2.6: Solving Absolute Value Equations and Inequalities (2024)

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    • 2.6: Solving Absolute Value Equations and Inequalities (1)
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    Learning Objectives

    • Review the definition of absolute value.
    • Solve absolute value equations.
    • Solve absolute value inequalities.

    Absolute Value Equations

    Recall that the absolute value63 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 \(|3|=3\).

    2.6: Solving Absolute Value Equations and Inequalities (2)

    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 \(\{±3\}\). In general, given any algebraic expression \(X\) and any positive number \(p\):

    \(\text{If}\: | X | = p \text { then } X = - p \text { or } X = p\)

    In other words, the argument of the absolute value64 \(X\) can be either positive or negative \(p\). Use this theorem to solve absolute value equations algebraically.

    Example \(\PageIndex{1}\):

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

    Solution

    In this case, the argument of the absolute value is \(x+2\) and must be equal to \(3\) or \(−3\).

    2.6: Solving Absolute Value Equations and Inequalities (3)

    Therefore, to solve this absolute value equation, set \(x+2\) equal to \(±3\) and solve each linear equation as usual.

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

    Answer:

    The solutions are \(−5\) and \(1\).

    To visualize these solutions, graph the functions on either side of the equal sign on the same set of coordinate axes. In this case, \(f (x) = |x + 2|\) is an absolute value function shifted two units horizontally to the left, and \(g (x) = 3\) is a constant function whose graph is a horizontal line. Determine the \(x\)-values where \(f (x) = g (x)\).

    2.6: Solving Absolute Value Equations and Inequalities (4)

    From the graph we can see that both functions coincide where \(x = −5\) and \(x = 1\). The solutions correspond to the points of intersection.

    Example \(\PageIndex{2}\):

    Solve: \(| 2 x + 3 | = 4\).

    Solution

    Here the argument of the absolute value is \(2x+3\) and can be equal to \(-4\) or \(4\).

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

    Check to see if these solutions satisfy the original equation.

    Check \(x=-\frac{7}{2}\) Check \(x=\frac{1}{2}\)
    \(\begin{aligned} | 2 x + 3 | & = 4 \\ \left| 2 \left( \color{Cerulean}{- \frac { 7 } { 2 }} \right) + 3 \right| & = 4 \\ | - 7 + 3 | & = 4 \\ | - 4 | & = 4 \\ 4 & = 4 \:\:\color{Cerulean}{✓} \end{aligned}\) \(\begin{array} { r } { | 2 x + 3 | = 4 } \\ { \left| 2 \left( \color{Cerulean}{\frac { 1 } { 2 }} \right) + 3 \right| = 4 } \\ { | 1 + 3 | = 4 } \\ { | 4 | = 4 } \\ { 4 = 4 } \:\:\color{Cerulean}{✓} \end{array}\)
    Table \(\PageIndex{1}\)

    Answer:

    The solutions are \(-\frac{7}{2}\) and \(\frac{1}{2}\).

    To apply the theorem, the absolute value must be isolated. The general steps for solving absolute value equations are outlined in the following example.

    Example \(\PageIndex{3}\):

    Solve: \(2 |5x − 1| − 3 = 9\).

    Solution

    Step 1: Isolate the absolute value to obtain the form \(|X| = p\).

    \(\begin{aligned} 2 | 5 x - 1 | - 3 & = 9 \:\:\:\color{Cerulean} { Add\: 3\: to\: both\: sides. } \\ 2 | 5 x - 1 | & = 12 \:\:\color{Cerulean} { Divide\: both\: sides\: by\: 2 } \\ | 5 x - 1 | & = 6 \end{aligned}\)

    Step 2: Set the argument of the absolute value equal to \(±p\). Here the argument is \(5x − 1\) and \(p = 6\).

    \(5 x - 1 = - 6 \text { or } 5 x - 1 = 6\)

    Step 3: Solve each of the resulting linear equations.

    \(\begin{array} { r l } { 5 x - 1 = - 6 \quad\:\:\text { or } \quad\quad5 x - 1 } & { \:\:\:\:\:\:= 6 } \\ { 5 x = - 5 }\quad\:\quad\quad\quad\quad\quad\quad\: & { 5 x = 7 } \\ { x = - 1 } \quad\quad\quad\quad\quad\quad\quad\:\:& { x = \frac { 7 } { 5 } } \end{array}\)

    Step 4: Verify the solutions in the original equation.

    Check \(x=-1\) Check \(x=\frac{7}{5}\)
    \(\begin{aligned} 2 | 5 x - 1 | - 3 & = 9 \\ 2 | 5 ( \color{Cerulean}{- 1}\color{Black}{ )} - 1 | - 3 & = 9 \\ 2 | - 5 - 1 | - 3 & = 9 \\ 2 | - 6 | - 3 & = 9 \\ 12 - 3 & = 9 \\ 9 & = 9 \color{Cerulean}{✓}\end{aligned}\) \(\begin{aligned} 2 | 5 x - 1 | - 3 & = 9 \\ 2 \left| 5 \left( \color{Cerulean}{\frac { 7 } { 5 }} \right) - 1 \right| - 3 & = 9 \\ 2 | 7 - 1 | - 3 & = 9 \\ 2 | 6 | - 3 & = 9 \\ 12 - 3 & = 9 \\ 9 & = 9 \color{Cerulean}{✓} \end{aligned}\)
    Table \(\PageIndex{2}\)

    Answer:

    The solutions are \(-1\) and \(\frac{7}{5}\)

    Exercise \(\PageIndex{1}\)

    Solve: \(2 - 7 | x + 4 | = - 12\).

    Answer

    \(-6, -2\)

    www.youtube.com/v/G0EjbqreYmU

    Not all absolute value equations will have two solutions.

    Example \(\PageIndex{4}\):

    Solve: \(| 7 x - 6 | + 3 = 3\).

    Solution

    Begin by isolating the absolute value.

    \(\begin{array} { l } { | 7 x - 6 | + 3 = 3 \:\:\:\color{Cerulean} { Subtract\: 3\: on\: both\: sides.} } \\ { \quad | 7 x - 6 | = 0 } \end{array}\)

    Only zero has the absolute value of zero, \(|0| = 0\). In other words, \(|X| = 0\) has one solution, namely \(X = 0\). Therefore, set the argument \(7x − 6\) equal to zero and then solve for \(x\).

    \(\begin{aligned} 7 x - 6 & = 0 \\ 7 x & = 6 \\ x & = \frac { 6 } { 7 } \end{aligned}\)

    Geometrically, one solution corresponds to one point of intersection.

    2.6: Solving Absolute Value Equations and Inequalities (5)

    Answer:

    The solution is \(\frac{6}{7}\).

    Example \(\PageIndex{5}\):

    Solve: \(|x+7|+5=4\).

    Solution

    Begin by isolating the absolute value.

    \(\begin{aligned} | x + 7 | + 5 & = 4 \:\:\color{Cerulean} { Subtract \: 5\: on\: both\: sides.} \\ | x + 7 | & = - 1 \end{aligned}\)

    In this case, we can see that the isolated absolute value is equal to a negative number. Recall that the absolute value will always be positive. Therefore, we conclude that there is no solution. Geometrically, there is no point of intersection.

    2.6: Solving Absolute Value Equations and Inequalities (6)

    Answer:

    There is no solution, \(Ø\).

    If given an equation with two absolute values of the form \(| a | = | b |\), then \(b\) must be the same as \(a\) or opposite. For example, if \(a=5\), then \(b = \pm 5\) and we have:

    \(| 5 | = | - 5 | \text { or } | 5 | = | + 5 |\)

    In general, given algebraic expressions \(X\) and \(Y\):

    \(\text{If} | X | = | Y | \text { then } X = - Y \text { or } X = Y\).

    In other words, if two absolute value expressions are equal, then the arguments can be the same or opposite.

    Example \(\PageIndex{6}\):

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

    Solution

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

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

    To check, we substitute these values into the original equation.

    Check \(x=1\) Check \(x=3\)
    \(\begin{aligned} | 2 x - 5 | & = | x - 4 | \\ | 2 ( \color{Cerulean}{1}\color{Black}{ )} - 5 | & = | ( \color{Cerulean}{1}\color{Black}{ )} - 4 | \\ | - 3 | & = | - 3 | \\ 3 & = 3 \color{Cerulean}{ ✓}\end{aligned}\) \(\begin{aligned} | 2 x - 5 | & = | x - 4 | \\ | 2 ( \color{Cerulean}{3}\color{Black}{ )} - 5 | & = | ( \color{Cerulean}{3}\color{Black}{ )} - 4 | \\ | 1 | & = | - 1 | \\ 1 & = 1 \color{Cerulean}{✓}\end{aligned}\)
    Table \(\PageIndex{3}\)

    As an exercise, use a graphing utility to graph both \(f(x)= |2x-5|\) and \(g(x)=|x-4|\) on the same set of axes. Verify that the graphs intersect where \(x\) is equal to \(1\) and \(3\).

    Answer:

    The solutions are \(1\) and \(3\).

    Exercise \(\PageIndex{2}\)

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

    Answer

    \(-2, 6\)

    www.youtube.com/v/CskWmsQCBMU

    Absolute Value Inequalities

    We begin by examining the solutions to the following inequality:

    \(| x | \leq 3\)

    The absolute value of a number represents the distance from the origin. Therefore, this equation 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.

    2.6: Solving Absolute Value Equations and Inequalities (7)

    Certainly we can see that there are infinitely many solutions to \(|x|≤3\) bounded by \(−3\) and \(3\). Express this solution set 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 this text, we will choose to express solutions in interval notation. In general, given any algebraic expression \(X\) and any positive number \(p\):

    \(\text{If} | X | \leq p \text { then } - p \leq X \leq p\).

    This theorem holds true for strict inequalities as well. In other words, we can convert any absolute value inequality involving "less than" into a compound inequality which can be solved as usual.

    Example \(\PageIndex{7}\):

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

    Solution

    Bound the argument \(x+2\) by \(−3\) and \(3\) 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.

    2.6: Solving Absolute Value Equations and Inequalities (8)

    Answer:

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

    The solution to \(| x + 2 | < 3\) can be interpreted graphically if we let \(f ( x ) = | x + 2 |\) and \(g(x)=3\) and then determine where \(f ( x ) < g ( x )\) by graphing both \(f\) and \(g\) on the same set of axes.

    2.6: Solving Absolute Value Equations and Inequalities (9)

    The solution consists of all \(x\)-values where the graph of \(f\) is below the graph of \(g\). In this case, we can see that \(|x + 2| < 3\) where the \(x\)-values are between \(−5\) and \(1\). To apply the theorem, we must first isolate the absolute value.

    Example \(\PageIndex{8}\):

    Solve: \(4 |x + 3| − 7 ≤ 5\).

    Solution

    Begin by isolating the absolute value.

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

    Next, apply the theorem and rewrite the absolute value inequality as a compound inequality.

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

    Solve.

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

    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:

    2.6: Solving Absolute Value Equations and Inequalities (10)

    Answer:

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

    Exercise \(\PageIndex{3}\)

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

    Answer

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

    2.6: Solving Absolute Value Equations and Inequalities (11)

    www.youtube.com/v/sX6ppL2Fbq0

    Next, we examine the solutions to an inequality that involves "greater than," as in the following example:

    \(| x | \geq 3\)

    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.

    2.6: Solving Absolute Value Equations and Inequalities (12)

    There are infinitely many solutions that can be expressed using set notation and interval notation as follows:

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

    In general, given any algebraic expression \(X\) and any positive number \(p\):

    \(\text{If} | X | \geq p \text { then } X \leq - p \text { or } X \geq p\).

    The theorem holds true for strict inequalities as well. In other words, we can convert any absolute value inequality involving “greater than” into a compound inequality that describes two intervals.

    Example \(\PageIndex{9}\):

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

    Solution

    The argument \(x+2\) must be less than \(−3\) or greater than \(3\).

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

    2.6: Solving Absolute Value Equations and Inequalities (13)

    Answer:

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

    The solution to \(|x + 2| > 3\) can be interpreted graphically if we let \(f (x) = |x + 2|\) and \(g (x) = 3\) and then determine where \(f(x) > g (x)\) by graphing both \(f\) and \(g\) on the same set of axes.

    2.6: Solving Absolute Value Equations and Inequalities (14)

    The solution consists of all \(x\)-values where the graph of \(f\) is above the graph of \(g\). In this case, we can see that \(|x + 2| > 3\) where the \(x\)-values are less than \(−5\) or are greater than \(1\). To apply the theorem we must first isolate the absolute value.

    Example \(\PageIndex{10}\):

    Solve: \(3 + 2 |4x − 7| ≥ 13\).

    Solution

    Begin by isolating the absolute value.

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

    Next, apply the theorem and rewrite the absolute value inequality as a compound inequality.

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

    Solve.

    \(\begin{array} { l } { 4 x - 7 \leq - 5 \text { or } 4 x - 7 \geq 5 } \\ \quad\:\:\:\:{ 4 x \leq 2 } \quad\quad\quad\:\:\: 4x\geq 12\\ \quad\:\:\:\:{ x \leq \frac { 2 } { 4 } } \quad\quad\quad\quad x\geq 3 \\ \quad\quad{ x \leq \frac { 1 } { 2 } } \end{array}\)

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

    2.6: Solving Absolute Value Equations and Inequalities (15)

    Answer:

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

    Exercise \(\PageIndex{4}\)

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

    Answer

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

    2.6: Solving Absolute Value Equations and Inequalities (16)

    www.youtube.com/v/P6HjRz6W4F4

    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 \(\PageIndex{11}\):

    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.

    2.6: Solving Absolute Value Equations and Inequalities (17)

    Geometrically, we can see that \(f(x)=|2x−1|+5\) is always greater than \(g(x)=2\).

    2.6: Solving Absolute Value Equations and Inequalities (18)

    Answer:

    All real numbers, \(ℝ\).

    Example \(\PageIndex{12}\):

    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.

    Geometrically, we can see that \(f(x)=|x+1|+4\) is never less than \(g(x)=3\).

    2.6: Solving Absolute Value Equations and Inequalities (19)

    Answer: \(Ø\)

    In summary, there are three cases for absolute value equations and inequalities. The relations \(=, <, \leq, > \) and \(≥\) determine which theorem to apply.

    Case 1: An absolute value equation:

    \(\begin{array} { c } { \text { If } | X | = p } \\ { \text { then } X = - p \text { or } X = p } \end{array}\)
    2.6: Solving Absolute Value Equations and Inequalities (20)

    Case 2: An absolute value inequality involving "less than."

    \(\begin{array} { c } { \text { If } | X | \leq p } \\ { \text { then } - p \leq X \leq p } \end{array}\)
    2.6: Solving Absolute Value Equations and Inequalities (21)

    Case 3: An absolute value inequality involving "greater than."

    \(\begin{array} { c } { \text { If } | X | \geq p } \\ { \text { then } X \leq - p \text { or } X \geq p } \end{array}\)
    2.6: Solving Absolute Value Equations and Inequalities (22)

    Key Takeaways

    • To solve an absolute value equation, such as \(|X| = p\), replace it with the two equations \(X = −p\) and \(X = p\) and then solve each as usual. Absolute value equations can have up to two solutions.
    • To solve an absolute value inequality involving “less than,” such as \(|X| ≤ p\), replace it with the compound inequality \(−p ≤ X ≤ p\) and then solve as usual.
    • To solve an absolute value inequality involving “greater than,” such as \(|X| ≥ p\), replace it with the compound inequality \(X ≤ −p\) or \(X ≥ p\) and then solve as usual.
    • Remember to isolate the absolute value before applying these theorems.

    Exercise \(\PageIndex{5}\)

    1. \(|x| = 9\)
    2. \(|x| = 1\)
    3. \(|x − 7| = 3\)
    4. \(|x − 2| = 5\)
    5. \(|x + 12| = 0\)
    6. \(|x + 8| = 0\)
    7. \(|x + 6| = −1\)
    8. \(|x − 2| = −5\)
    9. \(|2y − 1| = 13\)
    10. \(|3y − 5| = 16\)
    11. \(|−5t + 1| = 6\)
    12. \(|−6t + 2| = 8\)
    13. \(\left| \frac { 1 } { 2 } x - \frac { 2 } { 3 } \right| = \frac { 1 } { 6 }\)
    14. \(\left| \frac { 2 } { 3 } x + \frac { 1 } { 4 } \right| = \frac { 5 } { 12 }\)
    15. \(|0.2x + 1.6| = 3.6\)
    16. \(|0.3x − 1.2| = 2.7\)
    17. \(| 5 (y − 4) + 5| = 15\)
    18. \(| 2 (y − 1) − 3y| = 4\)
    19. \(|5x − 7| + 3 = 10\)
    20. \(|3x − 8| − 2 = 6\)
    21. \(9 + |7x + 1| = 9\)
    22. \(4 − |2x − 3| = 4\)
    23. \(3 |x − 8| + 4 = 25\)
    24. \(2 |x + 6| − 3 = 17\)
    25. \(9 + 5 |x − 1| = 4\)
    26. \(11 + 6 |x − 4| = 5\)
    27. \(8 − 2 |x + 1| = 4\)
    28. \(12 − 5 |x − 2| = 2\)
    29. \(\frac{1}{2} |x − 5| − \frac{2}{3} = −\frac{1}{6}\)
    30. \(\frac { 1 } { 3 } \left| x + \frac { 1 } { 2 } \right| + 1 = \frac { 3 } { 2 }\)
    31. \(−2 |7x + 1| − 4 = 2\)
    32. \(−3 |5x − 3| + 2 = 5\)
    33. \(1.2 |t − 2.8| − 4.8 = 1.2\)
    34. \(3.6 | t + 1.8| − 2.6 = 8.2\)
    35. \(\frac{1}{2} |2 (3x − 1) − 3| + 1 = 4\)
    36. \(\frac{2}{3} |4 (3x + 1) − 1| − 5 = 3\)
    37. \(|5x − 7| = |4x − 2|\)
    38. \(|8x − 3| = |7x − 12|\)
    39. \(|5y + 8| = |2y + 3|\)
    40. \(|7y + 2| = |5y − 2|\)
    41. \(|5 (x − 2)| = |3x|\)
    42. \(|3 (x + 1)| = |7x|\)
    43. \(\left| \frac { 2 } { 3 } x + \frac { 1 } { 2 } \right| = \left| \frac { 3 } { 2 } x - \frac { 1 } { 3 } \right|\)
    44. \(\left| \frac { 3 } { 5 } x - \frac { 5 } { 2 } \right| = \left| \frac { 1 } { 2 } x + \frac { 2 } { 5 } \right|\)
    45. \(|1.5t − 3.5| = |2.5t + 0.5|\)
    46. \(|3.2t − 1.4| = |1.8t + 2.8|\)
    47. \(|5 − 3 (2x + 1)| = |5x + 2|\)
    48. \(|3 − 2 (3x − 2)| = |4x − 1|\)
    Answer

    1. \(−9, 9\)

    3. \(4, 10\)

    5. \(−12\)

    7. \(Ø\)

    9. \(−6, 7\)

    11. \(−1, \frac{7}{5}\)

    13. \(1, \frac{5}{3}\)

    15. \(−26, 10\)

    17. \(0, 6\)

    19. \(0, \frac{14}{5}\)

    21. \(−\frac{1}{7}\)

    23. \(1, 15\)

    25. \(Ø\)

    27. \(−3, 1\)

    29. \(4, 6\)

    31. \(Ø\)

    33. \(−2.2, 7.8\)

    35. \(−\frac{1}{6}, \frac{11}{6}\)

    37. \(1, 5\)

    39. \(−\frac{5}{3}, −\frac{11}{7}\)

    41. \(\frac{5}{4} , 5\)

    43. \(−\frac{1}{13} , 1\)

    45. \(−4, 0.75\)

    47. \(0, 4\)

    Exercise \(\PageIndex{6}\)

    Solve and graph the solution set. In addition, give the solution set in interval notation.

    1. Solve for \(x: p |ax + b| − q = 0\)
    2. Solve for \(x: |ax + b| = |p + q|\)
    Answer

    1. \(x = \frac { - b q \pm q } { a p }\)

    Exercise \(\PageIndex{7}\)

    Solve and graph the solution set. In addition, give the solution set in interval notation.

    1. \(|x| < 5\)
    2. \(|x| ≤ 2\)
    3. \(|x + 3| ≤ 1\)
    4. \(|x − 7| < 8\)
    5. \(|x − 5| < 0\)
    6. \(|x + 8| < −7\)
    7. \(|2x − 3| ≤ 5\)
    8. \(|3x − 9| < 27\)
    9. \(|5x − 3| ≤ 0\)
    10. \(|10x + 5| < 25\)
    11. \(\left| \frac { 1 } { 3 } x - \frac { 2 } { 3 } \right| \leq 1\)
    12. \(\left| \frac { 1 } { 12 } x - \frac { 1 } { 2 } \right| \leq \frac { 3 } { 2 }\)
    13. \(|x| ≥ 5\)
    14. \(|x| > 1\)
    15. \(|x + 2| > 8\)
    16. \(|x − 7| ≥ 11\)
    17. \(|x + 5| ≥ 0\)
    18. \(|x − 12| > −4\)
    19. \(|2x − 5| ≥ 9\)
    20. \(|2x + 3| ≥ 15\)
    21. \(|4x − 3| > 9\)
    22. \(|3x − 7| ≥ 2\)
    23. \(\left| \frac { 1 } { 7 } x - \frac { 3 } { 14 } \right| > \frac { 1 } { 2 }\)
    24. \(\left| \frac { 1 } { 2 } x + \frac { 5 } { 4 } \right| > \frac { 3 } { 4 }\)
    Answer

    1. \(( - 5,5 )\);

    2.6: Solving Absolute Value Equations and Inequalities (23)

    3. \([ - 4 , - 2 ]\);

    2.6: Solving Absolute Value Equations and Inequalities (24)

    5. \(\emptyset\);

    2.6: Solving Absolute Value Equations and Inequalities (25)

    7. \([ - 1,4 ]\);

    2.6: Solving Absolute Value Equations and Inequalities (26)

    9. \(\left\{ \frac { 3 } { 5 } \right\}\);

    2.6: Solving Absolute Value Equations and Inequalities (27)

    11. \([ - 1,5 ]\);

    2.6: Solving Absolute Value Equations and Inequalities (28)

    13. \(( - \infty , - 5 ] \cup [ 5 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (29)

    15. \(( - \infty , - 10 ) \cup ( 6 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (30)

    17. \(\mathbb { R }\);

    2.6: Solving Absolute Value Equations and Inequalities (31)

    19. \(( - \infty , - 2 ] \cup [ 7 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (32)

    21. \(\left( - \infty , - \frac { 3 } { 2 } \right) \cup ( 3 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (33)

    23. \(( - \infty , - 2 ) \cup ( 5 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (34)

    Exercise \(\PageIndex{8}\)

    Solve and graph the solution set.

    1. \(|3 (2x − 1)| > 15\)
    2. \(|3 (x − 3)| ≤ 21\)
    3. \(−5 |x − 4| > −15\)
    4. \(−3 |x + 8| ≤ −18\)
    5. \(6 − 3 |x − 4| < 3\)
    6. \(5 − 2 |x + 4| ≤ −7\)
    7. \(6 − |2x + 5| < −5\)
    8. \(25 − |3x − 7| ≥ 18\)
    9. \(|2x + 25| − 4 ≥ 9\)
    10. \(|3 (x − 3)| − 8 < −2\)
    11. \(2 |9x + 5| + 8 > 6\)
    12. \(3 |4x − 9| + 4 < −1\)
    13. \(5 |4 − 3x| − 10 ≤ 0\)
    14. \(6 |1 − 4x| − 24 ≥ 0\)
    15. \(3 − 2 |x + 7| > −7\)
    16. \(9 − 7 |x − 4| < −12\)
    17. \(|5 (x − 4) + 5| > 15\)
    18. \(|3 (x − 9) + 6| ≤ 3\)
    19. \(\left| \frac { 1 } { 3 } ( x + 2 ) - \frac { 7 } { 6 } \right| - \frac { 2 } { 3 } \leq - \frac { 1 } { 6 }\)
    20. \(\left| \frac { 1 } { 10 } ( x + 3 ) - \frac { 1 } { 2 } \right| + \frac { 3 } { 20 } > \frac { 1 } { 4 }\)
    21. \(12 + 4 |2x − 1| ≤ 12\)
    22. \(3 − 6 |3x − 2| ≥ 3\)
    23. \(\frac{1}{2} |2x − 1| + 3 < 4\)
    24. 2 |\frac{1}{2} x + \frac{2}{3} | − 3 ≤ −1\)
    25. \(7 − |−4 + 2 (3 − 4x)| > 5\)
    26. \(9 − |6 + 3 (2x − 1)| ≥ 8\)
    27. \(\frac { 3 } { 2 } - \left| 2 - \frac { 1 } { 3 } x \right| < \frac { 1 } { 2 }\)
    28. \(\frac { 5 } { 4 } - \left| \frac { 1 } { 2 } - \frac { 1 } { 4 } x \right| < \frac { 3 } { 8 }\)
    Answer

    1. \(( - \infty , - 2 ) \cup ( 3 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (35)

    3. \(( 1,7 )\);

    2.6: Solving Absolute Value Equations and Inequalities (36)

    5. \(( - \infty , 3 ) \cup ( 5 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (37)

    7. \(( - \infty , - 8 ) \cup ( 3 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (38)

    9. \(( - \infty , - 19 ] \cup [ - 6 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (39)

    11. \(\mathbb { R }\);

    2.6: Solving Absolute Value Equations and Inequalities (40)

    13. \(\left[ \frac { 2 } { 3 } , 2 \right]\);

    2.6: Solving Absolute Value Equations and Inequalities (41)

    15. \(( - 12 , - 2 )\);

    2.6: Solving Absolute Value Equations and Inequalities (42)

    17. \(( - \infty , 0 ) \cup ( 6 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (43)

    19. \([ 0,3 ]\);

    2.6: Solving Absolute Value Equations and Inequalities (44)

    21. \(\frac { 1 } { 2 }\);

    2.6: Solving Absolute Value Equations and Inequalities (45)

    23. \(\left( - \frac { 1 } { 2 } , \frac { 3 } { 2 } \right)\);

    2.6: Solving Absolute Value Equations and Inequalities (46)

    25. \(\left( 0 , \frac { 1 } { 2 } \right)\);

    2.6: Solving Absolute Value Equations and Inequalities (47)

    27. \(( - \infty , 3 ) \cup ( 9 , \infty )\);

    2.6: Solving Absolute Value Equations and Inequalities (48)

    Exercise \(\PageIndex{9}\)

    Assume all variables in the denominator are nonzero.

    1. Solve for \(x\) where \(a, p > 0: p |ax + b| − q ≤ 0\)
    2. Solve for \(x\) where \(a, p > 0: p |ax + b| − q ≥ 0\)
    Answer

    1. \(\frac { - q - b p } { a p } \leq x \leq \frac { q - b p } { a p }\)

    Exercise \(\PageIndex{10}\)

    Given the graph of \(f\) and \(g\), determine the \(x\)-values where:

    (a) \(f ( x ) = g ( x )\)

    (b) \(f ( x ) > g ( x )\)

    (c) \(f ( x ) < g ( x )\)

    1.

    2.6: Solving Absolute Value Equations and Inequalities (49)

    2.

    2.6: Solving Absolute Value Equations and Inequalities (50)

    3.

    2.6: Solving Absolute Value Equations and Inequalities (51)

    4.

    2.6: Solving Absolute Value Equations and Inequalities (52)
    Answer

    1. (a) \(−6, 0\); (b) \((−∞, −6) ∪ (0, ∞)\); (c) \((−6, 0)\)

    3. (a) \(Ø\); (b) \(ℝ\); (c) \(Ø\)

    Exercise \(\PageIndex{11}\)

    1. Make three note cards, one for each of the three cases described in this section. On one side write the theorem, and on the other write a complete solution to a representative example. Share your strategy for identifying and solving absolute value equations and inequalities on the discussion board.
    2. Make your own examples of absolute value equations and inequalities that have no solution, at least one for each case described in this section. Illustrate your examples with a graph.
    Answer

    1. Answer may vary

    Footnotes

    63The distance from the graph of a number \(a\) to zero on a number line, denoted \(|a|\).

    64The number or expression inside the absolute value.

    2.6: Solving Absolute Value Equations and Inequalities (2024)

    FAQs

    2.6: Solving Absolute Value Equations and Inequalities? ›

    To solve an absolute value inequality involving “less than,” such as |X|≤p, replace it with the compound inequality −p≤X≤p and then solve as usual. To solve an absolute value inequality involving “greater than,” such as |X|≥p, replace it with the compound inequality X≤−p or X≥p and then solve as usual.

    How do I solve an inequality with absolute value? ›

    To solve inequalities with absolute values, use a number line to see how far the absolute value is from zero. Split into two cases: when it is positive or negative. Solve each case with algebra. The answer is both cases together, in intervals or words.

    How do you solve absolute value equations? ›

    Follow these steps to solve an absolute value equality which contains one absolute value:
    1. Isolate the absolute value on one side of the equation.
    2. Is the number on the other side of the equation negative? ...
    3. Write two equations without absolute values. ...
    4. Solve the two equations.

    What is the absolute value in algebra 2? ›

    Every real number is either positive or negative. The two numerals +7 and -7 have different signs but the numerical part of each is 7. This part of the numeral designates the absolute value of the number. The absolute value of a is denoted by | a | (a vertical bar on each side of the quantity).

    How to solve inequality equations? ›

    When solving an inequality: • you can add the same quantity to each side • you can subtract the same quantity from each side • you can multiply or divide each side by the same positive quantity If you multiply or divide each side by a negative quantity, the inequality symbol must be reversed. So the solution is x > −1.

    What is the rule for absolute value? ›

    Absolute value equations can yield two solutions because the absolute value of any number and its opposite is equal to the same value. This follows the rule |x| = k is equivalent to x = k or x = -k if and only if k is greater than or equal to 0.

    How to find absolute value? ›

    The absolute value (or modulus) | x | of a real number x is the non-negative value of x without regard to its sign. For example, the absolute value of 5 is 5, and the absolute value of −5 is also 5. The absolute value of a number may be thought of as its distance from zero along real number line.

    How do you simplify absolute value? ›

    An absolute value expression is an expression that uses the absolute value symbol. To simplify an absolute value expression, write the expression in its simplest form. To do this, the absolute value symbol will be temporarily treated as a set of parentheses. All the regular order of operations rules are to be followed.

    How do you enter the formula for absolute value? ›

    The “=ABS(number)” formula in Excel converts a negative number to a positive number, whereas a positive number remains unchanged. In other words, the ABS function in Excel is used to calculate the absolute value of any given number. For example: the absolute value of -4 is 4 and the absolute value of 4 is 4.

    How to solve an equation? ›

    To solve any equation we need to perform arithmetic operations, to separate the variable, such as: Adding the same number on both sides. Subtracting same number on both sides. Multiplying with the same number on both sides.

    Why are there 2 answers for absolute value? ›

    Absolute values must be solved for values greater than or equal to zero, and less than zero. Two equations must be performed, one for positive values and the other for negative values. Absolute values will have two solutions when they are equations, functions, in the inequalities that will give a set of results.

    What is an example of an absolute value function in math? ›

    Some of the examples of absolute value functions are: f(x) = |x| g(x) = |3x - 7| f(x) = |-x + 9|

    What are 2 things about absolute value? ›

    If a real number a is positive or zero, its absolute value is itself. The absolute value of −a is a. Absolute value is symbolized by vertical bars, as in |x|, |z|, or |v|, and obeys certain fundamental properties, such as |a · b| = |a| · |b| and |a + b| ≤ |a| + |b|.

    What is an example of an inequality? ›

    For example, 9<11, 18>17 are examples of numerical inequalities and x+7>y, y<10-x, x ≥ y > 11 are examples of algebraic inequalities. The symbols '<' and '>' represent the strict inequalities and the symbols '≤' and '≥' represent slack inequalities.

    How to read inequalities? ›

    Inequality Signs

    Sometimes it's easy to get tangled up in inequalities, just remember to read them from left to right. 2≠8 2 ≠ 8 , 2 is not equal to 8. The inequality x>y can also be written as y<x . The sides of any inequality can be switched as long as the inequality symbol between them is also reversed.

    How to know if an absolute value inequality has no solution? ›

    If the absolute value is less than or less than or equal to a negative number, there is no solution. The absolute value of something will never be less than or equal to a negative number. f. If the absolute value is greater than or greater than or equal to a negative number, the solution is all real numbers.

    How do you find the absolute value inequality of a word problem? ›

    When given a word problem relating to absolute value inequalities, first translate the words to math terms. In other words, represent the word problem as a mathematical equation or expression, and use variables to stand in for unknown quantities. Then, simply solve for the variables to solve the word problem!

    How to simplify absolute value? ›

    So, |x| means the absolute value of x, or modulus x. An absolute value expression is an expression that uses the absolute value symbol. To simplify an absolute value expression, write the expression in its simplest form. To do this, the absolute value symbol will be temporarily treated as a set of parentheses.

    What to do with a number outside the absolute value? ›

    When you have a number outside of the absolute value symbols, you need to distribute that number to both the positive and negative values inside the absolute value. If the expression inside the absolute value is positive, you will have two results: one with the number added and one with the number subtracted.

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