Absolute value as distance on a number line, solving an absolute value equation by splitting it into two, rejecting extraneous solutions, the and/or rule for absolute value inequalities, and writing a tolerance as one.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 1 — Equations and Inequalities
Solve Absolute Value Equations and Inequalities
Objectives
Five outcomes. The third exists because absolute value is the first place in this book where an answer can be a fake.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 51-57 — the lesson these objectives are drawn from
Warm-up
Lesson 1.6 gave you and-inequalities and or-inequalities. This lesson gives you a symbol that decides which one you get.
Discussion prompt
Name every number that is exactly 7 units away from 5 on a number line. How many are there, and why is that number of answers not a coincidence?
Hint: Distance has no direction, so a distance can be walked two ways.
Answer:
\[ 5 - 7 = -2 \qquad 5 + 7 = 12 \]
Two: negative two and twelve. It is never a coincidence — from any centre there are exactly two points at any positive distance, one on each side. That pair is the whole reason an absolute value equation splits into two equations.
Concept
The absolute value of a number is how far it sits from zero on the number line, with no regard for direction. Once you read the bars as a distance, every rule in this lesson is something you already know about number lines.
absolute value — The distance of a number from zero on the number line. It is never negative, and two different numbers can share the same absolute value.
The bars around an expression that is not just x measure distance from a different centre: the bars around x minus five measure distance from five.
Figure (svg): A number line showing that the absolute value of x minus five equals seven means x is seven units from five, giving two answers
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 51-51
Section
Section 1
Concept
For a positive number, the absolute value is the number itself. For zero it is zero. For a negative number it is the opposite of the number, which is what makes the result positive.
\[ \lvert x \rvert = \begin{cases} x, & x > 0 \\ 0, & x = 0 \\ -x, & x < 0 \end{cases} \]
The third case is the one that reads wrongly at first glance. The minus sign there does not make the answer negative; it undoes a negative that was already there.
Figure (svg): The three-case definition of absolute value, with the negative case highlighted as the one that surprises people
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 51-51 — the definition and the Interpreting Absolute Value Equations box
Picture it
The definition is written in cases so that it can be applied without a picture.
Figure (svg): The three-case definition of absolute value, with the negative case highlighted as the one that surprises people
Test it: the absolute value of negative six falls in the third case, so it equals the opposite of negative six, which is six. The rule and the distance picture always agree.
Worked example
The Interpreting Absolute Value Equations box, worked through.
\[ \text{What does } \lvert x - 5 \rvert = 7 \text{ say about } x? \]
Identify the centre
Why: The expression inside is x minus five, so the distance is being measured from five, not from zero.
\[ \text{centre is } 5 \]
Identify the distance
Why: The right side is seven, so x sits seven units from the centre.
\[ \text{distance is } 7 \]
Go seven units in each direction
Why: Five minus seven is negative two; five plus seven is twelve.
\[ -2\text{ and } 12 \]
Write it as two ordinary equations
Why: The distance picture and the two-equation method are the same statement.
\[ x - 5 = -7\text{ or } x - 5 = 7 \]
Figure (svg): The solution to Worked example read the bars as a distance shown as a ladder of expressions, one row per algebraic move
\[ x = -2 \quad \text{or} \quad x = 12 \]
Verify: substitute both into the original
Why: For 12: the inside is 7, and the absolute value of 7 is 7 — correct. For -2: the inside is -7, and the absolute value of -7 is also 7 — correct. Both are genuine solutions, which is what the two-directions picture predicted.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 51-51
Sorting
Decide the sign of each without computing the number.
Sort into buckets
Sort each expression by the sign of its value.
Everything here is settled by one question: is the minus sign inside the bars or outside them?
Worked example
Small numbers, so that the definition rather than the arithmetic is what is being tested.
\[ \text{Evaluate } \lvert 6 \rvert, \; \lvert 0 \rvert, \; \lvert -6 \rvert, \; -\lvert 6 \rvert, \; -\lvert -6 \rvert. \]
Apply the first case to the positive input
Why: Six is positive, so its absolute value is itself.
\[ | 6 | = 6 \]
Apply the second case to zero
Why: Zero is zero units from zero.
\[ | 0 | = 0 \]
Apply the third case to the negative input
Why: Negative six is negative, so its absolute value is its opposite, which is six.
\[ | - 6 | = 6 \]
Handle the minus signs that sit OUTSIDE the bars
Why: Those are not part of the absolute value at all; they act on the result afterwards, exactly like a minus sign outside a power.
\[ -| 6 | = -6\text{ and } -| - 6 | = -6 \]
Figure (svg): The solution to Worked example evaluate the three cases shown as a ladder of expressions, one row per algebraic move
\[ \lvert 6 \rvert = 6, \;\; \lvert 0 \rvert = 0, \;\; \lvert -6 \rvert = 6, \;\; -\lvert 6 \rvert = -6, \;\; -\lvert -6 \rvert = -6 \]
Verify: check that no absolute value came out negative
Why: The three expressions with bars alone all gave non-negative results, as the definition requires. The two negative answers both have a minus sign outside the bars, which is a separate operation applied after the absolute value was taken — the same distinction as brackets around a power in Lesson 1.2.
Trap
\[ \text{Evaluate } \lvert -6 \rvert. \]
Say the answer is negative six, because the input is negative
Why: The bars are read as decoration rather than as an operation.
\[ \lvert -6 \rvert = -6 \quad \text{(wrong)} \]
An absolute value is a distance, and no distance is negative.
\[ \text{Evaluate } \lvert -6 \rvert. \]
Apply the third case of the definition
Why: Negative six is negative, so its absolute value is the opposite of it.
\[ \lvert -6 \rvert = -(-6) = 6 \]
Compare with minus the absolute value of negative six, which is -6. The position of the minus sign relative to the bars decides everything.
Notation
Every part of this line says something specific.
Annotate
On: \( \lvert x - b \rvert = k \)
The plus-or-minus structure of the answer is not a rule to memorise. It is what the picture forces.
Prediction
Commit before you reason.
Predict first
How many solutions does the absolute value of x plus 3 equal negative 5 have?
Correct: None.
\[ \lvert x + 3 \rvert \geq 0 \; \text{for every } x, \quad \text{so it can never equal } -5 \]
Why: An absolute value is a distance, and no distance is negative, so no value of x can make the left side equal negative five. Recognising this before starting saves the whole two-equation procedure, which would otherwise produce two apparent solutions that both fail the check. Whenever the right side of an absolute value equation is negative, stop and report no solution.
Matching
Four absolute value statements, each saying something about distance.
Match the pairs
Why: The centre is whatever value makes the inside zero, which is why x plus two measures distance from negative two rather than from positive two. The sign flips when you read the centre off the expression, and that is the single most common misreading here. The last one is the edge case: a distance of zero has only one solution.
Section
Section 2
Concept
To solve an absolute value equation, write the two ordinary equations that say the inside equals the positive value or the negative of it. Solve each. Then check both answers in the original — a step that is genuinely necessary, as the next section shows.
\[ \lvert ax + b \rvert = c \;\Longrightarrow\; ax + b = c \;\text{ or }\; ax + b = -c \]
The procedure is stated in the textbook for the case where c is positive. If c is negative there is no solution, and if c is zero there is exactly one.
Figure (svg): A branching diagram showing one absolute value equation splitting into two ordinary equations
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 52-52 — Solving an Absolute Value Equation
Picture it
Example 2: the absolute value of five x minus ten equals forty-five.
Figure (svg): A branching diagram showing one absolute value equation splitting into two ordinary equations
Both branches are ordinary linear equations of exactly the kind Lesson 1.3 handled. Nothing new is needed to finish either one.
Worked example
Example 2. Both branches are solved and both are checked.
\[ \text{Solve } \lvert 5x - 10 \rvert = 45. \]
Write the two equations
Why: The inside expression can equal 45 or negative 45; those are the only two numbers whose distance from zero is 45.
\[ 5 x - 10 = 45\text{ or } 5 x - 10 = -45 \]
Solve the first branch
Why: Add ten, then divide by five.
\[ 5 x = 55,\text{ so } x = 11 \]
Solve the second branch
Why: Add ten to negative forty-five, then divide by five.
\[ 5 x = -35,\text{ so } x = -7 \]
Check both in the original
Why: Substituting 11 gives the absolute value of 45; substituting -7 gives the absolute value of -45. Both equal 45.
Figure (svg): The solution to Worked example split and solve shown as a ladder of expressions, one row per algebraic move
\[ x = 11 \quad \text{or} \quad x = -7 \]
Verify: substitute both values into the original equation
Why: At x equal to 11 the inside is 55 minus 10, which is 45, and the absolute value of 45 is 45 — correct. At x equal to -7 the inside is -35 minus 10, which is -45, and the absolute value of -45 is also 45 — correct. Both survive, so both are reported.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 52-52
Fill the middle
Guided Practice 3. One branch is given.
Fill in the blanks
\lvert x + 2 \rvert = 7 \;\Longrightarrow\; x + 2 = 7 \;\textx + 2 = -7\; ___
Why: The two branches set the inside expression equal to positive seven and to negative seven, because those are the only two numbers seven units from zero. Solving gives x equal to five and x equal to negative nine, and both check: the inside becomes 7 and -7 respectively, and both have absolute value 7.
Worked example
Example 1 and Guided Practice 4, done the same way.
\[ \text{Solve } \lvert 3x - 2 \rvert = 13. \]
Write the two equations
Why: Thirteen and negative thirteen are the two values the inside can take.
\[ 3 x - 2 = 13\text{ or } 3 x - 2 = -13 \]
Solve the first branch
Why: Add two, then divide by three.
\[ 3 x = 15,\text{ so } x = 5 \]
Solve the second branch
Why: Add two to negative thirteen, giving negative eleven, then divide by three.
\[ 3 x = -11,\text{ so } x = -\frac{11}{3} \]
Check both
Why: Neither branch had a variable on the right side, so neither can be extraneous — but checking costs nothing.
Figure (svg): The solution to Worked example a shifted centre shown as a ladder of expressions, one row per algebraic move
\[ x = 5 \quad \text{or} \quad x = -\tfrac{11}{3} \]
Verify: substitute both values
Why: At x equal to 5 the inside is 15 minus 2, which is 13 — correct. At x equal to negative eleven thirds the inside is -11 minus 2, which is -13, whose absolute value is 13 — also correct. A fractional answer is not a warning sign here; it just means the coefficient did not divide the constant evenly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 53-53
Trap
\[ \lvert 5x - 10 \rvert = 45 \]
Remove the bars and solve the equation that is left
Why: The bars are treated as brackets that can simply be erased.
\[ 5x - 10 = 45 \;\Longrightarrow\; x = 11 \]
Eleven is a genuine solution, which is exactly what makes this error survive: the answer is not wrong, it is incomplete.
\[ \lvert 5x - 10 \rvert = 45 \]
Write BOTH equations before solving either
Why: Two numbers have an absolute value of 45, so the inside expression has two possible values.
\[ 5x - 10 = 45 \;\text{ or }\; 5x - 10 = -45 \]
\[ x = 11 \;\text{ or }\; x = -7 \]
Half an answer scores zero on a test that asks for the solutions. Write both branches on the page before touching either of them.
Elimination
Solve the absolute value of x minus 3 equals 10.
Eliminate the wrong options
Which is the complete solution?
Survives elimination: A
Why: The centre is 3 and the distance is 10, so the answers are 3 minus 10 and 3 plus 10, giving negative seven and thirteen. Both check: the inside becomes -10 and 10 respectively, and both have absolute value 10. The distance picture gets this right faster than the algebra does.
Reverse engineer
An absolute value equation has solutions 4 and 10.
Fill in the blanks
\lvert x - 7 \rvert = 3
Why: Two solutions of an absolute value equation are always symmetric about the centre, so the centre is their average: four plus ten over two, which is seven. The distance is then how far each answer sits from seven, which is three. Reading a pair of answers back into an equation like this is exactly the skill Section 5 needs for writing tolerances.
Step zero
Look at these three equations. Do not solve any of them.
\[ \lvert 2x + 1 \rvert = 9, \qquad \lvert 2x + 1 \rvert = 0, \qquad \lvert 2x + 1 \rvert = -9 \]
Discussion prompt
For each, say how many solutions it has and how you know, before doing any algebra. What single feature of each equation decided your answer?
Hint: Look only at the right-hand side.
Answer:
The right-hand side decides all three. A positive value gives two solutions, one on each side of the centre. Zero gives one, because there is only one point at zero distance. A negative value gives none, because no distance is negative.
\[ \lvert 2x + 1 \rvert = 9 \;\Rightarrow\; x = 4 \text{ or } x = -5 \]
\[ \lvert 2x + 1 \rvert = 0 \;\Rightarrow\; x = -\tfrac{1}{2} \qquad \lvert 2x + 1 \rvert = -9 \;\Rightarrow\; \text{no solution} \]
Section
Section 3
Concept
When the right side of an absolute value equation contains a variable, the two-equation method can produce a value that fails the original. It is not a mistake in the algebra — it is a consequence of the method, and the check is what catches it.
extraneous solution — An apparent solution produced correctly by the solving method that must nevertheless be rejected, because it does not satisfy the original equation.
\[ \lvert 2x + 12 \rvert = 4x \]
The reason is that the right side must itself be non-negative, since it equals an absolute value. Any candidate that makes it negative is disqualified no matter how the algebra went.
Figure (svg): A branching diagram showing one absolute value equation splitting into two ordinary equations
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 52-52 — Check for extraneous solutions
Picture it
The method does not change; only the checking becomes compulsory.
Figure (svg): A branching diagram showing one absolute value equation splitting into two ordinary equations
Every branch is solved as before. The difference is that afterwards each candidate is tested, and one of them may be discarded.
Worked example
Example 3. The check is the point of this example, not the algebra.
\[ \text{Solve } \lvert 2x + 12 \rvert = 4x \text{ and check for extraneous solutions.} \]
Write the two equations
Why: The inside can equal 4x or its negative, exactly as before — the variable on the right changes nothing about the split.
\[ 2 x + 12 = 4 x\text{ or } 2 x + 12 = -4 x \]
Solve the first branch
Why: Subtract 2x from each side, then divide by two.
\[ 12 = 2 x,\text{ so } x = 6 \]
Solve the second branch
Why: Add 4x to each side, then subtract twelve and divide.
\[ 12 = -6 x,\text{ so } x = -2 \]
Check x equal to 6
Why: The inside is 24, whose absolute value is 24, and the right side is 4 times 6, also 24. It survives.
\[ 24 = 24,\text{ keep} \]
Check x equal to -2
Why: The inside is 8, whose absolute value is 8, but the right side is 4 times -2, which is -8. Eight does not equal negative eight.
\[ 8\text{ is not } -8,\text{ reject} \]
Figure (svg): The solution to Worked example one solution is a fake shown as a ladder of expressions, one row per algebraic move
\[ x = 6 \]
Verify: explain why the rejected value had to fail
Why: The right side of the original is 4x, which must be non-negative because it equals an absolute value. That forces x to be at least zero, and -2 is negative, so it could never have worked. Reading that condition off the equation before solving would have predicted the rejection.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 52-52
Discrimination
Do not solve. Just say whether an extraneous solution is possible.
Sort into buckets
Sort each equation by whether checking could reject a candidate.
Worked example
Guided Practice 6. A variable on the right does not guarantee an extraneous solution.
\[ \text{Solve } \lvert 4x - 1 \rvert = 2x + 9 \text{ and check for extraneous solutions.} \]
Write the two equations
Why: The inside equals the right side, or its negative.
\[ 4 x - 1 = 2 x + 9\text{ or } 4 x - 1 = -(2 x + 9) \]
Solve the first branch
Why: Subtract 2x, add one, divide by two.
\[ 2 x = 10,\text{ so } x = 5 \]
Solve the second branch
Why: Distribute the minus sign first, then collect.
\[ 6 x = -8,\text{ so } x = -\frac{4}{3} \]
Check x equal to 5
Why: The inside is 19, and the right side is 10 plus 9, also 19.
\[ 19 = 19,\text{ keep} \]
Check x equal to negative four thirds
Why: The inside is -19/3, whose absolute value is 19/3, and the right side is -8/3 plus 27/3, which is 19/3.
\[ \frac{19}{3} = \frac{19}{3},\text{ keep} \]
Figure (svg): The solution to Worked example both survive shown as a ladder of expressions, one row per algebraic move
\[ x = 5 \quad \text{or} \quad x = -\tfrac{4}{3} \]
Verify: confirm the right side is non-negative at both values
Why: At x equal to 5 the right side is 19, and at x equal to negative four thirds it is 19/3 — both positive, so neither candidate was ever in danger. Compare the previous example, where the rejected value made the right side negative. That single test predicts the outcome every time.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 53-53
Error analysis
A student solves an absolute value equation with a variable on the right and reports both answers.
Annotate
On: \( \lvert 2x + 5 \rvert = 3x \;\Longrightarrow\; x = 5 \;\text{ or }\; x = -1 \)
When the right side of an absolute value equation contains a variable, checking is not good practice - it is part of the method.
Explain it to yourself
The algebra was correct at every step, and yet one answer is false.
\[ \lvert 2x + 12 \rvert = 4x \;\Longrightarrow\; 2x + 12 = -4x \;\Longrightarrow\; x = -2 \]
Discussion prompt
Explain how a correct chain of steps can produce a value that does not solve the original equation. What did the second branch actually assume, and was that assumption checked anywhere?
Hint: Ask what has to be true about the inside expression for the second branch to be the right one.
Answer:
The second branch assumes the inside expression is negative, so that its absolute value is the opposite of it. That assumption is never tested during the solving — it is smuggled in when the branch is written.
At x equal to -2 the inside is actually positive 8, so the second branch was not the applicable one for that value. The check is what tests the assumption after the fact.
This is the first of several places in Algebra 2 where a method introduces a candidate that has to be verified. Squaring both sides in Chapter 6 does the same thing for the same reason.
Prediction
Commit before you solve.
Predict first
In the equation absolute value of x plus 4 equals x minus 1, which candidates could possibly be genuine?
Correct: Only candidates with x at least 1.
\[ x - 1 \geq 0 \;\Longrightarrow\; x \geq 1 \]
\[ x = -\tfrac{3}{2} < 1 \quad \Longrightarrow \quad \text{rejected; no solution} \]
Why: The right side must be non-negative because it equals an absolute value, so x minus 1 must be at least zero, which forces x to be at least 1. Solving gives the two branches x plus 4 equals x minus 1, which is impossible, and x plus 4 equals 1 minus x, giving x equal to negative three halves — which is less than 1 and therefore rejected. The equation has no solution, and the condition predicted it.
Explain it
A classmate says checking is for people who do not trust their algebra.
Discussion prompt
In three sentences, explain why checking an absolute value equation with a variable on the right is part of the method rather than a safety net, and give them the one-glance test for when it matters.
Hint: The key word is assumption.
Answer:
Writing the second branch assumes the inside is negative, and nothing in the solving verifies that assumption. So the method can produce a value that solves the branch without solving the original — the check is what tests the assumption.
The one-glance test: look at the right-hand side. If it contains a variable it can go negative, and then checking can reject something; if it is a fixed positive number, checking is only arithmetic insurance.
Section
Section 4
Concept
An absolute value inequality asks which values are close to the centre or far from it. Close is a single band, which is an and. Far is two rays going opposite ways, which is an or.
\[ \begin{aligned} \lvert ax + b \rvert < c &\;\Longleftrightarrow\; -c < ax + b < c \\ \lvert ax + b \rvert > c &\;\Longleftrightarrow\; ax + b < -c \;\text{ or }\; ax + b > c \end{aligned} \]
The inclusive versions work identically, with at-most and at-least symbols throughout and solid endpoints on the graph.
Figure (svg): Two number lines contrasting a less-than absolute value inequality as a band with a greater-than one as two rays
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 53-53 — Absolute Value Inequalities
Picture it
The same centre and the same distance, with the symbol reversed.
Figure (svg): Two number lines contrasting a less-than absolute value inequality as a band with a greater-than one as two rays
You never have to memorise which gives and and which gives or. Ask whether the inequality describes being near the centre or far from it, and the picture answers.
Worked example
Example 4. Two rays, and each is solved on its own.
\[ \text{Solve } \lvert 4x + 5 \rvert > 13 \text{ and graph the solution.} \]
Recognise it as a far-from-centre statement, so an or
Why: Greater than means the inside is more than thirteen units from zero, in either direction.
\[ 4 x + 5 < -13\text{ or } 4 x + 5 > 13 \]
Solve the first inequality
Why: Subtract five, then divide by four. Four is positive, so nothing reverses.
\[ 4 x < -18,\text{ so } x < -\frac{9}{2} \]
Solve the second inequality
Why: Subtract five, then divide by four.
\[ 4 x > 8,\text{ so } x > 2 \]
Graph both rays
Why: Both symbols are strict, so both endpoints are open.
\[ \text{open at } -\frac{9}{2}\text{ and at } 2 \]
Figure (svg): The solution to Worked example a greater-than inequality shown as a ladder of expressions, one row per algebraic move
\[ x < -\tfrac{9}{2} \quad \text{or} \quad x > 2 \]
Verify: test one value in each ray and one in the gap
Why: At x equal to 3 the inside is 17, whose absolute value is 17, which is greater than 13 — correct, and 3 is in the right ray. At x equal to -6 the inside is -19, absolute value 19, also greater than 13 — correct. At x equal to 0 the inside is 5, absolute value 5, which is not greater than 13 — correct, and 0 sits in the unshaded gap.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 53-53
Sorting
Do not solve. Decide the shape of each answer.
Sort into buckets
Sort each inequality by the shape of its solution set.
The symbol alone decides it, before any algebra. Less-than points inward; greater-than points outward.
Worked example
Example 5's structure, applied to the baseball specification.
\[ \text{Solve } \lvert w - 5.125 \rvert \leq 0.125. \]
Recognise it as a near-the-centre statement, so an and
Why: At most means the distance from the centre does not exceed the tolerance, which is one band.
\[ -0.125 \le w - 5.125 \le 0.125 \]
Add 5.125 to all three parts
Why: Every move in a three-part inequality applies to all three parts, as in Lesson 1.6.
\[ 5.000 \le w \le 5.250 \]
Read the endpoints
Why: Both symbols are inclusive, so both endpoints belong to the range.
\[ \text{solid dots at } 5\text{ and } 5.25 \]
Figure (svg): The solution to Worked example a less-than-or-equal inequality shown as a ladder of expressions, one row per algebraic move
\[ 5 \leq w \leq 5.25 \]
Verify: check both endpoints and the centre
Why: At w equal to 5 the distance from 5.125 is exactly 0.125, which is at most 0.125 — so the endpoint is included, matching the solid dot. At w equal to 5.25 the distance is again exactly 0.125. At the centre the distance is zero, comfortably inside. And a ball at 5.3 is 0.175 away, correctly excluded.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 54-54
Trap
\[ \lvert 2x - 7 \rvert < 1 \]
Split it into two separate inequalities joined by or
Why: The two-branch habit from absolute value EQUATIONS is carried over without checking the direction.
\[ 2x - 7 < -1 \;\text{ or }\; 2x - 7 < 1 \]
This gives x less than 4, which admits x equal to 0 — but at zero the inside is -7, whose absolute value is 7, not less than 1.
\[ \lvert 2x - 7 \rvert < 1 \]
Ask whether the statement means near the centre or far from it
Why: Less than means near, and near is a single band with the expression in the middle.
\[ -1 < 2x - 7 < 1 \]
\[ 6 < 2x < 8 \;\Longrightarrow\; 3 < x < 4 \]
Testing 3.5: the inside is 0, whose absolute value is 0, which is less than 1. Testing 0: the absolute value is 7, correctly excluded.
Comparison
Fill the blanks. This is the table worth having in your head.
Comparison matrix
| Inequality | Equivalent form | Endpoints |
|---|---|---|
| |ax + b| < c | -c < ax + b < c | open |
| |ax + b| <= c | -c <= ax + b <= c | solid |
| |ax + b| > c | ax + b < -c or ax + b > c | open |
| |ax + b| >= c | ax + b <= -c or ax + b >= c | solid |
Two rows give bands and two give rays, and the split is entirely by direction. The strict-or-inclusive choice only decides the dots.
Fill the middle
Guided Practice 9, rewritten as a less-than for this exercise.
Fill in the blanks
\lvert 3x + 5 \rvert < 10 \;\Longleftrightarrow\; -10 < 3x + 5 < 10
Why: Less than ten means the inside is within ten units of zero, so it lies strictly between negative ten and ten. Subtracting five from all three parts gives -15 less than 3x less than 5, and dividing by three gives -5 less than x less than 5 over 3. Testing x equal to 0: the inside is 5, whose absolute value is 5, which is indeed less than 10.
Edge cases
Absolute value inequalities behave strangely at the extremes, and knowing how saves time on a test.
Discussion prompt
What is the solution set of the absolute value of x being less than zero? What about less than a negative number? And what about greater than a negative number? Answer all three by thinking about distance, not algebra.
Hint: Remember that an absolute value is never negative.
Answer:
Less than zero: no solutions, because no distance is negative and a distance of exactly zero is not less than zero.
Less than a negative number: also no solutions, for the same reason and more strongly.
Greater than a negative number: every real number, because every absolute value is at least zero and zero already exceeds any negative number.
\[ \lvert x \rvert < 0 : \; \varnothing \qquad \lvert x \rvert < -3 : \; \varnothing \qquad \lvert x \rvert > -3 : \; \text{all reals} \]
Section
Section 5
Concept
Manufacturing specifications, safe ranges and margins of error all say the same thing: a measured value must be within a stated distance of a target. That is an absolute value inequality, and building one from a stated range is a two-step calculation.
\[ \lvert \text{actual} - \text{target} \rvert \leq \text{tolerance} \]
When the range is given by its two ends instead of a target and a tolerance, the target is the mean of the two ends and the tolerance is the distance from the mean to either end.
Figure (svg): A baseball weight specification drawn as a centre value with a tolerance band on each side
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 54-54 — Write a range as an absolute value inequality
Picture it
Example 5: a baseball should weigh 5.125 ounces, with a tolerance of 0.125 ounce.
Figure (svg): A baseball weight specification drawn as a centre value with a tolerance band on each side
The centre and the two ends are the same information written two ways. Being able to move between them is the whole of this section.
Worked example
Example 5. A target and a tolerance are given; the acceptable weights are wanted.
\[ \text{A baseball should weigh } 5.125 \text{ oz with a tolerance of } 0.125 \text{ oz. Find the acceptable weights.} \]
Write the specification as an absolute value inequality
Why: The distance between the actual weight and the target is at most the tolerance.
\[ | w - 5.125 | \le 0.125 \]
Rewrite the at-most form as a band
Why: Less than or equal means near the centre, which is a three-part inequality.
\[ -0.125 \le w - 5.125 \le 0.125 \]
Add 5.125 to all three parts
Why: Five point one two five minus 0.125 is exactly 5, and plus 0.125 is exactly 5.25.
\[ 5 \le w \le 5.25 \]
State the answer in words
Why: Any weight from 5 to 5.25 ounces inclusive is acceptable.
\[ 5\text{ to } 5.25\text{ ounces} \]
Figure (svg): The solution to Worked example from tolerance to range shown as a ladder of expressions, one row per algebraic move
\[ 5 \leq w \leq 5.25 \]
Verify: check the centre and the two extremes
Why: The midpoint of 5 and 5.25 is 5.125, the stated target, and each end is exactly 0.125 from it, the stated tolerance. Recovering both original numbers from the range confirms the conversion in both directions.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 54-54
Real world
A medication dose for a patient must be between 180 and 220 milligrams, inclusive.
Discussion prompt
Write this as an absolute value inequality, then say what the two numbers in your answer mean to a pharmacist. What would change about your inequality if the range were stated as strictly between 180 and 220?
Hint: The centre is the mean of the two ends.
Answer:
\[ \lvert d - 200 \rvert \leq 20 \]
The 200 is the target dose and the 20 is the largest acceptable deviation from it in either direction — which is exactly how a pharmacist would read a prescription written as two hundred milligrams plus or minus twenty.
Strictly between would change the at-most to a strict less-than, making both endpoints open. In a clinical context that distinction is real: it decides whether a dose of exactly 220 milligrams is dispensed or refused.
Worked example
Example 6. Now the two ends are given and the absolute value inequality is wanted.
\[ \text{Gymnastics mats must be between } 7.5 \text{ and } 8.25 \text{ inches thick, inclusive. Write the specification.} \]
Find the mean of the two extremes
Why: Seven point five plus 8.25 is 15.75, and half of that is 7.875. This is the target the range is centred on.
\[ \text{mean } = 7.875 \]
Find the tolerance as the distance from the mean to an end
Why: Eight point two five minus 7.875 is 0.375. The distance to the lower end is the same, which is what makes the mean the right centre.
\[ \text{tolerance } = 0.375 \]
Write the verbal model
Why: The distance between the actual thickness and the mean is at most the tolerance.
Write it in symbols
Why: Using t for the thickness.
\[ | t - 7.875 | \le 0.375 \]
Figure (svg): The solution to Worked example from range to tolerance shown as a ladder of expressions, one row per algebraic move
\[ \lvert t - 7.875 \rvert \leq 0.375 \]
Verify: expand the inequality back into a range
Why: The band form is -0.375 at most t minus 7.875 at most 0.375, and adding 7.875 throughout gives 7.5 at most t at most 8.25 — exactly the range stated in the problem. Recovering the original specification is the strongest possible check on the two-step conversion.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 54-54
Error analysis
A student writes the mat specification and uses the wrong distance.
Annotate
On: \( \lvert t - 7.875 \rvert \leq 0.75 \)
Always expand your finished inequality back into a range and compare it with the range you were given. That one step catches every error in this conversion.
Translation
Four everyday specifications.
Match the pairs
Why: Three of these give the target and tolerance directly, so the inequality can be written straight down. Only the second gives the two ends instead, which is why it needs the mean-and-half-width calculation first. Recognising which form you have been handed is the first step every time.
Estimation
A range runs from 7.5 to 8.25.
Predict first
Roughly where is the centre, and roughly what is the tolerance?
Correct: Centre near 7.9, tolerance near 0.4.
\[ \text{centre} = \frac{7.5 + 8.25}{2} = 7.875 \qquad \text{tolerance} = 8.25 - 7.875 = 0.375 \]
Why: The centre is roughly halfway between the ends, which is a little under 7.9 — the exact value is 7.875. The tolerance is the half-width, and the full width is 0.75, so the half-width is a little under 0.4, exactly 0.375. Estimating first catches the most common error in this conversion, which is using the full width as the tolerance and getting 0.75.
Counterexample
A classmate offers a shortcut for writing tolerances.
\[ \text{the tolerance is always the smaller end subtracted from the larger} \]
Discussion prompt
Show with the mat example that this gives the wrong answer, and state the correct rule in one sentence. Then say when the shortcut would accidentally be right.
Hint: Expand both versions back into ranges and compare.
Answer:
\[ 8.25 - 7.5 = 0.75 \quad \text{gives} \quad 7.125 \leq t \leq 8.625 \quad \text{(too wide)} \]
The correct rule is that the tolerance is the distance from the centre to either end, which is half the full width. The shortcut would only be right if the range had zero width, in which case both are zero — so never, in any useful case.
Comparison
Fill the blanks. The symbol decides the shape before you do any algebra.
Comparison matrix
| Statement | Means | Answer shape |
|---|---|---|
| |x - b| = k, k positive | exactly k from b | two points |
| |x - b| = 0 | exactly at b | one point |
| |x - b| < k | closer than k to b | one band |
| |x - b| > k | further than k from b | two rays |
| |x - b| = k, k negative | impossible | no solutions |
Every row is read straight off the distance picture. None of it needs memorising if you keep the number line in mind.
Pattern
One routine handles every absolute value problem in this lesson.
Step two is the one that saves the most time. Ten seconds spent reading the right-hand side can turn a five-line problem into a one-line answer.
OpenStax Algebra and Trigonometry 2e, §2.7 Linear Inequalities and Absolute Value Inequalities §2.7
Check
An absolute value equation. Write both branches before solving either.
Check your understanding
Solve the absolute value of 3x minus 2 equals 13.
Answer: A
Why: The two branches are 3x - 2 = 13 and 3x - 2 = -13, giving 3x = 15 and 3x = -11. So x = 5 or x = -11/3, and both check.
Check
A variable on the right. Check before you report.
Check your understanding
Solve the absolute value of 2x plus 5 equals 3x.
Answer: A
Why: The branches give x = 5 and x = -1. Checking x = -1: the left side is the absolute value of 3, which is 3, and the right side is -3. Since 3 does not equal -3, that candidate is extraneous and only x = 5 survives.
Check
An inequality. Decide the shape before you solve.
Check your understanding
Solve the absolute value of 4x plus 5 is greater than 13.
Answer: A
Why: Greater than means far from the centre, so it is an or: 4x + 5 < -13 or 4x + 5 > 13. Solving gives x < -9/2 or x > 2. Testing x = 0 gives an absolute value of 5, which is not greater than 13, so 0 is correctly excluded.
Real world
A poll reports that 47 percent of voters favour a proposal, with a margin of error of 3 percentage points.
Discussion prompt
Write the reported range as an absolute value inequality and expand it into a plain range. Then say what the poll can and cannot rule out, and what would change if the margin of error were 6 points instead.
Hint: The margin of error is a tolerance around the reported figure.
Answer:
\[ \lvert p - 47 \rvert \leq 3 \;\Longrightarrow\; 44 \leq p \leq 50 \]
The poll is consistent with any true figure from 44 to 50 percent. It therefore cannot rule out that the proposal is short of a majority, since 44 through 49 are all inside the range — a point worth making whenever a poll is described as showing something.
With a 6-point margin the range becomes 41 to 53, which now includes figures above half. A wider tolerance makes the claim weaker, not stronger, which is the opposite of how margins of error are often reported.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the absolute value of x minus 3 the same as the absolute value of 3 minus x?
Correct: Yes, always.
\[ \lvert x - 3 \rvert = \lvert 3 - x \rvert \quad \text{for every } x \]
\[ x = 10: \; \lvert 7 \rvert = 7 \;\text{ and }\; \lvert -7 \rvert = 7 \]
Why: The two expressions inside the bars are opposites, and a number and its opposite are the same distance from zero, so their absolute values are equal. In distance terms it is obvious: the distance from x to 3 is the same as the distance from 3 to x, because distance has no direction. This symmetry is worth knowing because it lets you write a tolerance either way round without changing the meaning.
Explain it
They can solve linear equations but have never seen absolute value bars in an equation.
Discussion prompt
In four sentences or fewer, explain what the bars mean, why an absolute value equation usually has two answers, and how to tell whether an absolute value inequality gives one band or two rays.
Hint: Lead with distance, not with the two-equation rule.
Answer:
The bars measure how far a number is from a centre, ignoring which side it is on. Because there are two points at any given distance from a centre, an absolute value equation normally has two answers.
For inequalities, ask whether the statement says close to the centre or far from it. Close gives a single band around the centre; far gives two rays with a gap in the middle.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the centre, find the value that makes the inside zero — for x plus two that is negative two, not two. For extraneous solutions, glance at the right side and check whenever it contains a variable. For band or rays, ask whether the statement means near the centre or far from it. For tolerances, the centre is the mean of the ends and the tolerance is half the width. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Draw a long number line across the top of a page. Mark a centre, mark the two points a fixed distance from it, and write beside them the absolute value equation they solve. Below that, draw two more number lines: shade one to show a less-than inequality as a band and the other to show a greater-than inequality as two rays, writing the equivalent and-form and or-form beside each. In the middle of the page write the three-case definition of absolute value and, next to the third case, work out the absolute value of negative six line by line. At the bottom, take a real specification — a shoe size, a room temperature you like, a budget — write it as a target and a tolerance, turn it into an absolute value inequality, and then expand it back into a range to check that you recover what you started with.
That last check, expanding back to the range you began with, is the one habit from this lesson worth carrying into every tolerance problem you ever meet.
Recap
Five things, and the third is the first appearance of an idea that returns in Chapters 6 and 8.
| If you see | Then |
|---|---|
| Bars equal a positive number | Two branches, two solutions |
| Bars equal a negative number | No solution; stop |
| A variable on the right | Checking is compulsory |
| Bars less than a number | One band, an and |
| Bars greater than a number | Two rays, an or |
That completes Chapter 1. Chapter 2 takes the linear expressions you have been solving and puts them on a coordinate plane, where solving becomes graphing and every equation becomes a line.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.7 Solve Absolute Value Equations and Inequalities §1.7, pp. 51-57 — everything on these slides traces back here
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