Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

2.7: Solve Absolute Value Inequalities

  • Page ID
    17391
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    Summary

    By the end of this section, you will be able to:

    • Solve absolute value equations
    • Solve absolute value inequalities with “less than”
    • Solve absolute value inequalities with “greater than”
    • Solve applications with absolute value

    Before you get started, take this readiness quiz.

    1. Evaluate: \(−|7|\).
      If you missed this problem, review [link].
    2. Fill in \(<,>,<,>,\) or \(=\) for each of the following pairs of numbers.
      ⓐ \(|−8|\text{___}−|−8|\) ⓑ \(12\text{___}−|−12|\) ⓒ \(|−6|\text{___}−6\) ⓓ \(−(−15)\text{___}−|−15|\)
      If you missed this problem, review [link].
    3. Simplify: \(14−2|8−3(4−1)|\).
      If you missed this problem, review [link].

    Solve Absolute Value Equations

    As we prepare to solve absolute value equations, we review our definition of absolute value.

    ABSOLUTE VALUE

    The absolute value of a number is its distance from zero on the number line.

    The absolute value of a number n is written as \(|n|\) and \(|n|\geq 0\) for all numbers.

    Absolute values are always greater than or equal to zero.

    We learned that both a number and its opposite are the same distance from zero on the number line. Since they have the same distance from zero, they have the same absolute value. For example:

    • \(−5\) is 5 units away from 0, so \(|−5|=5\).
    • \(5\) is 5 units away from 0, so \(|5|=5\).

    Figure \(\PageIndex{1}\) illustrates this idea.

    The figure is a number line with tick marks at negative 5, 0, and 5. The distance between negative 5 and 0 is given as 5 units, so the absolute value of negative 5 is 5. The distance between 5 and 0 is 5 units, so the absolute value of 5 is 5.
    Figure \(\PageIndex{1}\): The numbers 5 and \(−5\) are both five units away from zero.

    For the equation |x|=5,|x|=5, we are looking for all numbers that make this a true statement. We are looking for the numbers whose distance from zero is 5. We just saw that both 5 and −5−5 are five units from zero on the number line. They are the solutions to the equation.

    \(\begin{array} {ll} {\text{If}} &{|x|=5} \\ {\text{then}} &{x=−5\text{ or }x=5} \\ \end{array}\)

    The solution can be simplified to a single statement by writing \(x=\pm 5\). This is read, “x is equal to positive or negative 5”.

    We can generalize this to the following property for absolute value equations.

    ABSOLUTE VALUE EQUATIONS

    For any algebraic expression, u, and any positive real number, a,

    \[\begin{array} {ll} {\text{if}} &{|u|=a} \\ {\text{then}} &{u=−a \text{ or }u=a} \\ \nonumber \end{array}\]

    Remember that an absolute value cannot be a negative number.

    Example \(\PageIndex{1}\)

    Solve:

    1. \(|x|=8\)
    2. \(|y|=−6\)
    3. \(|z|=0\)
    Answer a

    \(\begin{array} {ll} {} &{|x|=8} \\ {\text{Write the equivalent equations.}} &{x=−8 \text{ or } x=8} \\ {} &{x=\pn 8} \\ \end{array}\)

    Answer b

    \(\begin{array} {ll} {} &{|y|=−6} \\ {} &{\text{No solution}} \\ \end{array}\)
    Since an absolute value is always positive, there are no solutions to this equation.

    Answer c

    \(\begin{array} {ll} {} &{|z|=0} \\ {\text{Write the equivalent equations.}} &{z=−0\text{ or }z=0} \\ {\text{Since }−0=0,} &{z=0} \\ \end{array}\)
    Both equations tell us that z=0z=0 and so there is only one solution.

    Example \(\PageIndex{2}\)

    Solve:

    1. \(|x|=2\)
    2. \(|y|=−4\)
    3. \(|z|=0\)
    Answer a

    \(\pm 2\)

    Answer b

    no solution

    Answer c

    0

    Example \(\PageIndex{3}\)

    Solve:

    1. \(|x|=11\)
    2. \(|y|=−5\)
    3. \(|z|=0\)
    Answer a

    \(\pm 11\)

    Answer b

    no solution

    Answer c

    0

    To solve an absolute value equation, we first isolate the absolute value expression using the same procedures we used to solve linear equations. Once we isolate the absolute value expression we rewrite it as the two equivalent equations.

    How to Solve Absolute Value Equations

    Example \(\PageIndex{4}\)

    Solve \(|5x−4|−3=8\).

    Answer

    Step 1 is to isolate the absolute value expression. The difference between the absolute value of the quantity 5 x minus 4 and 3 is equal to 8. Add 3 to both sides. The result is the absolute value of the quantity 5 x minus 4 is equal to 11.Step 2 is to write the equivalent equations, 5 x minus 4 is equal to negative 11 and 5 x minus 4 is equal to 11.Step 3 is to solve each equation. Add 4 to each side. 5 x is equal to negative 7 or 5 x is equal to 15. Divide each side by 5. The result is x is equal to negative seven-fifths or x is equal to 3.Step 4 is to check each solution. Substitute 3 and negative seven-fifths into the original equation, the difference between the absolute value of the quantity 5 x minus 4 and 3 is equal to 8. Substitute 3 for x. Is the difference between the absolute value of the quantity 5 times 3 minus 4 and 3 equal to 8? Is the difference between the absolute value of the quantity 15 minus 4 and 3 equal to 8? Is the difference between the absolute value of the 11 and 3 equal to 8? Is 11 minus 3 equal to 8? 8 is equal to 8, so the solution x is equal to 3 checks. Substitute negative seven-fifths for x. Is the difference between the absolute value of the quantity 5 times negative seven-fifths minus 4 and 3 equal to 8? Is the difference between the absolute value of the quantity negative 7 minus 4 and 3 equal to 8? Is the difference between the absolute value of the negative 11 and 3 equal to 8? Is 11 minus 3 equal to 8? 8 is equal to 8, so the solution x is equal to negative seven-fifths checks.

    Example \(\PageIndex{5}\)

    Solve: \(|3x−5|−1=6\).

    Answer

    \(x=4, \space x=−\frac{2}{3}\)

    Example \(\PageIndex{6}\)

    Solve: \(|4x−3|−5=2\).

    Answer

    \(x=−1,\space x=\frac{5}{2}\)

    The steps for solving an absolute value equation are summarized here.

    SOLVE ABSOLUTE VALUE EQUATIONS.

    1. Isolate the absolute value expression.
    2. Write the equivalent equations.
    3. Solve each equation.
    4. Check each solution.

    Example \(\PageIndex{7}\)

    Solve \(2|x−7|+5=9\).

    Answer
    \(2|x−7|+5=9\)
    Isolate the absolute value expression. \(2|x−7|=4\)
    \(|x−7|=2\)
    Write the equivalent equations. \(x−7=−2\) or \(x−7=2\)
    Solve each equation. \(x=5\) or \(x=9\)
    Check:
    .

    Example \(\PageIndex{8}\)

    Solve: \(3|x−4|−4=8\).

    Answer

    \(x=8,\space x=0\)

    Example \(\PageIndex{9}\)

    Solve: \(2|x−5|+3=9\).

    Answer

    \(x=8,\space x=2\)

    Remember, an absolute value is always positive!

    Example \(\PageIndex{10}\)

    Solve: \(|\frac{2}{3}x−4|+11=3\).

    Answer

    \(\begin{array} {ll} {} &{|\frac{2}{3}x−4|=−8} \\ {\text{Isolate the absolute value term.}} &{|\frac{2}{3}x−4|=−8} \\ {\text{An absolute value cannot be negative.}} &{\text{No solution}} \\ \end{array}\)

    Example \(\PageIndex{11}\)

    Solve: \(|\frac{3}{4}x−5|+9=4\).

    Answer

    No solution

    Example \(\PageIndex{12}\)

    Solve: \(|\frac{5}{6}x+3|+8=6\).

    Answer

    No solution

    Some of our absolute value equations could be of the form \(|u|=|v|\) where u and v are algebraic expressions. For example, \(|x−3|=|2x+1|\).

    How would we solve them? If two algebraic expressions are equal in absolute value, then they are either equal to each other or negatives of each other. The property for absolute value equations says that for any algebraic expression, u, and a positive real number, a, if \(|u|=a\), then \(u=−a\) or \(u=a\).

    This tell us that

    \(\begin{array} {llllll} {\text{if}} &{|u|=|v|} &{} &{} &{} &{} \\ {\text{then]} &{u=−v} &{\text{or}} &{u=v} &{\text{or}} &{u=v} \\ \end{array}\)

    This leads us to the following property for equations with two absolute values.

    EQUATIONS WITH TWO ABSOLUTE VALUES

    For any algebraic expressions, u and v,

    \[\begin{array} {ll} {\text{if}} &{|u|=|v|} \\ {\text{then}} &{u=−v\text{ or }u=v} \\ \nonumber \end{array}\]

    When we take the opposite of a quantity, we must be careful with the signs and to add parentheses where needed.

    Example \(\PageIndex{13}\)

    Solve: \(|5x−1|=|2x+3|\).

    Answer

    \(\begin{array} {ll} {} &{} &{|5x−1|=|2x+3|} &{} \\ {} &{} &{} &{} \\ {\text{Write the equivalent equations.}} &{5x−1=−(2x+3)} &{\text{or}} &{5x−1=2x+3} \\ {} &{5x−1=−2x−3} &{\text{or}} &{3x−1=3} \\ {\text{Solve each equation.}} &{7x−1=−3} &{} &{3x=4} \\ {} &{7x=−2} &{} &{x=43} \\ {} &{x=−27} &{\text{or}} &{x=43} \\ {\text{Check.}} &{} &{} &{} \\ {\text{We leave the check to you.}} &{} &{} &{} \\ \end{array}\)

    Example \(\PageIndex{14}\)

    Solve: \(|7x−3|=|3x+7|\).

    Answer

    \(x=−\frac{2}{5}, \space x=\frac{5}{2}\)

    Example \(\PageIndex{15}\)

    Solve: \(|6x−5|=|3x+4|\).

    Answer

    \(x=3, x=19\)

    Solve Absolute Value Inequalities with “Less Than”

    Let’s look now at what happens when we have an absolute value inequality. Everything we’ve learned about solving inequalities still holds, but we must consider how the absolute value impacts our work. Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line. For the equation \(|x|=5\), we saw that both 5 and \(−5\) are five units from zero on the number line. They are the solutions to the equation.

    \[\begin{array} {lll} {} &{|x|=5} &{} \\ {x=−5} &{\text{or}} &{x=5} \\ \nonumber \end{array}\]

    What about the inequality \(|x|\leq 5\)? Where are the numbers whose distance is less than or equal to 5? We know \(−5\) and 5 are both five units from zero. All the numbers between \(−5\) and 5 are less than five units from zero (Figure \(\PageIndex{2}\)).

    The figure is a number line with negative 5, 0, and 5 displayed. There is a left bracket at negative 5 and a right bracket at 5. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x which is less than or equal to 5.

    Figure \(\PageIndex{2}\).

    In a more general way, we can see that if \(|u|\leq a\), then \(−a\leq u\leq a\) (Figure \(\PageIndex{3}\)).

    The figure is a number line with negative a 0, and a displayed. There is a left bracket at negative a and a right bracket at a. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is less than or equal to a, then negative a is less than or equal to u which is less than or equal to a.

    Figure \(\PageIndex{3}\).

    This result is summarized here.

    ABSOLUTE VALUE INEQUALITIES WITH \(<\) OR \(\leq\)

    For any algebraic expression, u, and any positive real number, a,

    \[ \text{if} \quad |u|<a, \quad \text{then} \space −a<u<a \\ \text{if} \quad |u|\leq a, \quad \text{then} \space−a\leq u\leq a \nonumber\]

    After solving an inequality, it is often helpful to check some points to see if the solution makes sense. The graph of the solution divides the number line into three sections. Choose a value in each section and substitute it in the original inequality to see if it makes the inequality true or not. While this is not a complete check, it often helps verify the solution.

    Example \(\PageIndex{16}\)

    Solve \(|x|<7\). Graph the solution and write the solution in interval notation.

    Answer
    .
    Write the equivalent inequality. .
    Graph the solution. .
    Write the solution using interval notation. .

    Check:

    To verify, check a value in each section of the number line showing the solution. Choose numbers such as −8,−8, 1, and 9.

    The figure is a number line with a left parenthesis at negative 7, a right parenthesis at 7 and shading between the parentheses. The values negative 8, 1, and 9 are marked with points. The absolute value of negative 8 is less than 7 is false. It does not satisfy the absolute value of x is less than 7. The absolute value of 1 is less than 7 is true. It does satisfy the absolute value of x is less than 7. The absolute value of 9 is less than 7 is false. It does not satisfy the absolute value of x is less than 7.

    Example \(\PageIndex{17}\)

    Graph the solution and write the solution in interval notation: \(|x|<9\).

    Answer

    The solution is negative 9 is less than x which is less than 9. The number line shows open circles at negative 9 and 9 with shading in between the circles. The interval notation is negative 9 to 9 within parentheses.

    Example \(\PageIndex{18}\)

    Graph the solution and write the solution in interval notation: \(|x|<1\).

    Answer

    The solution is negative 1 is less than x which is less than 1. The number line shows open circles at negative 1 and 1 with shading in between the circles. The interval notation is negative 1 to 1 within parentheses.

    Example \(\PageIndex{19}\)

    Solve \(|5x−6|\leq 4\). Graph the solution and write the solution in interval notation.

    Answer
    Step 1. Isolate the absolute value expression.
    It is isolated.
    \(|5x−6|\leq 4\)
    Step 2. Write the equivalent compound inequality. \(−4\leq 5x−6\leq 4\)
    Step 3. Solve the compound inequality. \(2\leq 5x\leq 10\)
    \(\frac{2}{5}\leq x\leq 2\)
    Step 4. Graph the solution. .
    Step 5. Write the solution using interval notation. \([\frac{2}{5}, 2]\)
    Check:
    The check is left to you.

    Example \(\PageIndex{20}\)

    Solve \(|2x−1|\leq 5\). Graph the solution and write the solution in interval notation:

    Answer

    The solution is negative 2 is less than or equal to x which is less than or equal to 3. The number line shows closed circles at negative 2 and 3 with shading between the circles. The interval notation is negative 2 to 3 within brackets.

    Example \(\PageIndex{21}\)

    Solve \(|4x−5|\leq 3\). Graph the solution and write the solution in interval notation:

    Answer

    The solution is one-half is less than or equal to x which is less than or equal to 2. The number line shows closed circles at one-half and 2 with shading between the circles. The interval notation is one-half to 2 within brackets.

    SOLVE ABSOLUTE VALUE INEQUALITIES WITH \(<\) OR \(\leq\)

    1. Isolate the absolute value expression.
    2. Write the equivalent compound inequality.

      \[\begin{array} {lll} {|u|<a} &{\quad \text{is equivalent to}} &{−a<u<a} \\ {|u|\leq a} &{\quad \text{is equivalent to}} &{−a\leq u\leq a} \\ \nonumber \end{array}\]

    3. Solve the compound inequality.
    4. Graph the solution
    5. Write the solution using interval notation.

    Solve Absolute Value Inequalities with “Greater Than”

    What happens for absolute value inequalities that have “greater than”? Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line.

    We started with the inequality \(|x|\leq 5\). We saw that the numbers whose distance is less than or equal to five from zero on the number line were \(−5\) and 5 and all the numbers between \(−5\) and 5 (Figure \(\PageIndex{4}\)).

    The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its right and a right bracket at 5 with shading to its left. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x is less than or equal to 5.

    Figure \(\PageIndex{4}\).

    Now we want to look at the inequality \(|x|\geq 5\). Where are the numbers whose distance from zero is greater than or equal to five?

    Again both \(−5\) and 5 are five units from zero and so are included in the solution. Numbers whose distance from zero is greater than five units would be less than \(−5\) and greater than 5 on the number line (Figure \(\PageIndex{5}\)).

    The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its left and a left bracket at 5 with shading to its right. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is greater than or equal to 5, then x is less than or equal to negative 5 or x is greater than or equal to 5.

    Figure \(\PageIndex{5}\).

    In a more general way, we can see that if \(|u|\geq a\), then \(u\leq −a\) or \(u\leq a\). See Figure.

    The figure is a number line with negative a, 0, and a displayed. There is a right bracket at negative a that has shading to its left and a left bracket at a with shading to its right. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is greater than or equal to a, then u is less than or equal to negative a or u is greater than or equal to a.

    Figure \(\PageIndex{6}\).

    This result is summarized here.

    ABSOLUTE VALUE INEQUALITIES WITH \(>\) OR \(\geq\)

    For any algebraic expression, u, and any positive real number, a,

    \[\begin{array} {lll} {\text{if}} &{\quad |u|>a,} &{\quad \text{then } u<−a \text{ or } u>a} \\ {\text{if}} &{\quad |u|\geq a,} &{\quad \text{then } u\leq −a \text{ or } u\geq a} \\ \nonumber \end{array}\]

    Example \(\PageIndex{22}\)

    Solve \(|x|>4\). Graph the solution and write the solution in interval notation.

    Answer
    \(|x|>4\)
    Write the equivalent inequality. \(x<−4\) or \(x>4\)
    Graph the solution. .
    Write the solution using interval notation. \((−\inf ,−4)\cup (4,\inf )\)
    Check:

    To verify, check a value in each section of the number line showing the solution. Choose numbers such as −6,−6, 0, and 7.

    The figure is a number line with a right parenthesis at negative 4 with shading to its left and a left parenthesis at 4 shading to its right. The values negative 6, 0, and 7 are marked with points. The absolute value of negative 6 is greater than negative 4 is true. It does not satisfy the absolute value of x is greater than 4. The absolute value of 0 is greater than 4 is false. It does not satisfy the absolute value of x is greater than 4. The absolute value of 7 is less than 4 is true. It does satisfy the absolute value of x is greater than 4.

    Example \(\PageIndex{23}\)

    Solve \(|x|>2\). Graph the solution and write the solution in interval notation.

    Answer

    The solution is x is less than negative 2 or x is greater than 2. The number line shows an open circle at negative 2 with shading to its left and an open circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 2 to infinity within parentheses.

    Example \(\PageIndex{24}\)

    Solve \(|x|>1\). Graph the solution and write the solution in interval notation.

    Answer

    The solution is x is less than negative 1 or x is greater than 1. The number line shows an open circle at negative 1 with shading to its left and an open circle at 1 with shading to its right. The interval notation is the union of negative infinity to negative 1 within parentheses and 1 to infinity within parentheses.

    Example \(\PageIndex{25}\)

    Solve \(|2x−3|\geq 5\). Graph the solution and write the solution in interval notation.

    Answer
    \(|2x−3|\geq 5\)
    Step 1. Isolate the absolute value expression. It is isolated.
    Step 2. Write the equivalent compound inequality. \(2x−3\leq −5\) or \(2x−3\geq 5\)
    Step 3. Solve the compound inequality. \(2x\leq −2\) or \(2x\geq 8\)
    \(x\leq −1\) or \(x\geq 4\)
    Step 4. Graph the solution. .
    Step 5. Write the solution using interval notation. \((−\inf ,−1]\cup [4,\inf )\)
    Check:
    The check is left to you.

    Example \(\PageIndex{26}\)

    Solve \(|4x−3|\geq 5\). Graph the solution and write the solution in interval notation.

    Answer

    The solution is x is less than or equal to negative one-half or x is greater than or equal 2. The number line shows a closed circle at negative one-half with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative one-half within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis

    Example \(\PageIndex{27}\)

    Solve \(|3x−4|\geq 2\). Graph the solution and write the solution in interval notation.

    Answer

    The solution is x is less than or equal to two-thirds or x is greater than or equal 2. The number line shows a closed circle at two-thirds with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to two-thirds within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.

    SOLVE ABSOLUTE VALUE INEQUALITIES WITH \(>\) OR \(\geq\).

    1. Isolate the absolute value expression.
    2. Write the equivalent compound inequality.

      |u|>ais equivalent tou<−aoru>a|u|\geq ais equivalent tou\leq −aoru\geq a|u|>ais equivalent tou<−aoru>a|u|\geq ais equivalent tou\leq −aoru\geq a

    3. Solve the compound inequality.
    4. Graph the solution
    5. Write the solution using interval notation.

    Solve Applications with Absolute Value

    Absolute value inequalities are often used in the manufacturing process. An item must be made with near perfect specifications. Usually there is a certain tolerance of the difference from the specifications that is allowed. If the difference from the specifications exceeds the tolerance, the item is rejected.

    \[|\text{actual-ideal}|\leq \text{tolerance} \nonumber\]

    Example \(\PageIndex{28}\)

    The ideal diameter of a rod needed for a machine is 60 mm. The actual diameter can vary from the ideal diameter by \(0.075\) mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

    Answer

    \(\begin{array} {ll} {} &{\text{Let }x=\text{ the actual measurement}} \\ {\text{Use an absolute value inequality to express this situation.}} &{|\text{actual-ideal}|\leq \text{tolerance}} \\ {} &{|x−60|\leq 0.075} \\ {\text{Rewrite as a compound inequality.}} &{−0.075\leq x−60\leq 0.075} \\ {\text{Solve the inequality.}} &{59.925\leq x\leq 60.075} \\ {\text{Answer the question.}} &{\text{The diameter of the rod can be between}} \\ {} &{59.925 mm \text{ and } 60.075 mm.} \\ \end{array}\)

    Example \(\PageIndex{29}\)

    The ideal diameter of a rod needed for a machine is 80 mm. The actual diameter can vary from the ideal diameter by 0.009 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

    Answer

    The diameter of the rod can be between 79.991 and 80.009 mm.

    Example \(\PageIndex{30}\)

    The ideal diameter of a rod needed for a machine is 75 mm. The actual diameter can vary from the ideal diameter by 0.05 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

    Answer

    The diameter of the rod can be between 74.95 and 75.05 mm.

    Access this online resource for additional instruction and practice with solving linear absolute value equations and inequalities.

    Key Concepts

    • Absolute Value
      The absolute value of a number is its distance from 0 on the number line.
      The absolute value of a number n is written as \(|n|\) and \(|n|\geq 0\) for all numbers.
      Absolute values are always greater than or equal to zero.
    • Absolute Value Equations
      For any algebraic expression, u, and any positive real number, a,
      \(\begin{array} {ll} {\text{if}} &{\quad |u|=a} \\ {\text{then}} &{\quad u=−a \text{ or } u=a} \\ \end{array}\)
      Remember that an absolute value cannot be a negative number.
    • How to Solve Absolute Value Equations
      1. Isolate the absolute value expression.
      2. Write the equivalent equations.
      3. Solve each equation.
      4. Check each solution.
    • Equations with Two Absolute Values
      For any algebraic expressions, u and v,
      \(\begin{array} {ll} {\text{if}} &{\quad |u|=|v|} \\ {\text{then}} &{\quad u=−v \text{ or } u=v} \\ \end{array}\)
    • Absolute Value Inequalities with \(<\) or \(\leq\)
      For any algebraic expression, u, and any positive real number, a,
      \(\begin{array} {llll} {\text{if}} &{\quad |u|=a} &{\quad \text{then}} &{−a<u<a} \\ {\text{if}} &{\quad |u|\leq a} &{\quad \text{then}} &{−a\leq u\leq a} \\ \end{array}\)
    • How To Solve Absolute Value Inequalities with \(<\) or \(\leq\)
      1. Isolate the absolute value expression.
      2. Write the equivalent compound inequality.
        \(\begin{array} {lll} {|u|<a} &{\quad \text{is equivalent to}} &{\quad −a<u<a} \\ {|u|\leq a} &{\quad \text{is equivalent to}} &{\quad −a\leq u\leq a} \\ \end{array}\)
      3. Solve the compound inequality.
      4. Graph the solution
      5. Write the solution using interval notation
    • Absolute Value Inequalities with \(>\) or \(\geq\)
      For any algebraic expression, u, and any positive real number, a,
      \(\begin{array} {lll} {\text{if}} &{\quad |u|>a,} &{\text{then } u<−a\text{ or }u>a} \\ {\text{if}} &{\quad |u|\geq a,} &{\text{then } u\leq −a\text{ or }u\geq a} \\ \end{array}\)
    • How To Solve Absolute Value Inequalities with \(>\) or \(\geq\)
      1. Isolate the absolute value expression.
      2. Write the equivalent compound inequality.
        \(\begin{array} {lll} {|u|>a} &{\quad \text{is equivalent to}} &{\quad u<−a\text{ or }u>a} \\ {|u|\geq a} &{\quad \text{is equivalent to}} &{\quad u\leq −a\text{ or }u\geq a} \\ \end{array}\)
      3. Solve the compound inequality.
      4. Graph the solution
      5. Write the solution using interval notation

    Section Exercises

    Practice Makes Perfect

    Solve Absolute Value Equations

    In the following exercises, solve.

    ⓐ \(|x|=6\) ⓑ \(|y|=−3\)
    ⓒ \(|z|=0\)

    ⓐ \( |x|=4\) ⓑ \(|y|=−5\)
    ⓒ \(|z|=0\)

    Answer

    ⓐ \(x=4,x=−4\) ⓑ no solution ⓒ \(z=0\)

    ⓐ \(|x|=7 ⓑ \(|y|=−11\)
    ⓒ \(|z|=0\)

    ⓐ \(|x|=3\) ⓑ \(|y|=−1\)
    ⓒ \(|z|=0\)

    Answer

    ⓐ \(x=3,x=−3\) ⓑ no solution ⓒ \(z=0\)

    \(|2x−3|−4=1\)

    \(|4x−1|−3=0\)

    Answer

    \(x=1,x=−\frac{1}{2}\)

    \(|3x−4|+5=7\)

    \(|4x+7|+2=5\)

    Answer

    \(x=−1,x=−\frac{5}{2}\)

    \(4|x−1|+2=10\)

    \(3|x−4|+2=11\)

    Answer

    \(x=7,x=1\)

    \(3|4x−5|−4=11\)

    \(3|x+2|−5=4\)

    Answer

    \(x=1,x=−5\)

    \(−2|x−3|+8=−4\)

    \(−3|x−4|+4=−5\)

    Answer

    \(x=7,x=1\)

    \(|34x−3|+7=2\)

    \(|35x−2|+5=2\)

    Answer

    no solution

    \(|12x+5|+4=1\)

    \(|14x+3|+3=1\)

    Answer

    no solution

    \(|3x−2|=|2x−3|\)

    \(|4x+3|=|2x+1|\)

    Answer

    \(x=−1,x=−\frac{2}{3}\)

    \(|6x−5|=|2x+3|\)

    \(|6−x|=|3−2x|\)

    Answer

    \(x=−3,x=3\)

    Solve Absolute Value Inequalities with “less than”

    In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation.

    \(|x|<5\)

    \(|x|<1\)

    Answer

    The solution is negative 1 is less than x which is less than 1. The number line shows an open circle at negative 1, an open circle at 1, and shading between the circles. The interval notation is negative 1 to 1 within parentheses.

    \(|x|\leq 8\)

    \(|x|\leq 3\)

    Answer

    The solution is negative 3 is less than or equal to x which is less than or equal to 3. The number line shows a closed circle at negative 3, a closed circle at 3, and shading between the circles. The interval notation is negative 3 to 3 within brackets.

    \(|3x−3|\leq 6\)

    \(|2x−5|\leq 3\)

    Answer

    The solution is 1 is less than or equal to x which is less than or equal to 4. The number line shows a closed circle at 1, a closed circle at 4, and shading between the circles. The interval notation is 1 to 4 within brackets.

    \(|2x+3|+5<4\)

    \(|3x−7|+3<1\)

    Answer

    The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

    \(|4x−3|<1\)

    \(|6x−5|<7\)

    Answer

    The solution is negative one-third is less than x which is less than 2. The number line shows an open circle at negative one-half, an open circle at 2, and shading between the circles. The interval notation is negative one-third to 2 within parentheses.

    \(|x−4|\leq −1\)

    \(|5x+1|\leq −2\)

    Answer

    The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

    Solve Absolute Value Inequalities with “greater than”

    In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation.

    \(|x|>3\)

    \(|x|>6\)

    Answer

    The solution is x is less than negative 6 or x is greater than 6. The number line shows an open circle at negative 6 with shading to its left and an open circle at 6 with shading to its right. The interval notation is the union of negative infinity to negative 6 within parentheses and 6 to infinity within parentheses

    \(|x|\geq 2\)

    \(|x|\geq 5\)

    Answer

    The solution is x is less than negative 5 or x is greater than 5. The number line shows an open circle at negative 5 with shading to its left and an open circle at 5 with shading to its right. The interval notation is the union of negative infinity to negative 5 within parentheses and 5 to infinity within parentheses.

    \(|3x−8|>−1\)

    \(|x−5|>−2\)

    Answer

    The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

    \(|3x−2|>4\)

    \(|2x−1|>5\)

    Answer

    The solution is x is less than negative 2 or x is greater than 3. The number line shows an open circle at negative 2 with shading to its left and an open circle at 3 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 3 to infinity within parentheses.

    \(|x+3|\geq 5\)

    \(|x−7|\geq 1\)

    Answer

    The solution is x is less than or equal to 6 or x is greater than or equal to 8. The number line shows a closed circle at 6 with shading to its left and a closed circle at 8 with shading to its right. The interval notation is the union of negative infinity to 6 within parenthesis and a bracket and 8 to infinity within a bracket and a parenthesis.

    \(3|x|+4\geq 1\)

    \(5|x|+6\geq 1\)

    Answer

    The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

    In the following exercises, solve. For each inequality, also graph the solution and write the solution in interval notation.

    \(2|x+6|+4=8\)

    \(|3x−4|\geq 2\)

    Answer

    \(x=4,x=27\)

    \(|6x−5|=|2x+3|\)

    \(|4x−3|<5\)

    Answer

    \(x=3,x=2\)

    \(|2x−5|+2=3\)

    \(|3x+1|−3=7\)

    Answer

    \(x=3,x=−\frac{11}{3}\)

    \(|7x+2|+8<4\)

    \(5|2x−1|−3=7\)

    Answer

    \(x=\frac{3}{2},x=−\frac{1}{2}\)

    \(|x−7|>−3\)

    \(|8−x|=|4−3x|\)

    Answer

    The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

    Solve Applications with Absolute Value

    In the following exercises, solve.

    A chicken farm ideally produces 200,000 eggs per day. But this total can vary by as much as 25,000 eggs. What is the maximum and minimum expected production at the farm?

    An organic juice bottler ideally produces 215,000 bottle per day. But this total can vary by as much as 7,500 bottles. What is the maximum and minimum expected production at the bottling company?

    Answer

    The minimum to maximum expected production is 207,500 to 2,225,000 bottles

    In order to insure compliance with the law, Miguel routinely overshoots the weight of his tortillas by 0.5 gram. He just received a report that told him that he could be losing as much as $100,000 per year using this practice. He now plans to buy new equipment that guarantees the thickness of the tortilla within 0.005 inches. If the ideal thickness of the tortilla is 0.04 inches, what thickness of tortillas will be guaranteed?

    At Lilly’s Bakery, the ideal weight of a loaf of bread is 24 ounces. By law, the actual weight can vary from the ideal by 1.5 ounces. What range of weight will be acceptable to the inspector without causing the bakery being fined?

    Answer

    The acceptable weight is 22.5 to 25.5 ounces.

    Writing Exercises

    Write a graphical description of the absolute value of a number

    In your own words, explain how to solve the absolute value inequality, \(|3x−2|\geq 4\).

    Answer

    Answers will vary.

    Self Check

    ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

    This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve absolute value equations. In row 3, the I can was solve absolute value inequalities with “less than.” In row 4, the I can was solve absolute value inequalities with “greater than.” In row 5, the I can was solve applications with absolute value.

    ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?