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2.6: Solving Absolute Value Equations and Inequalities
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- Page ID 6238
Learning Objectives
- Review the definition of absolute value.
- Solve absolute value equations.
- Solve absolute value inequalities.
Absolute Value Equations
Recall that the absolute value 63 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 \(|3|=3\).
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 value 64 \(X\) can be either positive or negative \(p\). Use this theorem to solve absolute value equations algebraically.
Example \(\PageIndex{1}\):
Solve: \(|x+2|=3\).
In this case, the argument of the absolute value is \(x+2\) and must be equal to \(3\) or \(−3\).
Therefore, to solve this absolute value equation, set \(x+2\) equal to \(±3\) and solve each linear equation as usual.
\(\begin{array} { c } { | x + 2 | = 3 } \\ { x + 2 = - 3 \quad \quad\text { or } \quad\quad x + 2 = 3 } \\ { x = - 5 \quad\quad\quad\quad\quad\quad\quad x = 1 } \end{array}\)
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 \(f (x) = g (x)\).
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: \(| 2 x + 3 | = 4\).
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.
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: \(2 |5x − 1| − 3 = 9\).
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.
The solutions are \(-1\) and \(\frac{7}{5}\)
Exercise \(\PageIndex{1}\)
Solve: \(2 - 7 | x + 4 | = - 12\).
www.youtube.com/v/G0EjbqreYmU
Not all absolute value equations will have two solutions.
Example \(\PageIndex{4}\):
Solve: \(| 7 x - 6 | + 3 = 3\).
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.
The solution is \(\frac{6}{7}\).
Example \(\PageIndex{5}\):
Solve: \(|x+7|+5=4\).
\(\begin{aligned} | x + 7 | + 5 & = 4 \:\:\color{Cerulean} { Subtract \: 5\: on\: both\: sides.} \\ | x + 7 | & = - 1 \end{aligned}\)
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.
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: \(| 2 x - 5 | = | x - 4 |\).
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.
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 \(3\).
The solutions are \(1\) and \(3\).
Exercise \(\PageIndex{2}\)
Solve: \(| x + 10 | = | 3 x - 2 |\).
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.
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: \(|x+2|<3\).
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.
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.
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: \(4 |x + 3| − 7 ≤ 5\).
\(\begin{array} { c } { 4 | x + 3 | - 7 \leq 5 } \\ { 4 | x + 3 | \leq 12 } \\ { | x + 3 | \leq 3 } \end{array}\)
Next, apply the theorem and rewrite the absolute value inequality as a compound inequality.
\(\begin{array} { c } { | x + 3 | \leq 3 } \\ { - 3 \leq x + 3 \leq 3 } \end{array}\)
\(\begin{aligned} - 3 \leq x + 3 \leq & 3 \\ - 3 \color{Cerulean}{- 3} \color{Black}{ \leq} x + 3 \color{Cerulean}{- 3} & \color{Black}{ \leq} 3 \color{Cerulean}{- 3} \\ - 6 \leq x \leq 0 \end{aligned}\)
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:
Using interval notation, \([−6,0]\)
Exercise \(\PageIndex{3}\)
Solve and graph the solution set: \(3 + | 4 x - 5 | < 8\).
Interval notation: \((0, \frac{5}{2})\)
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.
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: \(|x+2|>3\).
The argument \(x+2\) must be less than \(−3\) or greater than \(3\).
\(\begin{array} { c } { | x + 2 | > 3 } \\ { x + 2 < - 3 \quad \text { or } \quad x + 2 > 3 } \\ { x < - 5 }\quad\quad\quad\quad\quad\: x>1 \end{array}\)
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.
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: \(3 + 2 |4x − 7| ≥ 13\).
\(\begin{array} { r } { 3 + 2 | 4 x - 7 | \geq 13 } \\ { 2 | 4 x - 7 | \geq 10 } \\ { | 4 x - 7 | \geq 5 } \end{array}\)
\(\begin{array} &\quad\quad\quad\quad\:\:\:|4x-7|\geq 5 \\ 4 x - 7 \leq - 5 \quad \text { or } \quad 4 x - 7 \geq 5 \end{array}\)
\(\begin{array} { l } { 4 x - 7 \leq - 5 \text { or } 4 x - 7 \geq 5 } \\ \quad\:\:\:\:{ 4 x \leq 2 } \quad\quad\quad\:\:\: 4x\geq 12\\ \quad\:\:\:\:{ x \leq \frac { 2 } { 4 } } \quad\quad\quad\quad x\geq 3 \\ \quad\quad{ x \leq \frac { 1 } { 2 } } \end{array}\)
Shade the solutions on a number line and present the answer using interval notation.
Using interval notation, \((−∞,\frac { 1 } { 2 }]∪[3,∞)\)
Exercise \(\PageIndex{4}\)
Solve and graph: \(3 | 6 x + 5 | - 2 > 13\).
Using interval notation, \(\left( - \infty , - \frac { 5 } { 3 } \right) \cup ( 0 , \infty )\)
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: \(|2x−1|+5>2\).
\(\begin{array} { c } { | 2 x - 1 | + 5 > 2 } \\ { | 2 x - 1 | > - 3 } \end{array}\)
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.
Geometrically, we can see that \(f(x)=|2x−1|+5\) is always greater than \(g(x)=2\).
All real numbers, \(ℝ\).
Example \(\PageIndex{12}\):
Solve and graph: \(|x+1|+4≤3\).
\(\begin{array} { l } { | x + 1 | + 4 \leq 3 } \\ { | x + 1 | \leq - 1 } \end{array}\)
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 \(g(x)=3\).
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:
Case 2: an absolute value inequality involving " less than .", case 3: an absolute value inequality involving " greater than .", 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 \(\PageIndex{5}\)
- \(|x| = 9\)
- \(|x| = 1\)
- \(|x − 7| = 3\)
- \(|x − 2| = 5\)
- \(|x + 12| = 0\)
- \(|x + 8| = 0\)
- \(|x + 6| = −1\)
- \(|x − 2| = −5\)
- \(|2y − 1| = 13\)
- \(|3y − 5| = 16\)
- \(|−5t + 1| = 6\)
- \(|−6t + 2| = 8\)
- \(\left| \frac { 1 } { 2 } x - \frac { 2 } { 3 } \right| = \frac { 1 } { 6 }\)
- \(\left| \frac { 2 } { 3 } x + \frac { 1 } { 4 } \right| = \frac { 5 } { 12 }\)
- \(|0.2x + 1.6| = 3.6\)
- \(|0.3x − 1.2| = 2.7\)
- \(| 5 (y − 4) + 5| = 15\)
- \(| 2 (y − 1) − 3y| = 4\)
- \(|5x − 7| + 3 = 10\)
- \(|3x − 8| − 2 = 6\)
- \(9 + |7x + 1| = 9\)
- \(4 − |2x − 3| = 4\)
- \(3 |x − 8| + 4 = 25\)
- \(2 |x + 6| − 3 = 17\)
- \(9 + 5 |x − 1| = 4\)
- \(11 + 6 |x − 4| = 5\)
- \(8 − 2 |x + 1| = 4\)
- \(12 − 5 |x − 2| = 2\)
- \(\frac{1}{2} |x − 5| − \frac{2}{3} = −\frac{1}{6}\)
- \(\frac { 1 } { 3 } \left| x + \frac { 1 } { 2 } \right| + 1 = \frac { 3 } { 2 }\)
- \(−2 |7x + 1| − 4 = 2\)
- \(−3 |5x − 3| + 2 = 5\)
- \(1.2 |t − 2.8| − 4.8 = 1.2\)
- \(3.6 | t + 1.8| − 2.6 = 8.2\)
- \(\frac{1}{2} |2 (3x − 1) − 3| + 1 = 4\)
- \(\frac{2}{3} |4 (3x + 1) − 1| − 5 = 3\)
- \(|5x − 7| = |4x − 2|\)
- \(|8x − 3| = |7x − 12|\)
- \(|5y + 8| = |2y + 3|\)
- \(|7y + 2| = |5y − 2|\)
- \(|5 (x − 2)| = |3x|\)
- \(|3 (x + 1)| = |7x|\)
- \(\left| \frac { 2 } { 3 } x + \frac { 1 } { 2 } \right| = \left| \frac { 3 } { 2 } x - \frac { 1 } { 3 } \right|\)
- \(\left| \frac { 3 } { 5 } x - \frac { 5 } { 2 } \right| = \left| \frac { 1 } { 2 } x + \frac { 2 } { 5 } \right|\)
- \(|1.5t − 3.5| = |2.5t + 0.5|\)
- \(|3.2t − 1.4| = |1.8t + 2.8|\)
- \(|5 − 3 (2x + 1)| = |5x + 2|\)
- \(|3 − 2 (3x − 2)| = |4x − 1|\)
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.
- Solve for \(x: p |ax + b| − q = 0\)
- Solve for \(x: |ax + b| = |p + q|\)
1. \(x = \frac { - b q \pm q } { a p }\)
Exercise \(\PageIndex{7}\)
- \(|x| < 5\)
- \(|x| ≤ 2\)
- \(|x + 3| ≤ 1\)
- \(|x − 7| < 8\)
- \(|x − 5| < 0\)
- \(|x + 8| < −7\)
- \(|2x − 3| ≤ 5\)
- \(|3x − 9| < 27\)
- \(|5x − 3| ≤ 0\)
- \(|10x + 5| < 25\)
- \(\left| \frac { 1 } { 3 } x - \frac { 2 } { 3 } \right| \leq 1\)
- \(\left| \frac { 1 } { 12 } x - \frac { 1 } { 2 } \right| \leq \frac { 3 } { 2 }\)
- \(|x| ≥ 5\)
- \(|x| > 1\)
- \(|x + 2| > 8\)
- \(|x − 7| ≥ 11\)
- \(|x + 5| ≥ 0\)
- \(|x − 12| > −4\)
- \(|2x − 5| ≥ 9\)
- \(|2x + 3| ≥ 15\)
- \(|4x − 3| > 9\)
- \(|3x − 7| ≥ 2\)
- \(\left| \frac { 1 } { 7 } x - \frac { 3 } { 14 } \right| > \frac { 1 } { 2 }\)
- \(\left| \frac { 1 } { 2 } x + \frac { 5 } { 4 } \right| > \frac { 3 } { 4 }\)
1. \(( - 5,5 )\);
3. \([ - 4 , - 2 ]\);
5. \(\emptyset\);
7. \([ - 1,4 ]\);
9. \(\left\{ \frac { 3 } { 5 } \right\}\);
11. \([ - 1,5 ]\);
13. \(( - \infty , - 5 ] \cup [ 5 , \infty )\);
15. \(( - \infty , - 10 ) \cup ( 6 , \infty )\);
17. \(\mathbb { R }\);
19. \(( - \infty , - 2 ] \cup [ 7 , \infty )\);
21. \(\left( - \infty , - \frac { 3 } { 2 } \right) \cup ( 3 , \infty )\);
23. \(( - \infty , - 2 ) \cup ( 5 , \infty )\);
Exercise \(\PageIndex{8}\)
Solve and graph the solution set.
- \(|3 (2x − 1)| > 15\)
- \(|3 (x − 3)| ≤ 21\)
- \(−5 |x − 4| > −15\)
- \(−3 |x + 8| ≤ −18\)
- \(6 − 3 |x − 4| < 3\)
- \(5 − 2 |x + 4| ≤ −7\)
- \(6 − |2x + 5| < −5\)
- \(25 − |3x − 7| ≥ 18\)
- \(|2x + 25| − 4 ≥ 9\)
- \(|3 (x − 3)| − 8 < −2\)
- \(2 |9x + 5| + 8 > 6\)
- \(3 |4x − 9| + 4 < −1\)
- \(5 |4 − 3x| − 10 ≤ 0\)
- \(6 |1 − 4x| − 24 ≥ 0\)
- \(3 − 2 |x + 7| > −7\)
- \(9 − 7 |x − 4| < −12\)
- \(|5 (x − 4) + 5| > 15\)
- \(|3 (x − 9) + 6| ≤ 3\)
- \(\left| \frac { 1 } { 3 } ( x + 2 ) - \frac { 7 } { 6 } \right| - \frac { 2 } { 3 } \leq - \frac { 1 } { 6 }\)
- \(\left| \frac { 1 } { 10 } ( x + 3 ) - \frac { 1 } { 2 } \right| + \frac { 3 } { 20 } > \frac { 1 } { 4 }\)
- \(12 + 4 |2x − 1| ≤ 12\)
- \(3 − 6 |3x − 2| ≥ 3\)
- \(\frac{1}{2} |2x − 1| + 3 < 4\)
- 2 |\frac{1}{2} x + \frac{2}{3} | − 3 ≤ −1\)
- \(7 − |−4 + 2 (3 − 4x)| > 5\)
- \(9 − |6 + 3 (2x − 1)| ≥ 8\)
- \(\frac { 3 } { 2 } - \left| 2 - \frac { 1 } { 3 } x \right| < \frac { 1 } { 2 }\)
- \(\frac { 5 } { 4 } - \left| \frac { 1 } { 2 } - \frac { 1 } { 4 } x \right| < \frac { 3 } { 8 }\)
1. \(( - \infty , - 2 ) \cup ( 3 , \infty )\);
3. \(( 1,7 )\);
5. \(( - \infty , 3 ) \cup ( 5 , \infty )\);
7. \(( - \infty , - 8 ) \cup ( 3 , \infty )\);
9. \(( - \infty , - 19 ] \cup [ - 6 , \infty )\);
11. \(\mathbb { R }\);
13. \(\left[ \frac { 2 } { 3 } , 2 \right]\);
15. \(( - 12 , - 2 )\);
17. \(( - \infty , 0 ) \cup ( 6 , \infty )\);
19. \([ 0,3 ]\);
21. \(\frac { 1 } { 2 }\);
23. \(\left( - \frac { 1 } { 2 } , \frac { 3 } { 2 } \right)\);
25. \(\left( 0 , \frac { 1 } { 2 } \right)\);
27. \(( - \infty , 3 ) \cup ( 9 , \infty )\);
Exercise \(\PageIndex{9}\)
Assume all variables in the denominator are nonzero.
- Solve for \(x\) where \(a, p > 0: p |ax + b| − q ≤ 0\)
- Solve for \(x\) where \(a, p > 0: p |ax + b| − q ≥ 0\)
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. (a) \(−6, 0\); (b) \((−∞, −6) ∪ (0, ∞)\); (c) \((−6, 0)\)
3. (a) \(Ø\); (b) \(ℝ\); (c) \(Ø\)
Exercise \(\PageIndex{11}\)
- 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.
- 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.
1. Answer may vary
63 The distance from the graph of a number \(a\) to zero on a number line, denoted \(|a|\).
64 The number or expression inside the absolute value.
Absolute Value in Algebra
Absolute value means ....
... how far a number is from zero:
"6" is 6 away from zero, and "−6" is also 6 away from zero.
So the absolute value of 6 is 6 , and the absolute value of −6 is also 6
Absolute Value Symbol
To show we want the absolute value we put "|" marks either side (called "bars"), like these examples:
More Formal
More formally we have:
Which says the absolute value of x equals:
- x when x is greater than zero
- 0 when x equals 0
- −x when x is less than zero (this "flips" the number back to positive)
So when a number is positive or zero we leave it alone, when it is negative we change it to positive using −x.
Example: what is |−17| ?
Well, it is less than zero, so we need to calculate "−x":
− ( −17 ) = + 17
(Because two minuses make a plus )
Useful Properties
Here are some properties of absolute values that can be useful:
That makes sense ... |a| can never be less than zero.
Squaring a makes it positive or zero (for a as a Real Number). Then taking the square root will "undo" the squaring, but leave it positive or zero.
Means these are the same:
- the absolute value of (a times b), and
- (the absolute value of a) times (the absolute value of b)
Which can also be useful when solving
Which is often the key to solving most absolute value questions.
Example: Solve |x+2| = 5
Using "|u| = a is the same as u = ±a ":
Which has two solutions:
Graphically
Let us graph that example:
It is easier to graph when we have an "=0" equation, so subtract 5 from both sides:
|x+2| − 5 = 0
Now let's plot y=|x+2|−5 and find where it equals zero.
Here is the plot of y=|x+2|−5 , but just for fun let's make the graph by shifting it around :
And the two solutions (circled) are −7 and +3 .
Absolute Value Inequalities
Mixing Absolute Values and Inequalites needs a little care!
There are 4 inequalities:
Less Than, Less Than or Equal To
With " < " and " ≤ " we get one interval centered on zero:
Example: Solve |x| < 3
This means the distance from x to zero must be less than 3:
Everything in between (but not including) −3 and 3
It can be rewritten as:
−3 < x < 3
As an interval it can be written as:
The same thing works for "Less Than or Equal To":
Example: Solve |x| ≤ 3
Everything in between and including -3 and 3
How about a bigger example?
Example: Solve |3x-6| ≤ 12
Rewrite it as:
−12 ≤ 3x−6 ≤ 12
−6 ≤ 3x ≤ 18
Lastly, multiply by (1/3). Because we are multiplying by a positive number, the inequalities will not change:
−2 ≤ x ≤ 6
Greater Than, Greater Than or Equal To
This is different ... we get two separate intervals :
Example: Solve |x| > 3
It looks like this:
less than -3 or from 3 onwards
It can be rewritten as
x < −3 or x > 3
(−∞, −3) U (3, +∞)
Note: U means "Union" of the two intervals.
Careful! Do not write it as
Because x cannot be less than -3 and greater than 3 at the same time
It is really:
x is less than −3 or greater than 3
The same thing works for "Greater Than or Equal To":
Example: Solve |x| ≥ 3
Can be rewritten as
x ≤ −3 or x ≥ 3
(−∞, −3] U [3, +∞)
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Solving Simpler Absolute-Value Equations
Simpler Harder Special Case
When we take the absolute value of a number, we always end up with a positive number (or zero). Whether the input was positive or negative (or zero), the output is always positive (or zero). For instance, | 3 | = 3 , and | −3 | = 3 also.
This property — that both the positive and the negative become positive — makes solving absolute-value equations a little tricky. But once you learn the "trick", they're not so bad. Let's start with something simple:
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Solving Absolute Value Equations
Solve | x | = 3
I've pretty much already solved this, in my discussion above:
| −3 | = 3
So then x must be equal to 3 or equal to −3 .
But how am I supposed to solve this if I don't already know the answer? I will use the positive / negative property of the absolute value to split the equation into two cases, and I will use the fact that the "minus" sign in the negative case indicates "the opposite sign", not "a negative number".
For example, if I have x = −6 , then " − x " indicates "the opposite of x " or, in this case, −(−6) = +6 , a positive number. The "minus" sign in " − x " just indicates that I am changing the sign on x . It does not indicate a negative number. This distinction is crucial!
Whatever the value of x might be, taking the absolute value of x makes it positive. Since x might originally have been positive and might originally have been negative, I must acknowledge this fact when I remove the absolute-value bars. I do this by splitting the equation into two cases. For this exercise, these cases are as follows:
a. If the value of x was non-negative (that is, if it was positive or zero) to start with, then I can bring that value out of the absolute-value bars without changing its sign, giving me the equation x = 3 .
b. If the value of x was negative to start with, then I can bring that value out of the absolute-value bars by changing the sign on x , giving me the equation − x = 3 , which solves as x = −3 .
Then my solution is
x = ±3
We can, by the way, verify the above solution graphically. When we attempt to solve the absolute-value equation | x | = 3 , we are, in effect, setting two line equations equal to each other and finding where they cross. For instance:
In the above, I've plotted the graph of y 1 = | x | (being the blue line that looks like a "V") and y 2 = 3 (being the green horizontal line). These two graphs cross at x = −3 and at x = +3 (being the two red dots).
If you're wanting to check your answers on a test (before you hand it in), it can be helpful to plug each side of the original absolute-value equation into your calculator as their own functions; then ask the calculator for the intersection points.
Of course, any solution can also be verified by plugging it back into the original exercise, and confirming that the left-hand side (LHS) of the equation simplifies to the same value as does the right-hand side (RHS) of the equation. For the equation above, here's my check:
x = −3
LHS: | x | = | −3 |
LHS: | x | = | +3 |
If you're ever in doubt about your solution to an equation, try graphing or else try plugging your solution back into the original question. Checking your work is always okay!
The step in the above, where the absolute-value equation was restated in two forms, one with a "plus" and one with a "minus", gives us a handy way to simplify things: When we have isolated the absolute value and go to take off the bars, we can split the equation into two cases; we will signify these cases by placing a "minus" on the opposite side of the equation (for one case) and a "plus" on the opposite side (for the other). Here's how this works:
Solve | x + 2 | = 7 , and check your solution(s).
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The absolute value is isolated on the left-hand side of the equation, so it's already set up for me to split the equation into two cases. To clear the absolute-value bars, I must split the equation into its two possible two cases, one each for if the contents of the absolute-value bars (that is, if the "argument" of the absolute value) is negative and if it's non-negative (that is, if it's positive or zero). To do this, I create two new equations, where the only difference between then is the sign on the right-hand side. First, I'll do the "minus" case:
x + 2 = −7
x = −9
Now I'll do the non-negative case, where I can just drop the bars and solve:
Now I need to check my solutions. I'll do this by plugging them back into the original equation, since the grader can't see me checking plots on my graphing calculator.
x = −9:
LHS: |(−9) + 2|
= |−7| = 7 = RHS
LHS: |(5) + 2|
= |7| = 7 = RHS
Both solutions check, so my answer is:
x = −9, 5
Solve | 2 x − 3 | − 4 = 3
First, I'll isolate the absolute-value part of the equation; that is, I'll get the absolute-value expression by itself on one side of the "equals" sign, with everything else on the other side:
| 2 x − 3 | − 4 = 3
| 2 x − 3 | = 7
Now I'll clear the absolute-value bars by splitting the equation into its two cases, one for each sign on the argument. First I'll do the negative case:
2 x − 3 = −7
2 x = −4
x = −2
And then I'll do the non-negative case:
2 x − 3 = 7
The exercise doesn't tell me to check, so I won't. (But, if I'd wanted to, I could have plugged "abs(2X−3)−4" and "3" into my calculator (as Y1 and Y2, respectively), and seen that the intersection points were at my x -values.) My answer is:
x = −2, 5
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How to Solve Absolute Value Equations
Last Updated: January 31, 2023 References
This article was co-authored by wikiHow Staff . Our trained team of editors and researchers validate articles for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. This article has been viewed 23,120 times.
Setting up the Problem
- For example, |9| = 9; |-9| = -(-9) = 9.
Calculating the Values
Check Your Work
Expert Q&A
Video . by using this service, some information may be shared with youtube..
- Remember that absolute value bars are distinct from parentheses and function differently. Thanks Helpful 0 Not Helpful 0
- Once you've solved for any variables, remember to simplify absolute values accordingly. Thanks Helpful 0 Not Helpful 0
You Might Also Like
- ↑ http://tutorial.math.lamar.edu/Classes/Alg/SolveAbsValueEqns.aspx
- ↑ https://www.mathsisfun.com/numbers/absolute-value.html
- ↑ http://www.varsitytutors.com/high_school_math-help/solving-absolute-value-equations
- ↑ http://www.purplemath.com/modules/solveabs.htm
- ↑ https://www.khanacademy.org/math/algebra/absolute-value-equations-functions/absolute-value-equations/v/absolute-value-equations
About this article
To solve absolute value equations, first isolate the absolute value terms by moving anything outside of the vertical bars to the other side of the equation. Next, solve for the positive value of the equation by isolating the variable. Since the absolute variable can represent 2 numbers, then solve for the negative value by putting a negative sign outside the vertical bars. Then, move the negative by dividing both sides by -1 and solve for the variable. If you want to learn how to check your answers for an absolute value equation, keep reading the article! Did this summary help you? Yes No
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How to Solve Integers and Absolute Value Problems? (+FREE Worksheet!)
Two vertical lines around a number or expression are used to indicate the absolute value of that number or expression. Here, you can learn how to find the absolute value of a number and how to solve math problems containing absolute values and integers.
The absolute value of the real number \(a\) Is written in the form of \(| a |\) and is a positive number. Two vertical lines around a number or expression are used to indicate the absolute value of that number or expression. The output value of the absolute value is always greater than or equal to zero. Absolute value is used to indicate the distance of a number from zero on the line of real numbers.
Related Topics
- How to Add and Subtract Integers
- How to Multiply and Divide Integers
- How to Use Order of Operations
- How to Order Integers and Numbers
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Step-by-step guide to solving integers and absolute value problems.
- The absolute value of a positive number is equal to the same positive number.
- The absolute value of zero is equal to zero.
- The absolute value of a negative number is the positive value of that number.
- Note: To find the absolute value of a number, just find its distance from \(0\) on a number line! For example, the distance of \(12\) and \(- \ 12\) from zero on number line is \(12\)!
Integers and Absolute Value – Example 1:
Solve. \(|8 \ – \ 2| \ × \ \frac{ |- \ 4 \ × \ 6|}{3}=\)
First solve \(|8 \ – \ 2|, →|8 \ – \ 2|=|6|\), the absolute value of \(6\) is \(6\), \(|6|=6\) \(6 \ × \ \frac{ |- \ 4 \ × \ 6|}{3}=\) Now solve \(|- \ 4 \ × \ 6|, → |- \ 4 \ × \ 6|=|- \ 24|\), the absolute value of \(- \ 24\) is \(24\), \(|- \ 24|=24\) Then: \(6 \ × \ \frac{ 24}{3}= 6 \ × \ 8=48 \)
Integers and Absolute Value – Example 2:
Solve. \(\frac{ |- \ 12|}{3} \ × \ |9 \ – \ 4|=\)
First find \(|- \ 12| , →\) the absolute value of \(- \ 12\) is \(12\), then: \(|- \ 12|=12\) \(\frac{12}{3} \ × \ |9 \ – \ 4|= \) Next, solve \(|9 \ – \ 4|, → |9 \ – \ 4|=| \ 5|\), the absolute value of \( \ 5\) is \(5\), \(| \ 5|=5\) Then: \(\frac{12}{3} \ × \ 5=4 \ × \ 5=20\)
Integers and Absolute Value – Example 3:
Solve. \(\frac{ |-18|}{9}×|5-8|=\)
First find \(|-18| , →\) the absolute value of \(-18\) is \(18\), then: \(|-18|=18\) \( \frac{18}{9}×|5-8|=\) Next, solve \(|5-8|, → |5-8|=|-3|\), the absolute value of \(-3\) is \(3\), \(|-3|=3\) Then: \(\frac{18}{9}×3=2×3=6\)
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Integers and absolute value – example 4:.
Solve. \(|10-5|×\frac{ |-2×6|}{3}=\)
First solve \(|10-5|, →|10-5|=|5|\), the absolute value of \(5\) is \(5, |5|=5\) \( 5×\frac{ |-2×6|}{3}= \) Now solve \(|-2×6|, → |-2×6|=|-12|\), the absolute value of \(-12\) is \(12, |-12|=12\) Then: \(5×\frac{ 12}{3}= 5×4=20\)
Exercises for Solving Integers and Absolute Value Problems
- \(\color{blue}{|-43| – |12| + 10}\)
- \(\color{blue}{76 + |-15-45| – |3|}\)
- \(\color{blue}{30 + |-62| – 46}\)
- \(\color{blue}{|32| – |-78| + 90}\)
- \(\color{blue}{|-35+4| + 6 – 4}\)
- \(\color{blue}{|-4| + |-11|}\)
Download Integers and Absolute Value Worksheet
- \(\color{blue}{41}\)
- \(\color{blue}{133}\)
- \(\color{blue}{46}\)
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Absolute Value Inequality
How to solve absolute value inequalities, general formula.
In the picture below, you can see generalized example of absolute value equation and also the topic of this web page: absolute value inequalities .
The diagram below illustrated the difference between an absolute value equation and two absolute value inequalities
Absolute Value Inequality Calculator
Enter any values for A,b and c for any absolute value equation |A x + b| = c into the text boxes below and this solver will calculate your answer and show all of the steps!
Practice Problems
Solve the absolute value inequality below: $$| x -1 | \ge 2 $$
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Solving absolute value equations is as easy as working with regular linear equations. The only additional key step that you need to remember is to separate the original absolute value equation into two parts: positive and negative ( ±) components. Below is the general approach on how to break them down into two equations:
To solve absolute value equations, find x values that make the expression inside the absolute value positive or negative the constant. To graph absolute value functions, plot two lines for the positive and negative cases that meet at the expression's zero. The graph is v-shaped. Created by Sal Khan and CK-12 Foundation. Questions Tips & Thanks
Here is a set of practice problems to accompany the Absolute Value Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.
Solving absolute value equations Solve absolute value equations Google Classroom What are the solutions of the following equation? − 3 | x + 5 | + 1 = 7 | x + 5 | + 8 Choose 1 answer: Choose 1 answer: (Choice A) x = − 57 5 or x = − 43 5 A x = − 57 5 or x = − 43 5 (Choice B) x = − 57 10 or x = 43 5 B x = − 57 10 or x = 43 5 (Choice C)
STEP ONE: Isolate the Absolute Value In this example, the absolute value is already isolated on one side of the equals sign, which means that there are no other terms outside of the absolute value, so you can move onto step two. STEP TWO: Solve for Positive AND Solve for Negative
• Intro How To Solve Absolute Value Equations The Organic Chemistry Tutor 7.41M subscribers Join Subscribe Subscribed 13K 707K views 5 years ago New Algebra Playlist This math video tutorial...
To solve absolute value problems with one solution, identify expression in absolute value bars. Recall absolute value is zero only if expression is zero. Make expression equal right-hand side. Use algebra to find x values that satisfy equation. Check solutions by plugging in. Graph solutions on number line. Mark points.
Step 2: Set the argument of the absolute value equal to ± p. Here the argument is 5x − 1 and p = 6. 5x − 1 = − 6 or 5x − 1 = 6. Step 3: Solve each of the resulting linear equations. 5x − 1 = − 6 or 5x − 1 = 6 5x = − 5 5x = 7 x = − 1 x = 7 5. Step 4: Verify the solutions in the original equation. Check x = − 1.
14K 1M views 6 years ago This algebra video tutorial provides a basic introduction into absolute value equations. it explains how to solve absolute value equations the easy way. It...
We can also give a strict mathematical/formula definition for absolute value. It is, |p| = {p if p ≥ 0 −p if p < 0 | p | = { p if p ≥ 0 − p if p < 0. This tells us to look at the sign of p p and if it's positive we just drop the absolute value bar. If p p is negative we drop the absolute value bars and then put in a negative in front ...
Steps The General Steps to solve an absolute value equation are: Rewrite the absolute value equation as two separate equations, one positive and the other negative Solve each equation separately After solving, substitute your answers back into original equation to verify that you solutions are valid
http://www.greenemath.com/Step-by-Step full solutions for several practice tests that involve absolute value. Here, we will look at problems that involve the...
More Formal. More formally we have: Which says the absolute value of x equals: x when x is greater than zero; 0 when x equals 0; −x when x is less than zero (this "flips" the number back to positive); So when a number is positive or zero we leave it alone, when it is negative we change it to positive using −x.
When we attempt to solve the absolute-value equation | x | = 3, we are, in effect, setting two line equations equal to each other and finding where they cross. For instance: In the above, I've plotted the graph of y1 = | x | (being the blue line that looks like a "V") and y2 = 3 (being the green horizontal line).
Try MathPapa Algebra Calculator Clear Absolute Value Equation Calculator » Show Keypad Shows you the step-by-step solutions solving absolute value equations! This calculator will solve your problems.
Solving the equation 8|x+7|+4 = -6|x+7|+6 which has two possible solutions. Created by Sal Khan. Questions Tips & Thanks Want to join the conversation? Sort by: Top Voted albertan123456789 8 years ago how does the absolute value thing separate the equation into 2 equations • ( 8 votes) Upvote Flag Hamda Khan 8 years ago
1. Set up the equation for the positive value. An equation involving absolute value will have two possible solutions. To set up the positive equation, simply remove the absolute value bars, and solve the equation as normal. [6] For example, the positive equation for. | 6 x − 2 | = 4 {\displaystyle |6x-2|=4} is.
Solution: First solve \ (|8 \ - \ 2|, →|8 \ - \ 2|=|6|\), the absolute value of \ (6\) is \ (6\), \ (|6|=6\) \ (6 \ × \ \frac { |- \ 4 \ × \ 6|} {3}=\) Now solve \ (|- \ 4 \ × \ 6|, → |- \ 4 \ × \ 6|=|- \ 24|\), the absolute value of \ (- \ 24\) is \ (24\), \ (|- \ 24|=24\) Then: \ (6 \ × \ \frac { 24} {3}= 6 \ × \ 8=48 \)
http://www.greenemath.com/In this lesson, we will learn how to solve more advanced absolute value equations. We will begin by learning how to solve a nested ...
Example 1 The diagram below illustrated the difference between an absolute value equation and two absolute value inequalities Example 2 Absolute Value Inequality Calculator Enter any values for A,b and c for any absolute value equation |A x + b| = c into the text boxes below and this solver will calculate your answer and show all of the steps!
• 2 comments ( 13 votes) Flag John Clint Russell 11 years ago No, this would result in the right numbers, but the wrong inequality. You would have gotten w <= 152.5 and w <= 147.5. Which is equal to W being less than 147.5 to -infinity. ( 6 votes) Flag Show more...
This problem demonstrates how to solve an absolute value inequality and how to represent the solution using interval notation.