2.6: Solving Absolute Value Equations and Inequalities (2024)

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 \(−3\).

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 \(1\).

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.

Answer:

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

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.

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

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.

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.

Answer:

There is no solution, \(Ø\).

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.

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

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.

Answer:

Using interval notation, \((−5,1)\).

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:

Answer:

Using interval notation, \([−6,0]\)

Exercise \(\PageIndex{3}\)

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

Answer

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

www.youtube.com/v/sX6ppL2Fbq0

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 \(3\).

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

Answer:

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

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.

Answer:

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

Exercise \(\PageIndex{4}\)

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

Answer

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

www.youtube.com/v/P6HjRz6W4F4

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.

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

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 \(g(x)=3\).

Answer: \(Ø\)

Case 1: An absolute value equation:

\(\begin{array} { c } { \text { If } | X | = p } \\ { \text { then } X = - p \text { or } X = p } \end{array}\)

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

\(\begin{array} { c } { \text { If } | X | \leq p } \\ { \text { then } - p \leq X \leq p } \end{array}\)

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

\(\begin{array} { c } { \text { If } | X | \geq p } \\ { \text { then } X \leq - p \text { or } X \geq p } \end{array}\)

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

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.

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

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.

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.

3.

4.

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