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Mathematics LibreTexts

1.7: Absolute Value Equations and Inequalities

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    38273
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    Solving an Absolute Value Equation

    Recall that the absolute value of a real number \(a\), denoted \(|a|\), is defined as the distance between zero (the origin) and the graph of that real number on the number line. For example, \(|−3|=3\) and \(|3|=3\).

    b4aaf730a9b49a73a73f1eee65bc4cc1.png

    In addition, the absolute value of a real number can be defined algebraically as a piecewise function.

    \(| a | = \left\{ \begin{array} { l } { a \text { if } a \geq 0 } \\ { - a \text { if } a < 0 } \end{array} \right.\)

    Given this definition, \(|3| = 3\) and \(|−3| = − (−3) = 3\).Therefore, the equation \(|x| = 3\) has two solutions for \(x\), namely \(\{±3\}\).

    Next, To solve an equation such as \(|2x−6|=8\), notice that the absolute value will be equal to \(8\) if the quantity inside the absolute value bars is \(8\) or \(−8\). This leads to two different equations can be solved independently.

    \(|2x−6|=8\)
    \( \begin{array} {r|r}
    2x-6= 8 & 2x-6= -8  \\
    2x= 14 &  2x= -2  \\
    x= 7    &   x= -1
    \end{array} \)

    Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

    Definition is a green note

    Definition: Absolute Value Equations

    The absolute value of any algebraic expression \(u\), is written as \(|u|\) and is never negative, although the argument \(u\) inside the absolute value bars can be either positive or negative.
    For any real number \(c\), the absolute value equation \(| u | = c\) has the following properties

    • If \(c<0\), then \(|u|=c\) has no solution.
    • If \(c=0\), then \(|u|=c\) has one solution.
    • If \(c>0\), then \(|u|=c\) has two solutions.

     

    Howto: Solve an Absolute Value Equation.

    1. Isolate the absolute value expression \(|u|\)  on one side of the equal sign, producing an equation of the form  \(|u|=c\)
    2. If \(c>0\), write and solve two equations: \(u=c\) and \(u=−c\).
      If \(c=0\), solve the single equation: \(u=0\)
      If  \(c<0\), the equation has no solution.

     

    Example \(\PageIndex{1}\): Solving Absolute Value Equations

    Solve the following absolute value equations:

    1. \(|6x+4|=8\)
    2. \( |3x+4| + 7=1 \)
    3. \(| 2 x + 3 |-1 = 3\)
    4. \(2 |5x − 1| − 3 = 9\).
    5. \(|−5x+10| + 3=3\)

    Solution:

    1. The absolute value expression is isolated already, so rewrite it as two separate equations and solve them. 

    \(|6x+4|=8\)
    \( \begin{array} {r|r}
    6x+4= 8  &  6x+4= -8 \\
    6x= 4  & 6x= -12  \\
    x= \dfrac{2}{3}   &  x= -2  \\
    \end{array} \)

    \( \quad \) The solution set is \( \left\{  \tfrac{2}{3}, -2 \right\} \)

    1. Isolate the absolute value expression in the equation.

     \( |3x+4| + 7=1 \) 
    \( |3x+4| =-6 \)

    \( \quad \) An absolute value cannot be negative. Therefore the equation does not have a solution.
    \( \quad \) The solution set is \( \{ \:\: \} \)

    1. Isolate the absolute value expression in the equation.

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

    \( \quad \) Rewrite the absolute value as two separate equations and solve them.

    \( \begin{array} {r|r}
    2 x + 3 = - 4   &  2 x + 3 = 4 \\
    2 x = - 7  & 2 x = 1  \\
    - \frac { 7 } { 2 }   &  x = \frac { 1 } { 2 } \\
    \end{array} \)

    \( \quad \) The solution set is \( \{  - \frac { 7 } { 2 }, \frac { 1 } { 2 } \} \)

    1. Isolate the absolute value expression in the equation.

    \(2 |5x − 1| − 3 = 9\)
    \(2 |5x − 1|  = 12\)
    \( |5x − 1|  = 6\)

    \( \quad \) Rewrite the absolute value as two separate equations and solve them.

    \( \begin{array} {r|r}
     5 x - 1 = - 6 &  5 x - 1 =  6 \\
    5 x = - 5  & 5 x = 7  \\
    x = - 1   & x = \frac { 7 } { 5 }  \\
    \end{array} \)

    \( \quad \) The solution set is \( \{  -1, \frac { 7 } { 5 } \} \)

    1. Isolate the absolute value expression in \(|−5x+10| + 3=3\). 

    \(|−5x+10| + 3=3\)
    \(|−5x+10| =0\)

    \( \quad \) The equation is set equal to zero, so we have to write only one equation.

    \[\begin{align*} -5x+10&= 0\\ -5x&= -10\\ x&= 2 \end{align*}\]

    \( \quad \) The solution set is \( \{ 2 \} \)

     Try It \(\PageIndex{1}\)

    Solve.

    1. \(|1−4x|+8=13\).
    2. \(2 - 7 | x + 4 | = - 12\).
    Answer

    a, \(x=−1, x=\dfrac{3}{2}\) \( \qquad \) b. \(-6, -2\)

    If two absolute value expressions are equal, then the arguments can either be the exact same value, or be the exact same value but with opposite signs.

    Example \(\PageIndex{2}\):

    Solve: \(| 2 x - 5 | = | x - 4 |\).

    Solution

    Set \(2x-5\) equal to \(\pm ( x - 4 )\) and then solve each linear equation.

    \(\begin{array} { c } { | 2 x - 5 | = | x - 4 | } \\ { 2 x - 5 = - ( x - 4 ) \:\: \text { or }\:\: 2 x - 5 = + ( x - 4 ) } \\ { 2 x - 5 = - x + 4 }\quad\quad\quad 2x-5=x-4 \\ { 3 x = 9 }\quad\quad\quad\quad\quad\quad \quad\quad x=1 \\ { x = 3 \quad\quad\quad\quad\quad\quad\quad\quad\quad\:\:\:\:} \end{array}\)

    The solution set is \( \{ 1, 3 \} \)

     Try It \(\PageIndex{2}\)

    Solve: \(| x + 10 | = | 3 x - 2 |\).

    Answer

    \(-2, 6\)

    Absolute Value Inequalities

    Consider the solutions to the inequality \(| x | \leq 3\). The absolute value of a number represents the distance from the origin. Therefore, this inequality describes all numbers whose distance from zero is less than or equal to \(3\). We can graph this solution set by shading all such numbers.

    54d7fb08641f26cbbcabbdf76cd2ada7.png

    This solution set is expressed using set notation or interval notation as follows:

    \(\begin{array} { c } { \{ x | - 3 \leq x \leq 3 \} \color{Cerulean} { Set\: Notation } } \\ { [ - 3,3 ] \quad \color{Cerulean}{ Interval \:Notation } } \end{array}\)

    In contrast, examine solutions to the inequality \(| x | \geq 3\). This inequality describes all numbers whose distance from the origin is greater than or equal to \(3\). On a graph, we can shade all such numbers.

    ee5443c82bae2d950182100b7ab9ff13.png

    This solution set is expressed using using set notation and interval notation as follows:

    \(\begin{array} { l } { \{ x | x \leq - 3 \text { or } x \geq 3 \} \:\:\color{Cerulean} { Set\: Notation } } \\ { ( - \infty , - 3 ] \cup [ 3 , \infty ) \:\:\color{Cerulean} { Interval\: Notation } } \end{array}\)

    Howto: Solve a Linear Absolute Value Inequality.

    Given an algebraic expression \( u \) and a positive number \(c \),

    1. Isolate the absolute value expression \(|u|\)  on left side of the inequality symbol
    2. The solution of \( | u |  <  c \) is a single interval composed of the numbers that satisfy the compound inequality \( -c < u < c \).
    3. The solution of \( | u |  >  c \) is a combination of two separate intervals that satisfy the two separate inequalities \( u < -c \) or \( u > c \). NOTE that  \( | u |  >  c \) CANNOT be written as \( -c < u > c \).
    4. These rules are valid if \( < \) is replaced with \( \le \) and \( > \) is replaced with \( \ge \).

    Example \(\PageIndex{3}\):

    Solve and graph the solution set: \(|x+2|<3\).

    Solution

    The absolute value expression is already isolated. Bound the argument \(x+2\) by \(−3\) and \(3\) in a compound inequality and solve.

    \(\begin{array} { c } { | x + 2 | < 3 } \\ { - 3 < x + 2 < 3 } \\ { - 3 \color{Cerulean}{- 2}\color{Black}{ <} x + 2 \color{Cerulean}{- 2}\color{Black}{ <} 3 \color{Cerulean}{- 2} } \\ { - 5 < x < 1 } \end{array}\)

    Here we use open dots to indicate strict inequalities on the graph as follows.

    d198f40ad4498592e3fb836345d5d10d.png

    Using interval notation, \((−5,1)\).

    Example \(\PageIndex{4}\):

    Solve: \(4 |x + 3| − 7 ≤ 5\).

    Solution

    Isolate the absolute value expression.

    \(\begin{array} { c } { 4 | x + 3 | - 7 \leq 5 } \\ { 4 | x + 3 | \leq 12 } \\ { | x + 3 | \leq 3 } \end{array}\)

    Rewrite the absolute value inequality as a compound inequality and solve.

    \(\begin{array} { c } { - 3 \leq x + 3 \leq 3 } \end{array}\)

    \(\begin{aligned}  - 3 \color{Cerulean}{- 3} \color{Black}{ \leq} x + 3 \color{Cerulean}{- 3} & \color{Black}{ \leq} 3 \color{Cerulean}{- 3} \\ - 6 \leq x \leq 0 \end{aligned}\)

    Shade the solutions on a number line and present the answer in interval notation. Here we use closed dots to indicate inclusive inequalities on the graph as follows:

    8abae6a57d4658e1e28b81d29a8ba467.png

    Using interval notation, \([−6,0]\)

     Try It \(\PageIndex{4}\)

    Solve and graph the solution set: \(3 + | 4 x - 5 | < 8\).

    Answer

    Interval notation: \((0, \frac{5}{2})\)

    53fbc057eea3477b5709c94560299f26.png

    Example \(\PageIndex{5}\):

    Solve and graph the solution set: \(|x+2|>3\).

    Solution

    The absolute value expression is already isolated. The argument \(x+2\) must be less than \(−3\) or greater than \(3\).

    \(\begin{array} { c } { | x + 2 | > 3 } \\ { x + 2 < - 3 \quad \text { or } \quad x + 2 > 3 } \\ { x < - 5 }\quad\quad\quad\quad\quad\: x>1 \end{array}\)

    12b84b5b3331a000842bb1191680226e.png

    Using interval notation, \((−∞,−5)∪(1,∞)\).

    Example \(\PageIndex{6}\):

    Solve: \(13 - 2 |4x − 7| ≤ 3\).

    Solution

    Isolate the absolute value expression. Notice in the last step that division by \(-2\) changes the inequality symbol.

    \(\begin{array} { r } { 13 - 2 | 4 x - 7 | \leq 3 } \\ { -2 | 4 x - 7 | \leq -10 } \\ { | 4 x - 7 | \geq 5 } \end{array}\)

    Next, apply the theorem and rewrite the absolute value inequality as a compound inequality with two separate intervals and solve.

    \(\begin{array}  4 x - 7 \leq - 5 \quad \text { or } \quad 4 x - 7 \geq 5 \end{array}\)

    \(\begin{array} { l }  \quad\:\:{ 4 x \leq 2 } \quad\quad\quad\:\:\: 4x\geq 12\\
    \quad\:\:{  x \leq \frac { 2 } { 4 } } \quad\quad\quad\quad x\geq \frac { 12 } { 4 } \\
    \quad\:\:{ x \leq \frac { 1 } { 2 } }   \quad\quad\quad\quad x\geq 3 \end{array}\)

    Shade the solutions on a number line and present the answer using interval notation.

    06fafde1bcd818b99ee494ab4c092fc0.png

    Using interval notation, \((−∞, \tfrac { 1 } { 2 }]∪[3,∞)\)

     Try It \(\PageIndex{6}\)

    Solve and graph: \(3 | 6 x + 5 | - 2 > 13\).

    Answer

    Using interval notation, \(\left( - \infty , - \frac { 5 } { 3 } \right) \cup ( 0 , \infty )\)

    8775b7be1a775881d4ce3e89ac626c6c.png

    Up to this point, the solution sets of linear absolute value inequalities have consisted of a single bounded interval or two unbounded intervals. This is not always the case.

    Example \(\PageIndex{7}\):

    Solve and graph: \(|2x−1|+5>2\).

    Solution

    Begin by isolating the absolute value.

    \(\begin{array} { c } { | 2 x - 1 | + 5 > 2 } \\ { | 2 x - 1 | > - 3 } \end{array}\)

    Notice that we have an absolute value greater than a negative number. For any real number x the absolute value of the argument will always be positive. Hence, any real number will solve this inequality.

    1ae423963434b72a27df0ffb19ce6b24.png

    All real numbers, \(ℝ\).

    Example \(\PageIndex{8}\):

    Solve and graph: \(|x+1|+4≤3\).

    Solution

    Begin by isolating the absolute value.

    \(\begin{array} { l } { | x + 1 | + 4 \leq 3 } \\ { | x + 1 | \leq - 1 } \end{array}\)

    In this case, we can see that the isolated absolute value is to be less than or equal to a negative number. Again, the absolute value will always be positive; hence, we can conclude that there is no solution.

    Answer: \(Ø\) or \( \{ \:\: \} \)

     

     

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