a \right| \]
The order of the subtraction does not matter. The two results are opposites, and opposites sit the same distance from zero.
Concept
Every single problem in this deck is a sentence about distance. Practice reading them out loud.
\[ \left| x - 4 \right| = 6 \]
Read it: the distance from x to 4 is exactly 6.
\[ \left| x - 4 \right| < 6 \]
Read it: x is closer than 6 units to 4.
\[ \left| x - 4 \right| > 6 \]
Read it: x is farther than 6 units from 4.
Worked example
\[ \left| x - 2 \right| = 5 \]
Read it as a distance sentence
Why: The expression x minus 2 measures the gap between x and 2, so the equation says that gap equals 5.
Walk 5 units to the right of 2
Why: Moving 5 to the right of 2 lands on 7, which is one number at that distance.
\[ 2 + 5 = 7 \]
Walk 5 units to the left of 2
Why: Distance works in both directions, so there is a second landing spot. Moving 5 to the left of 2 lands on negative 3.
\[ 2 - 5 = -3 \]
Figure (svg): Number line centered at 2 with dots at negative 3 and 7, each five units from 2
Verify both answers in the original equation
Why: For 7: 7 minus 2 is 5 and the absolute value of 5 is 5. For negative 3: negative 3 minus 2 is negative 5 and the absolute value of negative 5 is 5. Both sides agree in both cases.
\[ \left| 7 - 2 \right| = \left| 5 \right| = 5 \qquad \left| -3 - 2 \right| = \left| -5 \right| = 5 \]
Blank canvas
Draw it
Draw what Worked example: which numbers sit 5 units from 2? just did — the shape of it, not the line-by-line working. One picture, labels only where you need them. Then check it against the steps: anything you could not draw is a step you followed rather than understood.
Section
Section 2
Concept
If a quantity sits a fixed positive distance from zero, there are exactly two places it could be: that far right, or that far left.
\[ \left| X \right| = k \quad (k > 0) \;\Longleftrightarrow\; X = k \;\text{ or }\; X = -k \]
Here the capital letter stands for whatever is inside the bars - a single variable, a binomial, anything.
Picture it
Animation
Shows: One equation, two cases — a rendered Manim animation.
Rendered with Manim.
Takeaway: Distance five from three, in either direction.
Intuition
A missing sock is six feet from you. Do you know where it is? No - it could be six feet ahead or six feet behind. Distance alone leaves two possibilities.
So an absolute-value equation almost always has two answers. Reporting only one is the most common way to lose half the credit on this topic.
Worked example
\[ \left| x \right| = 7 \]
Read the sentence
Why: The distance from x to zero is 7. Two numbers on the line are 7 units from zero.
Write both cases
Why: The inside can equal the positive value or its opposite. Both give a distance of 7.
\[ x = 7 \qquad \text{or} \qquad x = -7 \]
Verify each answer in the original equation
Why: The absolute value of 7 is 7 and the absolute value of negative 7 is also 7. Both make the original statement true.
\[ \left| 7 \right| = 7 \;\checkmark \qquad \left| -7 \right| = 7 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the simplest split", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The absolute value of 7 is 7 and the absolute value of negative 7 is also 7. Both make the original statement true.
Worked example
\[ \left| 2x - 5 \right| = 9 \]
The absolute value is already alone on the left
Why: Nothing is multiplying it and nothing is added to it, so we may split immediately.
Case 1: set the inside equal to the positive value
Why: This is the branch where the quantity inside the bars is already 9 units to the right of zero.
\[ 2x - 5 = 9 \;\Longrightarrow\; 2x = 14 \;\Longrightarrow\; x = 7 \]
Case 2: set the inside equal to the negative value
Why: This is the branch where the quantity inside is 9 units to the left of zero, which is just as far away.
\[ 2x - 5 = -9 \;\Longrightarrow\; 2x = -4 \;\Longrightarrow\; x = -2 \]
Verify both solutions in the original equation
Why: At x equal to 7 the inside is 9 and its absolute value is 9. At x equal to negative 2 the inside is negative 9 and its absolute value is also 9. Both check out.
\[ \left| 2(7) - 5 \right| = \left| 9 \right| = 9 \;\checkmark \qquad \left| 2(-2) - 5 \right| = \left| -9 \right| = 9 \;\checkmark \]
Concept
The two-case rule only describes an absolute value standing alone on one side. If anything is multiplying it, dividing it, or added to it, undo that first.
\[ 3\left| x - 4 \right| + 2 = 14 \quad \text{is not yet ready to split} \]
Peel the outside layers off exactly the way you would to isolate a variable in a two-step equation.
Picture it
Animation
Shows: Isolate the bars before splitting — a rendered Manim animation.
Rendered with Manim.
Takeaway: Splitting too early doubles the algebra and the chance of error.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Splitting straight away, while the absolute value is still buried.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This silently pretends the 3 and the plus 2 are not there.
Strip the outside layers, then split.
Why: This silently pretends the 3 and the plus 2 are not there. The rule never said the inside equals the far side of the equation - it says the inside equals whatever the isolated absolute value equals.
Trap
Splitting straight away, while the absolute value is still buried.
\[ 3\left| x - 4 \right| + 2 = 14 \]
Set the inside equal to 14 and to negative 14
Why: This silently pretends the 3 and the plus 2 are not there. The rule never said the inside equals the far side of the equation - it says the inside equals whatever the isolated absolute value equals.
\[ x - 4 = 14 \;\text{or}\; x - 4 = -14 \;\Longrightarrow\; x = 18 \;\text{or}\; x = -10 \]
Both answers fail the check
Why: At x equal to 18 the left side is 3 times 14 plus 2, which is 44, not 14. At x equal to negative 10 it is 3 times 14 plus 2 again, which is 44. Neither works.
\[ 3\left| 18 - 4 \right| + 2 = 44 \ne 14 \]
Strip the outside layers, then split.
\[ 3\left| x - 4 \right| + 2 = 14 \]
Subtract 2, then divide by 3
Why: Now the absolute value stands alone, which is the only situation the two-case rule describes.
\[ 3\left| x - 4 \right| = 12 \;\Longrightarrow\; \left| x - 4 \right| = 4 \]
Now split and check
Why: The inside is 4 units from zero, so it equals 4 or negative 4, giving x equal to 8 or 0. At x equal to 8 the left side is 3 times 4 plus 2, which is 14. At x equal to 0 it is 3 times 4 plus 2, also 14.
\[ x = 8 \;\text{ or }\; x = 0 \]
Worked example
\[ 3\left| x - 4 \right| + 2 = 14 \]
Subtract 2 from both sides
Why: The plus 2 is the outermost layer, so it comes off first - same order you would use on a two-step linear equation.
\[ 3\left| x - 4 \right| = 12 \]
Divide both sides by 3
Why: Now the absolute value stands completely alone, which is the only form the two-case rule applies to.
\[ \left| x - 4 \right| = 4 \]
Split into the two cases
Why: The quantity inside is 4 units from zero, so it is either 4 or negative 4.
\[ x - 4 = 4 \qquad \text{or} \qquad x - 4 = -4 \]
Solve each case
Why: Add 4 to both sides in each branch. Each branch is now an ordinary one-step linear equation.
\[ x = 8 \qquad \text{or} \qquad x = 0 \]
Verify both solutions in the original equation
Why: At x equal to 8 the inside is 4, its absolute value is 4, and 3 times 4 plus 2 is 14. At x equal to 0 the inside is negative 4, its absolute value is 4, and again 3 times 4 plus 2 is 14. Both sides agree.
\[ 3\left| 8 - 4 \right| + 2 = 14 \;\checkmark \qquad 3\left| 0 - 4 \right| + 2 = 14 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "isolate, then split", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 8 the inside is 4, its absolute value is 4, and 3 times 4 plus 2 is 14. At x equal to 0 the inside is negative 4, its absolute value is 4, and again 3 times 4 plus 2 is 14. Both sides agree.
Step zero
Discussion prompt
Worked example: a negative coefficient outside — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Subtract 9 from both sides
Answer:
Worked example
\[ -2\left| x + 1 \right| + 9 = 3 \]
Subtract 9 from both sides
Why: Clear the added constant first so the absolute-value term is by itself.
\[ -2\left| x + 1 \right| = -6 \]
Divide both sides by negative 2
Why: This is an equation, not an inequality, so dividing by a negative changes nothing except the numbers. Negative 6 divided by negative 2 is positive 3.
\[ \left| x + 1 \right| = 3 \]
Pause and sanity-check the isolated value
Why: The isolated absolute value came out positive, so solutions exist. Had it come out negative there would be none.
\[ 3 > 0 \;\checkmark \]
Split and solve
Why: The inside sits 3 units from zero, so it is 3 or negative 3. Subtract 1 from both sides in each branch.
\[ x + 1 = 3 \;\Rightarrow\; x = 2 \qquad x + 1 = -3 \;\Rightarrow\; x = -4 \]
Verify both solutions in the original equation
Why: At x equal to 2 the inside is 3, so negative 2 times 3 plus 9 is 3. At x equal to negative 4 the inside is negative 3 whose absolute value is 3, so again negative 2 times 3 plus 9 is 3.
\[ -2\left| 2 + 1 \right| + 9 = 3 \;\checkmark \qquad -2\left| -4 + 1 \right| + 9 = 3 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a negative coefficient outside", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 2 the inside is 3, so negative 2 times 3 plus 9 is 3. At x equal to negative 4 the inside is negative 3 whose absolute value is 3, so again negative 2 times 3 plus 9 is 3.
Concept
Once the absolute value is alone, look at the number on the other side before you split.
\[ \left| X \right| = k \quad \text{with } k < 0 \;\Longrightarrow\; \text{no solution} \]
No distance is negative, so nothing can satisfy it. The solution set is empty.
\[ \text{Solution set: } \varnothing \]
Worked example
\[ \left| x - 4 \right| + 7 = 3 \]
Isolate the absolute value
Why: Subtract 7 from both sides. Do this before judging anything about solutions.
\[ \left| x - 4 \right| = -4 \]
Stop - do not split
Why: The equation now asks for a quantity whose distance from zero is negative 4. Distances are never negative, so no real number can do this.
Report the empty set
Why: Writing no solution is the complete answer here. Splitting anyway would manufacture two fake answers.
\[ \varnothing \]
Verify the reasoning by bounding the left side
Why: The absolute value part is at least 0 for every x, so the whole left side is at least 7. Since 7 is already bigger than 3, the left side can never equal 3. The smallest it ever gets, at x equal to 4, is exactly 7.
\[ \left| x - 4 \right| + 7 \ge 7 > 3 \quad \text{for every real } x \]
Concept
There is one more special value on the right: zero.
\[ \left| X \right| = 0 \;\Longleftrightarrow\; X = 0 \]
Only one number sits zero units from zero. The two cases collapse into the same equation, so you get a single solution.
Picture it
Animation
Shows: When there is no solution — a rendered Manim animation.
Rendered with Manim.
Takeaway: Distance is never negative, so check the right-hand side first.
Worked example
\[ \left| 3x - 6 \right| = 0 \]
Notice the right side is zero, not positive
Why: Zero is the one right-hand value that does not produce two branches, because zero and its opposite are the same number.
Set the inside equal to zero
Why: The only quantity whose distance from zero is zero is zero itself.
\[ 3x - 6 = 0 \]
Solve the linear equation
Why: Add 6 to both sides and divide by 3.
\[ x = 2 \]
Verify the solution in the original equation
Why: At x equal to 2 the inside is 3 times 2 minus 6, which is 0, and the absolute value of 0 is 0. That matches the right side exactly.
\[ \left| 3(2) - 6 \right| = \left| 0 \right| = 0 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the single-solution case", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 2 the inside is 3 times 2 minus 6, which is 0, and the absolute value of 0 is 0. That matches the right side exactly.
Concept
If the far side contains a variable, you still split into two cases - but now the far side might be negative for some of your candidates.
\[ \left| X \right| = E \;\Longrightarrow\; X = E \;\text{ or }\; X = -E \]
Because of that, every candidate must be substituted back into the original equation. Any candidate that makes the far side negative is thrown out.
Fill the middle
Fill in the blanks
From Trap: negating only the constant — finish the line. Write what belongs on the right of the equals sign before you look.
\left| 2x - 1 \right| = x + 4
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Only the 4 got a minus sign. The x was left alone, so this is not the opposite of the right side at all.
Trap
Writing the negative case by flipping only the number at the end.
\[ \left| 2x - 1 \right| = x + 4 \]
Negative case written as the inside equals x minus 4
Why: Only the 4 got a minus sign. The x was left alone, so this is not the opposite of the right side at all.
\[ 2x - 1 = x - 4 \;\Longrightarrow\; x = -3 \]
The candidate fails the check
Why: At x equal to negative 3 the left side is the absolute value of negative 7, which is 7, while the right side is 1. Seven does not equal one.
\[ \left| 2(-3) - 1 \right| = 7 \ne -3 + 4 = 1 \]
Negate the entire other side, parentheses and all.
\[ \left| 2x - 1 \right| = x + 4 \]
Negative case written as the inside equals the opposite of the whole right side
Why: Wrap the right side in parentheses before attaching the minus, then distribute. Every term changes sign.
\[ 2x - 1 = -(x + 4) = -x - 4 \;\Longrightarrow\; 3x = -3 \;\Longrightarrow\; x = -1 \]
This candidate passes the check
Why: At x equal to negative 1 the left side is the absolute value of negative 3, which is 3, and the right side is 3. Both sides agree.
\[ \left| 2(-1) - 1 \right| = 3 = -1 + 4 \;\checkmark \]
Fill the middle
Fill in the blanks
From Worked example: variable on both sides — finish the line. Write what belongs on the right of the equals sign before you look.
\left| 2x - 1 \right| = x + 4
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Subtract x from both sides and add 1 to both sides.
Worked example
\[ \left| 2x - 1 \right| = x + 4 \]
Case 1: the inside equals the right side
Why: Subtract x from both sides and add 1 to both sides.
\[ 2x - 1 = x + 4 \;\Longrightarrow\; x = 5 \]
Case 2: the inside equals the opposite of the entire right side
Why: Put the whole right side in parentheses first, then distribute the minus sign to both terms.
\[ 2x - 1 = -(x + 4) \;\Longrightarrow\; 2x - 1 = -x - 4 \]
Finish case 2
Why: Add x to both sides and add 1 to both sides, then divide by 3.
\[ 3x = -3 \;\Longrightarrow\; x = -1 \]
Verify both candidates in the original equation
Why: At x equal to 5 the left side is the absolute value of 9, which is 9, and the right side is 9. At x equal to negative 1 the left side is the absolute value of negative 3, which is 3, and the right side is 3. Both survive.
\[ \left| 2(5) - 1 \right| = 9 = 5 + 4 \;\checkmark \qquad \left| 2(-1) - 1 \right| = 3 = -1 + 4 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "variable on both sides", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 5 the left side is the absolute value of 9, which is 9, and the right side is 9. At x equal to negative 1 the left side is the absolute value of negative 3, which is 3, and the right side is 3. Both survive.
Ranking
Put in order
Put the moves of Worked example: a candidate that must be thrown out into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Subtract x from both sides, then subtract 10, then divide by 2.
Worked example
\[ \left| x - 2 \right| = 3x + 10 \]
Case 1: the inside equals the right side
Why: Subtract x from both sides, then subtract 10, then divide by 2.
\[ x - 2 = 3x + 10 \;\Longrightarrow\; -12 = 2x \;\Longrightarrow\; x = -6 \]
Case 2: the inside equals the opposite of the whole right side
Why: Negate both terms of the right side, then collect the variable terms.
\[ x - 2 = -3x - 10 \;\Longrightarrow\; 4x = -8 \;\Longrightarrow\; x = -2 \]
Test the first candidate
Why: At x equal to negative 6 the left side is the absolute value of negative 8, which is positive 8, but the right side is negative 8. A positive can never equal a negative, so this candidate is rejected.
\[ \left| -6 - 2 \right| = 8 \ne 3(-6) + 10 = -8 \]
Verify the surviving candidate in the original equation
Why: At x equal to negative 2 the left side is the absolute value of negative 4, which is 4, and the right side is negative 6 plus 10, which is 4. Both sides agree, so the only solution is x equal to negative 2.
\[ \left| -2 - 2 \right| = 4 = 3(-2) + 10 \;\checkmark \qquad x = -2 \]
Picture it
Animation
Shows: Each line of the worked example "a candidate that must be thrown out", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to negative 2 the left side is the absolute value of negative 4, which is 4, and the right side is negative 6 plus 10, which is 4. Both sides agree, so the only solution is x equal to negative 2.
Section
Section 3
Concept
Sometimes both sides wear bars. The picture still runs the show: two quantities sit the same distance from zero only when they are equal or opposites.
\[ \left| A \right| = \left| B \right| \;\Longleftrightarrow\; A = B \;\text{ or }\; A = -B \]
You still get exactly two branches. Only the second one changes: you negate the whole other inside, parentheses and all.
Picture it
Figure (svg): Number line with dots at negative seven and seven, both labeled seven units from zero
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Two hikers are the same number of miles from camp. Either they are standing together, or one went up the trail and the other went down it by the same amount.
Intuition
Two hikers are the same number of miles from camp. Either they are standing together, or one went up the trail and the other went down it by the same amount.
Figure (svg): Number line with dots at negative seven and seven, both labeled seven units from zero
There is no third possibility. That is why this rule has exactly two branches and never three.
Hypothesis
Predict first
Worked example: bars on both sides is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Case 1: drop both sets of bars as written
Why: This is the branch where the two insides are the very same number. Subtract x from both sides, then subtract 1 from both sides.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
\[ \left| x - 3 \right| = \left| 2x + 1 \right| \]
Case 1: drop both sets of bars as written
Why: This is the branch where the two insides are the very same number. Subtract x from both sides, then subtract 1 from both sides.
\[ x - 3 = 2x + 1 \;\Longrightarrow\; -4 = x \]
Case 2: negate the entire second inside
Why: This is the branch where the insides are opposites. Wrap the second inside in parentheses before attaching the minus sign, then distribute to both terms.
\[ x - 3 = -(2x + 1) = -2x - 1 \]
Finish case 2
Why: Add 2x to both sides and add 3 to both sides, which gives 3x equal to 2, then divide by 3.
\[ 3x = 2 \;\Longrightarrow\; x = \frac{2}{3} \]
Verify both solutions in the original equation
Why: At x equal to negative 4 the left side is the absolute value of negative 7, which is 7, and the right side is the absolute value of negative 7, also 7. At x equal to two thirds the left side is the absolute value of negative seven thirds and the right side is the absolute value of seven thirds, both seven thirds.
\[ \left| -4 - 3 \right| = 7 = \left| 2(-4) + 1 \right| \;\checkmark \qquad \left| \tfrac{2}{3} - 3 \right| = \tfrac{7}{3} = \left| 2\left(\tfrac{2}{3}\right) + 1 \right| \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "bars on both sides", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to negative 4 the left side is the absolute value of negative 7, which is 7, and the right side is the absolute value of negative 7, also 7. At x equal to two thirds the left side is the absolute value of negative seven thirds and the right side is the absolute value of seven thirds, both seven thirds.
Worked example
\[ \left| 2x + 5 \right| = \left| 2x - 3 \right| \]
Case 1: set the two insides equal
Why: Start the same way as always, even though the two insides start with the identical variable term.
\[ 2x + 5 = 2x - 3 \;\Longrightarrow\; 5 = -3 \]
Case 1 collapses to a false statement, so it gives nothing
Why: Subtracting 2x wipes out the variable and leaves a claim that is never true. That branch simply has no solutions - it is information, not a mistake.
Case 2: negate the whole second inside
Why: Parentheses first, then distribute the minus sign so that both the 2x and the negative 3 change sign.
\[ 2x + 5 = -(2x - 3) = -2x + 3 \]
Solve case 2
Why: Add 2x to both sides and subtract 5 from both sides to get 4x equal to negative 2, then divide by 4 and reduce.
\[ 4x = -2 \;\Longrightarrow\; x = -\frac{1}{2} \]
Verify the single solution in the original equation
Why: At x equal to negative one half the left inside is negative 1 plus 5, which is 4, and the right inside is negative 1 minus 3, which is negative 4. Their absolute values are both 4, so the sides agree and this equation has exactly one solution.
\[ \left| 2\left(-\tfrac{1}{2}\right) + 5 \right| = \left| 4 \right| = 4 = \left| -4 \right| = \left| 2\left(-\tfrac{1}{2}\right) - 3 \right| \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "when one case collapses", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to negative one half the left inside is negative 1 plus 5, which is 4, and the right inside is negative 1 minus 3, which is negative 4. Their absolute values are both 4, so the sides agree and this equation has exactly one solution.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Solving both cases and reporting both answers without testing either one.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Case 1 gives x equal to 6 and case 2 gives x equal to four thirds.
Solve both cases, then test every candidate in the original equation.
Why: Case 1 gives x equal to 6 and case 2 gives x equal to four thirds. Neither candidate was ever substituted back into the original equation.
Trap
Solving both cases and reporting both answers without testing either one.
\[ \left| x + 1 \right| = 2x - 5 \]
Both cases are solved and both are kept
Why: Case 1 gives x equal to 6 and case 2 gives x equal to four thirds. Neither candidate was ever substituted back into the original equation.
\[ x = 6 \quad \text{and} \quad x = \frac{4}{3} \]
The second answer forces the right side negative
Why: At x equal to four thirds the right side is eight thirds minus 5, which is negative seven thirds. The left side is a distance, so it can never equal a negative number - the two sides do not match.
\[ \left| \tfrac{4}{3} + 1 \right| = \tfrac{7}{3} \ne 2\left(\tfrac{4}{3}\right) - 5 = -\tfrac{7}{3} \]
Solve both cases, then test every candidate in the original equation.
\[ \left| x + 1 \right| = 2x - 5 \]
Test the first candidate
Why: At x equal to 6 the left side is the absolute value of 7, which is 7, and the right side is 12 minus 5, which is 7. Both sides agree, so this one survives.
\[ \left| 6 + 1 \right| = 7 = 2(6) - 5 \;\checkmark \]
Reject the extraneous candidate and report only what survived
Why: Four thirds makes the right side negative, so it cannot be a solution no matter how cleanly the algebra produced it. The complete answer is the single value 6.
\[ x = 6 \]
Translation
\( \left| x + 1 \right| = 2x - 5 \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Ranking
Put in order
These are the steps of Pattern: solving any absolute-value equation, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Every absolute-value equation in this deck yields to the same five moves, in this order.
1. Isolate the absolute value
Why: Undo every multiplication, division, addition, and subtraction sitting outside the bars. The two-case rule only describes an absolute value standing alone.
2. Read the other side before you split
Why: A negative there means no solution. A zero there means exactly one solution. A positive number or a variable expression means two cases.
\[ \left| X \right| = k: \quad k < 0 \Rightarrow \varnothing, \quad k = 0 \Rightarrow \text{one}, \quad k > 0 \Rightarrow \text{two} \]
3. Split into the two cases
Why: The inside equals the other side, or the inside equals the opposite of the entire other side. Parentheses around the other side before the minus sign, always.
\[ X = E \qquad \text{or} \qquad X = -E \]
4. Solve each case as an ordinary linear equation
Why: Nothing special happens here. Each branch is the kind of equation you already solve.
5. Check every candidate in the original equation
Why: Required whenever a variable appears on the other side, and never a waste of time otherwise. A candidate that makes the other side negative is thrown out.
Edge cases
Discussion prompt
Pattern: solving any absolute-value equation works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Every absolute-value equation in this deck yields to the same five moves, in this order.
Elimination
Eliminate the wrong options
What is the solution set of this equation?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Add 5 to both sides to get twice the absolute value equal to 12, then divide by 2 so the absolute value equals 6. The inside is then 6 or negative 6, giving x equal to 3 or x equal to negative 9. Verify: twice the absolute value of 6 minus 5 is 7, and twice the absolute value of negative 6 minus 5 is also 7.
Check
Peel the outside off first, then split. Work it on paper before you choose.
\[ 2\left| x + 3 \right| - 5 = 7 \]
Check your understanding
What is the solution set of this equation?
Answer: A
Why: Add 5 to both sides to get twice the absolute value equal to 12, then divide by 2 so the absolute value equals 6. The inside is then 6 or negative 6, giving x equal to 3 or x equal to negative 9. Verify: twice the absolute value of 6 minus 5 is 7, and twice the absolute value of negative 6 minus 5 is also 7.
Check
Isolate first, then look hard at the number that is left on the right.
\[ \left| 4x - 3 \right| + 8 = 5 \]
Check your understanding
What is the solution set of this equation?
Answer: A
Why: Subtracting 8 from both sides leaves the absolute value equal to negative 3. An absolute value is a distance from zero, so it is never negative, and no real number can satisfy it. The left side is in fact always at least 8, which is already larger than 5.
Section
Section 4
Picture it
Figure (svg): Number line with the segment between negative three and three shaded and hollow circles at both ends
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Swap the equals sign for a less-than sign and the sentence changes from a location to a neighborhood.
Concept
Swap the equals sign for a less-than sign and the sentence changes from a location to a neighborhood.
\[ \left| x \right| < 3 \]
Read it: x is closer than 3 units to zero.
Figure (svg): Number line with the segment between negative three and three shaded and hollow circles at both ends
Picture it
Animation
Shows: Less than means AND — a rendered Manim animation.
Rendered with Manim.
Takeaway: Close to zero, so the answer is a band around it.
Intuition
A dog is tied to a post with a three-foot leash. Where can it be? Anywhere within three feet of the post - one connected patch of grass around it.
It has to be near the post and on the correct side of both edges at once. That double requirement is why a less-than problem becomes an and, and why the answer is a single interval.
compound and — Two conditions that must both hold at the same time. On a number line the solution is the overlap of the two pieces, which for an absolute value is one interval.
Definition probe
Sort into buckets
Every line below is part of the definition of absolute value or of compound and — one or the other, never both. Put each where it belongs.
Concept
Write the whole inside between the negative and the positive version of the number. No cases, no splitting - one three-part inequality.
\[ \left| X \right| < k \quad (k > 0) \;\Longleftrightarrow\; -k < X < k \]
The same rule holds with less-than-or-equal-to on both sides; the endpoints just get included.
\[ \left| X \right| \le k \;\Longleftrightarrow\; -k \le X \le k \]
Step zero
Discussion prompt
Worked example: the plainest less-than — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Read it as a distance sentence
Answer:
Worked example
\[ \left| x \right| \le 3 \]
Read it as a distance sentence
Why: The distance from x to zero is at most 3, so x cannot get farther than 3 units away in either direction.
Sandwich the inside between negative 3 and 3
Why: Both edges apply at once: x must be at least negative 3 and at most 3.
\[ -3 \le x \le 3 \]
Write the answer in interval notation
Why: Square brackets because the endpoints satisfy the or-equal-to part of the symbol.
\[ [-3, 3] \]
Verify with a point inside, an endpoint, and a point outside
Why: At x equal to 0 the left side is 0, which is at most 3, so it belongs. At x equal to 3 the left side is exactly 3, which the or-equal-to part allows. At x equal to 4 the left side is 4, which is too big, so 4 is correctly left out.
\[ \left| 0 \right| = 0 \le 3 \;\checkmark \qquad \left| 3 \right| = 3 \le 3 \;\checkmark \qquad \left| 4 \right| = 4 \not\le 3 \]
Worked example
\[ \left| 2x - 1 \right| < 7 \]
Sandwich the entire inside
Why: The quantity 2x minus 1 is what must stay within 7 units of zero, so that whole expression goes in the middle.
\[ -7 < 2x - 1 < 7 \]
Add 1 to all three parts
Why: Whatever you do to the middle you must do to both outer parts, so the chain stays true.
\[ -6 < 2x < 8 \]
Divide all three parts by 2
Why: Two is positive, so the direction of both inequality symbols is unchanged.
\[ -3 < x < 4 \]
Write the interval
Why: Round parentheses on both ends because the original symbol was strict, so neither endpoint belongs.
\[ (-3, 4) \]
Verify with an inside point and both endpoints in the original inequality
Why: At x equal to 0 the left side is the absolute value of negative 1, which is 1, and 1 is less than 7. At x equal to 4 the left side is exactly 7, which is not less than 7, so the open endpoint is right. At x equal to negative 3 the left side is the absolute value of negative 7, again exactly 7, so that end is open too.
\[ \left| 2(0) - 1 \right| = 1 < 7 \;\checkmark \qquad \left| 2(4) - 1 \right| = 7 \not< 7 \qquad \left| 2(-3) - 1 \right| = 7 \not< 7 \]
Worked example
\[ \left| x - 5 \right| + 2 \le 6 \]
Subtract 2 from both sides
Why: The isolate-first rule is the same for inequalities as for equations. Nothing may sit outside the bars when you apply the sandwich.
\[ \left| x - 5 \right| \le 4 \]
Sandwich the inside between negative 4 and 4
Why: The isolated value is positive, so the rule applies and gives one connected interval.
\[ -4 \le x - 5 \le 4 \]
Add 5 to all three parts
Why: This isolates x in the middle while keeping the chain balanced.
\[ 1 \le x \le 9 \]
Figure (svg): Number line with the segment from one to nine shaded and solid dots at both ends
Verify the center, an endpoint, and a point just outside
Why: At x equal to 5 the left side is 0 plus 2, which is 2 and at most 6. At x equal to 9 it is 4 plus 2, exactly 6, so the closed endpoint is correct. At x equal to 10 it is 5 plus 2, which is 7 and too large, so 10 is properly excluded. The answer is the interval from 1 to 9 with both endpoints included.
\[ [1, 9] \]
Blank canvas
Draw it
Draw what Worked example: isolate, then sandwich just did — the shape of it, not the line-by-line working. One picture, labels only where you need them. Then check it against the steps: anything you could not draw is a step you followed rather than understood.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Splitting a less-than into two separate branches joined by the word or.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This copies the equation habit of writing two independent cases.
A less-than is an and: sandwich the inside between the two numbers.
Why: This copies the equation habit of writing two independent cases. For a less-than the two conditions must hold together, not separately.
Trap
Splitting a less-than into two separate branches joined by the word or.
\[ \left| x - 1 \right| < 3 \]
Two branches joined by or
Why: This copies the equation habit of writing two independent cases. For a less-than the two conditions must hold together, not separately.
\[ x - 1 < 3 \;\text{ or }\; x - 1 > -3 \;\Longrightarrow\; x < 4 \;\text{ or }\; x > -2 \]
The union swallows the whole number line
Why: Every real number satisfies at least one of those two, so this answer claims all real numbers. Test 10: the left side is the absolute value of 9, which is 9, and 9 is not less than 3.
\[ \left| 10 - 1 \right| = 9 \not< 3 \]
A less-than is an and: sandwich the inside between the two numbers.
\[ \left| x - 1 \right| < 3 \]
One three-part inequality, then add 1 everywhere
Why: Both edges apply at once. Adding 1 to all three parts isolates x in the middle.
\[ -3 < x - 1 < 3 \;\Longrightarrow\; -2 < x < 4 \]
The answer is one bounded interval
Why: Test 10 again: it is not between negative 2 and 4, so it is correctly excluded now. Test 0: the left side is the absolute value of negative 1, which is 1 and less than 3, and 0 does lie in the interval.
\[ (-2, 4) \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Both edges apply at once. Adding 1 to all three parts isolates x in the middle.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
This copies the equation habit of writing two independent cases. For a less-than the two conditions must hold together, not separately.
Check
Sandwich it, solve the three-part chain, then read off the interval.
\[ \left| x + 2 \right| \le 5 \]
Check your understanding
Which interval is the solution set?
Answer: A
Why: Sandwiching gives negative 5 at most x plus 2 at most 5. Subtracting 2 from all three parts gives negative 7 at most x at most 3, which is the interval from negative 7 to 3 with both endpoints included. Verify at the endpoint 3: the absolute value of 5 is 5, which satisfies the or-equal-to part.
Section
Section 5
Concept
Turn the symbol around and the neighborhood turns inside out. Now you want the numbers that are far from zero, and those live on two separate sides.
\[ \left| X \right| > k \quad (k > 0) \;\Longleftrightarrow\; X > k \;\text{ or }\; X < -k \]
There is no way to be far to the right and far to the left at the same time, so this one is genuinely an or.
Picture it
Animation
Shows: Greater than means OR — a rendered Manim animation.
Rendered with Manim.
Takeaway: Far from zero, so the answer is everything except a band.
Picture it
Figure (svg): Number line with shaded rays running left from negative four and right from four, hollow circles at both, and the middle unshaded
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Picture a fenced yard around the post. Less-than keeps the dog inside the fence. Greater-than puts it outside, and outside has two disconnected pieces: past the left fence or past the right fence.
Intuition
Picture a fenced yard around the post. Less-than keeps the dog inside the fence. Greater-than puts it outside, and outside has two disconnected pieces: past the left fence or past the right fence.
Figure (svg): Number line with shaded rays running left from negative four and right from four, hollow circles at both, and the middle unshaded
Two shaded pieces on the picture means two intervals in the answer, written with a union symbol between them.
Worked example
\[ \left| x \right| > 4 \]
Read it as a distance sentence
Why: The distance from x to zero is more than 4 units. That can happen by going far right or by going far left.
Write the two rays with the word or
Why: The right-hand branch keeps the number as it is; the left-hand branch flips the symbol and the sign together.
\[ x > 4 \qquad \text{or} \qquad x < -4 \]
Write the answer as a union of two intervals
Why: Two disconnected pieces cannot be written as a single interval, so the union symbol does the joining.
\[ (-\infty, -4) \cup (4, \infty) \]
Verify one point from each ray and one from the gap
Why: At x equal to 5 the left side is 5, which is more than 4. At x equal to negative 5 the left side is 5 as well, so that ray belongs too. At x equal to 0 the left side is 0, which is not more than 4, so the middle is correctly left out.
\[ \left| 5 \right| = 5 > 4 \;\checkmark \qquad \left| -5 \right| = 5 > 4 \;\checkmark \qquad \left| 0 \right| = 0 \not> 4 \]
Picture it
Animation
Shows: Each line of the worked example "the plainest greater-than", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 5 the left side is 5, which is more than 4. At x equal to negative 5 the left side is 5 as well, so that ray belongs too. At x equal to 0 the left side is 0, which is not more than 4, so the middle is correctly left out.
Worked example
\[ \left| 3x + 2 \right| \ge 8 \]
Branch 1: the inside is at least 8
Why: This is the far-to-the-right branch. Subtract 2 from both sides, then divide by 3.
\[ 3x + 2 \ge 8 \;\Longrightarrow\; 3x \ge 6 \;\Longrightarrow\; x \ge 2 \]
Branch 2: the inside is at most negative 8
Why: This is the far-to-the-left branch. Both the sign of the 8 and the direction of the symbol change, because being far to the left means being a very negative number.
\[ 3x + 2 \le -8 \;\Longrightarrow\; 3x \le -10 \;\Longrightarrow\; x \le -\frac{10}{3} \]
Join the two branches with a union
Why: Square brackets on the finite ends because the symbol includes equality; always round on an infinite end, since infinity is a direction and never a number you can reach.
\[ \left(-\infty, -\tfrac{10}{3}\right] \cup [2, \infty) \]
Verify both endpoints and a point in the gap
Why: At x equal to 2 the inside is 8, so the left side is exactly 8 and the or-equal-to part is satisfied. At x equal to negative ten thirds the inside is negative 10 plus 2, which is negative 8, and its absolute value is again exactly 8. At x equal to 0 the left side is 2, which is not at least 8, so the gap between the endpoints is correctly excluded.
\[ \left| 3(2) + 2 \right| = 8 \;\checkmark \qquad \left| 3\left(-\tfrac{10}{3}\right) + 2 \right| = \left| -8 \right| = 8 \;\checkmark \qquad \left| 3(0) + 2 \right| = 2 \not\ge 8 \]
Picture it
Animation
Shows: Each line of the worked example "a binomial inside a greater-than", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 2 the inside is 8, so the left side is exactly 8 and the or-equal-to part is satisfied. At x equal to negative ten thirds the inside is negative 10 plus 2, which is negative 8, and its absolute value is again exactly 8. At x equal to 0 the left side is 2, which is not at least 8, so the gap between the endpoints is correctly excluded.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Negating the number but leaving the inequality symbol pointing the same way.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Only the 8 got a minus sign. The symbol still says at least, which describes numbers to the right of negative 8 rather than to the left of it.
Negate the number and reverse the symbol, because far to the left means very negative.
Why: Only the 8 got a minus sign. The symbol still says at least, which describes numbers to the right of negative 8 rather than to the left of it.
Trap
Negating the number but leaving the inequality symbol pointing the same way.
\[ \left| 3x + 2 \right| \ge 8 \]
Second branch written with the symbol unchanged
Why: Only the 8 got a minus sign. The symbol still says at least, which describes numbers to the right of negative 8 rather than to the left of it.
\[ 3x + 2 \ge -8 \;\Longrightarrow\; x \ge -\frac{10}{3} \]
The answer now includes numbers that fail
Why: Combining with the first branch leaves everything from negative ten thirds up. But at x equal to 0 the left side is 2, which is nowhere near at least 8, so the answer is wrong.
\[ \left| 3(0) + 2 \right| = 2 \not\ge 8 \]
Negate the number and reverse the symbol, because far to the left means very negative.
\[ \left| 3x + 2 \right| \ge 8 \]
Second branch written with the symbol reversed
Why: The inside must be at most negative 8 to be at least 8 units away on the left side of zero.
\[ 3x + 2 \le -8 \;\Longrightarrow\; x \le -\frac{10}{3} \]
The union now excludes the middle
Why: Zero no longer appears in the answer, and the endpoint checks give exactly 8 on both ends. The two rays point away from each other, which is what a greater-than always looks like.
\[ \left(-\infty, -\tfrac{10}{3}\right] \cup [2, \infty) \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Prediction
Predict first
Which interval notation gives the solution set?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: (-infinity, 1) union (5, infinity)
Why: The first branch gives 2x minus 6 greater than 4, so x is greater than 5. The second branch gives 2x minus 6 less than negative 4, so x is less than 1. Verify at x equal to 6: the absolute value of 6 is 6, which is more than 4. At x equal to 3 the left side is 0, which fails, so the middle is excluded.
Check
Two branches, joined with a union. Watch the direction of the second symbol.
\[ \left| 2x - 6 \right| > 4 \]
Check your understanding
Which interval notation gives the solution set?
Answer: A
Why: The first branch gives 2x minus 6 greater than 4, so x is greater than 5. The second branch gives 2x minus 6 less than negative 4, so x is less than 1. Verify at x equal to 6: the absolute value of 6 is 6, which is more than 4. At x equal to 3 the left side is 0, which fails, so the middle is excluded.
Section
Section 6
Concept
An isolated absolute value is never negative. Put a negative number on the other side of an inequality and the answer is decided before you do any algebra at all.
\[ \left| X \right| < \text{negative} \;\Longrightarrow\; \varnothing \]
\[ \left| X \right| > \text{negative} \;\Longrightarrow\; (-\infty, \infty) \]
One is never true and the other is always true. Splitting or sandwiching either of them just wastes time and usually invents a wrong answer.
Explain it
Discussion prompt
Explain When the other side is negative to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
An isolated absolute value is never negative. Put a negative number on the other side of an inequality and the answer is decided before you do any algebra at all.
Intuition
Asking for a distance smaller than a negative number is asking for something shorter than impossible. Nothing qualifies.
Asking for a distance larger than a negative number is free: every distance, including a distance of zero, already beats it. Everything qualifies.
The tell is always the same - check the sign of the isolated right side before you pick a method.
Analogy
Discussion prompt
Explain Nothing is closer than nothing by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Asking for a distance smaller than a negative number is asking for something shorter than impossible. Nothing qualifies.
Worked example
\[ \left| x + 6 \right| > -2 \]
The absolute value is already isolated
Why: Nothing sits outside the bars, so we may read the right side immediately.
Compare the smallest possible left side to the right side
Why: An absolute value is at least zero for every input, and zero is already bigger than negative 2. Every real number therefore satisfies the inequality.
\[ \left| x + 6 \right| \ge 0 > -2 \]
Report all real numbers
Why: The solution set is the entire number line, written as one infinite interval.
\[ (-\infty, \infty) \]
Verify at the worst possible input
Why: The left side is smallest at x equal to negative 6, where the inside is zero and the left side is exactly 0. Even there, 0 is greater than negative 2, so no input can fail. A second test at x equal to 4 gives 10, which also passes.
\[ \left| -6 + 6 \right| = 0 > -2 \;\checkmark \qquad \left| 4 + 6 \right| = 10 > -2 \;\checkmark \]
Picture it
Animation
Shows: And when everything works — a rendered Manim animation.
Rendered with Manim.
Takeaway: Any distance beats a negative one, so the inequality is free.
Fill the middle
Fill in the blanks
From Worked example: an inequality with no solution — finish the line. Write what belongs on the right of the equals sign before you look.
\left| 4\left(\tfrac5 \not< 3___\right) - 7 \right| + 5 = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Subtract 5 from both sides. Judging the problem before isolating is how students talk themselves into the wrong special case.
Worked example
\[ \left| 4x - 7 \right| + 5 < 3 \]
Isolate the absolute value first
Why: Subtract 5 from both sides. Judging the problem before isolating is how students talk themselves into the wrong special case.
\[ \left| 4x - 7 \right| < -2 \]
Stop: a distance cannot be less than a negative number
Why: The smallest an absolute value ever gets is zero, and zero is not less than negative 2. No input can make this true.
Report the empty set
Why: There is nothing to graph and nothing to split. Writing no solution is the complete answer.
\[ \varnothing \]
Verify by bounding the original left side
Why: The absolute value part is at least 0, so the whole left side is at least 5, and 5 is already bigger than 3. The minimum happens at x equal to seven fourths, where the left side is exactly 5 - still too big.
\[ \left| 4\left(\tfrac{7}{4}\right) - 7 \right| + 5 = 5 \not< 3 \]
Picture it
Animation
Shows: Each line of the worked example "an inequality with no solution", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The absolute value part is at least 0, so the whole left side is at least 5, and 5 is already bigger than 3. The minimum happens at x equal to seven fourths, where the left side is exactly 5 - still too big.
Concept
Factories, labs, and scales all speak the same sentence: a measurement must land within some amount of a target. That is a distance, so it is an absolute value.
\[ \left| \text{measured} - \text{target} \right| \le \text{tolerance} \]
tolerance — The largest error allowed between a measured value and its target. The measurement minus the target is the error, and its absolute value must stay under the tolerance.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of absolute value, compound and, tolerance as Absolute Value Equations and Inequalities uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
Say a part should be 12 millimeters and the shop allows three hundredths of a millimeter of slop. The target is the center of the interval and the tolerance is the radius.
So the acceptable parts form one interval centered on the target - the less-than picture from Part 4, wearing a hard hat.
The rejected parts form the two rays outside it. Which one you want depends on whether the question asks what passes or what gets thrown away.
Counterexample
Discussion prompt
Say a part should be 12 millimeters and the shop allows three hundredths of a millimeter of slop. The target is the center of the interval and the tolerance is the radius.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The rejected parts form the two rays outside it. Which one you want depends on whether the question asks what passes or what gets thrown away.
Step zero
Discussion prompt
Worked example: a machine-shop tolerance — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the variable and write the sentence as an inequality
Answer:
Worked example
A bolt is specified at 12 millimeters in diameter, with a tolerance of three hundredths of a millimeter. Which diameters pass inspection?
Name the variable and write the sentence as an inequality
Why: Let d be the measured diameter in millimeters. The error is d minus 12, and its distance from zero must stay within the tolerance.
\[ \left| d - 12 \right| \le 0.03 \]
Sandwich the inside
Why: This is a less-than form, so the error sits between the negative and positive tolerance and the answer will be one interval.
\[ -0.03 \le d - 12 \le 0.03 \]
Add 12 to all three parts
Why: This converts a statement about the error into a statement about the actual diameter, which is what the inspector measures.
\[ 11.97 \le d \le 12.03 \]
State the answer with units
Why: Any diameter from 11.97 to 12.03 millimeters passes, endpoints included, because the tolerance was stated with an at-most.
\[ [11.97,\; 12.03] \text{ millimeters} \]
Verify an accepted edge case and a rejected one
Why: A bolt at 12.03 millimeters has an error of three hundredths, which is exactly the tolerance and therefore allowed. A bolt at 12.04 has an error of four hundredths, which exceeds the tolerance and is rejected - and 12.04 does lie outside the reported interval.
\[ \left| 12.03 - 12 \right| = 0.03 \le 0.03 \;\checkmark \qquad \left| 12.04 - 12 \right| = 0.04 \not\le 0.03 \]
Picture it
Animation
Shows: Each line of the worked example "a machine-shop tolerance", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A bolt at 12.03 millimeters has an error of three hundredths, which is exactly the tolerance and therefore allowed. A bolt at 12.04 has an error of four hundredths, which exceeds the tolerance and is rejected - and 12.04 does lie outside the reported interval.
Ranking
Put in order
Put the moves of Worked example: turning an interval back into an absolute value into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The center is the average of the two endpoints, and it plays the role of the target in a tolerance statement.
Worked example
A thermostat holds a room between 3 and 11 degrees above freezing. Write that as a single absolute-value inequality.
Find the center of the interval
Why: The center is the average of the two endpoints, and it plays the role of the target in a tolerance statement.
\[ \text{center} = \frac{3 + 11}{2} = 7 \]
Find the radius of the interval
Why: The radius is half the width, which is how far each endpoint sits from the center. It plays the role of the tolerance.
\[ \text{radius} = \frac{11 - 3}{2} = 4 \]
Write the distance sentence
Why: The temperature stays within 4 units of 7, and within-a-distance is exactly what an absolute value with a less-than says.
\[ \left| T - 7 \right| \le 4 \]
Verify by solving it back into an interval
Why: Sandwiching gives negative 4 at most T minus 7 at most 4, and adding 7 to all three parts returns 3 at most T at most 11 - the interval we started from. A spot check at 12 gives a distance of 5, correctly rejected.
\[ -4 \le T - 7 \le 4 \;\Longrightarrow\; 3 \le T \le 11 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "turning an interval back into an absolute value", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Sandwiching gives negative 4 at most T minus 7 at most 4, and adding 7 to all three parts returns 3 at most T at most 11 - the interval we started from. A spot check at 12 gives a distance of 5, correctly rejected.
Ranking
Put in order
These are the steps of Pattern: solving any absolute-value inequality, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
The inequality recipe starts the same way the equation recipe did, and then the symbol picks the shape of the answer.
1. Isolate the absolute value
Why: Undo everything outside the bars first. No rule below applies until the absolute value stands alone.
2. Check the sign of the other side
Why: Negative with a less-than means no solution. Negative with a greater-than means all real numbers. Otherwise continue.
\[ \left| X \right| < \text{neg} \Rightarrow \varnothing \qquad \left| X \right| > \text{neg} \Rightarrow (-\infty, \infty) \]
3. Let the symbol choose the shape
Why: Less-than means and: one sandwich, one interval. Greater-than means or: two branches, two rays joined by a union. Remember it as less-thand and greator.
\[ \left| X \right| < k \Rightarrow -k < X < k \qquad \left| X \right| > k \Rightarrow X > k \;\text{or}\; X < -k \]
4. Solve, and reverse the symbol on the negative branch
Why: In the greater-than case the left branch flips both the sign of the number and the direction of the symbol. In the sandwich case, do every operation to all three parts.
5. Match brackets to the symbol, then test points
Why: Strict symbols get round parentheses and hollow dots; or-equal-to symbols get square brackets and solid dots. Infinite ends are always round. Then test one point inside the answer and one outside it in the original inequality.
Notation
Annotate
From Pattern: solving any absolute-value inequality — read this one piece at a time. What is each part doing?
On: \( \left| X \right| < k \Rightarrow -k < X < k \qquad \left| X \right| > k \Rightarrow X > k \;\text{or}\; X < -k \)
Prediction
Predict first
Which statement describes the acceptable weights, and what interval does it give?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: The distance from the weight to 18 is at most 0.4, giving the interval [17.6, 18.4]
Why: The error is the weight minus the target 18, and it must stay within the tolerance of four tenths. Sandwiching gives negative 0.4 at most the error at most 0.4, and adding 18 to all three parts gives 17.6 to 18.4. Verify: a box at 18.4 has an error of exactly 0.4 and is accepted, while 18.5 has an error of 0.5 and is not.
Check
A cereal box is labeled 18 ounces, and the plant accepts any box whose weight is within four tenths of an ounce of the label. Let the weight be the variable.
Check your understanding
Which statement describes the acceptable weights, and what interval does it give?
Answer: A
Why: The error is the weight minus the target 18, and it must stay within the tolerance of four tenths. Sandwiching gives negative 0.4 at most the error at most 0.4, and adding 18 to all three parts gives 17.6 to 18.4. Verify: a box at 18.4 has an error of exactly 0.4 and is accepted, while 18.5 has an error of 0.5 and is not.
Elimination
Eliminate the wrong options
Which of these is true for every real number x?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: An absolute value is at least zero for every input, and zero is already greater than negative 3, so this holds no matter what x is. The solution set is all real numbers. The worst case is x equal to 5, where the left side is 0, and even that passes.
Check
Remember the two special cases: a negative on the other side settles everything before you solve.
Check your understanding
Which of these is true for every real number x?
Answer: A
Why: An absolute value is at least zero for every input, and zero is already greater than negative 3, so this holds no matter what x is. The solution set is all real numbers. The worst case is x equal to 5, where the left side is 0, and even that passes.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Part 1 - Distance from Zero · Part 2 - Absolute Value Equations · Part 3 - Bars on Both Sides · Part 4 - Less Than: One Interval · Part 5 - Greater Than: Two Rays · Part 6 - Special Cases and Real Tolerances. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
One picture ran the whole deck: an absolute value is a distance from zero, so it is never negative, and a distance can be reached from two directions.
| Form | What it says | Answer shape |
|---|---|---|
| Absolute value equals a positive number | distance is exactly that far | two solutions |
| Absolute value equals zero | distance is zero | one solution |
| Absolute value equals a negative number | impossible | no solution |
| Absolute value less than a positive number | close to the center | one interval, an and |
| Absolute value greater than a positive number | far from the center | two rays, an or |
Next stop: quadratic equations, where a squared term makes two answers appear for a very similar reason - two numbers can share the same square, just as two numbers can share the same distance from zero.
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