# Absolute Value Equations And Inequalities

Questions on solving equations and inequalities with absolute value are presented along with detailed solutions. You may want to work through the tutorials on solving equations with absolute value before solving the questions below.

## Review of Equations and Inequalities with Absolute ValueIn order to solve equations or inequalities with absolute value, we need to rewrite them without absolute value, solve them and if necessary check the solutions obtained.The most important about the absolute value of an algebraic expression, is that we can simplify them if we know the sign of the expression inside the absolute value symbol according to the following definition. | x | = x if x ≥ 0 and | x | = - x if x < 0 Some specific equations and inequalities with absolute value may be written without the absolute value symbol as follows: 1) |x| = k , where k is a positive real number, is equivalent to x = k or x = - k 2) |x| < k , where k is a positive real number, is equivalent to - k < x < k 3) |x| > k , where k is a positive real number, is equivalent to k < 0 or x < - k
## Absolute Value Equations## Question 1Find all real solutions to the equation with absolute value.2|-2x - 2| - 3 = 13Solution to Question 1Rewrite the equation with the absolute value expression on one side 2| - 2 x - 2| = 16 , add 3 to both sides | - 2 x - 2| = 8 , divide both sides by 2 The above equation is equivalent to stating that the expression - 2 x - 2 is equal to 8 or - 8. a) -2 x - 2 = 8 x = - 5 b) - 2 x - 2 = - 8 x = 3 As an exercise, check by substitution that x = - 5 and x = 3 are solutions to the given equation. Solution set: {-5,3}
## Question 2Find all real solutions to the equation with absolute value.|x - 1| = 2x + 1Solution to Question 2The signs of x - 1 and 2 x + 1 change with x and one way to solve this equation is to consider two cases a) assume x - 1 ≥ 0 and rewrite the equation as x - 1 = 2x + 1 solve for x x = -2 b) assume x - 1 ≤ 0 and rewrite the equation as -(x - 1) = 2x + 1 - x + 1 = 2x + 1 solve for x x = 0 Important: Because we assumed the sign of the expression inside the absolute value symbol, we need to check the two solutions found above.
Check x = - 2Substitute x by - 2 in both sides of the given equation. Left side: |(-2) - 1| = |-2 + 1| = 1 Right side: 2(-2) + 1 = - 3 The two sides are not equal and therefore x = -2 is not a solution to the given equation. Check x = 0Substitute x by 0 in both sides of the given equation. Left side: |(0) - 1| = 1 Right side: 2(0) + 1 = 1 The two sides are equal and therefore x = 0 is a solution to the given equation. Solution set: {0}
## Question 3Find all real solutions to the equation|2x - 1| = x^{2}Solution to Question 3x ^{2} is either 0 or positive but the sign of 2 x - 1 changes with x. Let us consider two cases
a) assume 2 x - 1 ≥ 0 and rewrite the equation as 2x - 1 = x ^{2}which may be written as x ^{2} - 2x + 1 = 0
solve for x x = 1 b) assume 2 x - 1 ≤ 0 and rewrite the equation as - (2x - 1) = x ^{2}which may be written as x ^{2} + 2 x - 1 = 0
solve for x x = - 1 - √ 2 , - 1 + √ 2 Check all solutions found above (detailed calculations will be left out for you as an exercise) All solutions found above (1 , - 1 - √ 2 and - 1 + √ 2) are solutions to the original equation given above.
## Question 4Find all real solutions to the equation with absolute value.-2|x + 1| -2 = 4Solution to Question 4Rewrite the given equation with the term containing the absolute value on one side. -2|x + 1| = 6 |x + 1| = - 3 The absolute value of an algebraic expression cannot be negative and therefore the given equation has no solutions.
## Question 5Find all real solutions to the equation with absolute value.|x - 1| = |x^{2} -2x + 1|Solution to Question 4Note that x ^{2} -2x + 1 can be written as
x ^{2} -2x + 1 = (x - 1)^{2} , an expression that is positive or equal to zero and our equation may be written as:
|x - 1| = |x - 1| ^{ 2}Rewrite the above as |x - 1| ^{ 2} - |x - 1| = 0
Factor |x - 1|(|x - 1| - 1) = 0 which gives |x - 1| = 0 and solution x = 1 or |x - 1| - 1 = 0 |x - 1| = 1 Solve the above to obtain two solutions x = 1 and x = 2. We did not make any assumptions as in the above questions and we therefore do not need to check our solutions (except perhaps for mistakes made). Solution set: {0,1,2}
## Absolute Value Inequalities## Question 6Solve the inequality.Solution to Question 6The above inequality is solved by writing a double inequality equivalent to the given inequality but without absolute value - 3 < x + 2 < 3 Solve the double inequality to obtain - 5 < x < 1 The above solution set is written in interval form as follows (-5 , 1)
## Question 7Solve the inequality with absolute value..- 3|-2x + 4| ≤ 4Solution to Question 7Divide both sides of the given inequality by - 3 and change the symbol of the inequality to obtain |-2x + 4| ≥ - 4 / 3 The absolute value of an expression is always positive or equal to zero. Hence any value of x in the above inequality will satisfy it and therefore the solution set of the given inequality is the set of all the real numbers given by \( (-\infty , \infty) \)
## Question 8Solve the inequality.Solution to Example 8Solving the above inequality is equivalent to solving -2 x - 4 > 9 or -2 x - 4 < - 9 Which gives \( x \lt -13 / 2\) or \( x \gt 5 / 2 \) The above solution set is written in interval form as follows \( (-\infty , -13 / 2) \cup (5 / 2 , + \infty) \)
## Question 9Solve the inequality with absolute values.|x + 1| ≤ |x^{2} + 2x + 1|Solution to Example 9Note that the algebraic expression |x ^{2} + 2x + 1| may be written as
|x ^{2} + 2x + 1| =
Hence the inequality may be written as |x + 1| - |x + 1| ^{ 2} ≤ 0
|x + 1| is positive or equal to zero, hence the left side of above inequality may be factored as |x + 1|(1 - |x + 1|) ≤ 0 We now use a table to study the sign of the two factors on the left side of the above inequality to determine the solution set. |x + 1| is positive for all values of x except at x = - 1 where it is equal to zero. The zeros of (1 - |x + 1|) are - 2 and 0. The table of signs is shown below. \( (-\infty , - 2] \cup {\{-1\}} \cup [0 , + \infty) \) Check the above answer to the inequality graphically using the graphs of the two sides (|x + 1| in blue and |x ^{2} + 2x + 1| in red) of the given inequality shown below.
## Question 10Solve the inequality.Solution to Question 10One way to take out the absolute value is to study the sign of the expression inside the absolute value symbol. There are two cases Case 1
For \( x^2 - 4 \ge 0 \), or \( x\) in the interval \( (-\infty , -2] \cup [2 , +\infty)\), we can write \( |x^2 - 4| = x^2 - 4 \) Substitute the expression with the absolute value in the given inequality and solve \( x + 2 \lt x^2 - 4\) \( x^2 - x - 6 \gt 0\) The solution set to above inequality is given by the interval \( (-\infty , -2) \cup (3 , +\infty)\) We need to select only the solutions within the interval \( (-\infty , -2] \cup [2 , +\infty)\). The intersection of intervals \((-\infty , -2] \cup [2 , +\infty)\) and \((-\infty , -2) \cup (3 , +\infty)\) gives the solution set \((-\infty , -2) \cup (3 , +\infty)\) Case 2
For \(x^2 - 4 \lt 0\), or \(x\) in the interval \((-2 , 2)\), we can write \( |x^2 - 4| = -(x^2 - 4) \) Substitute the expression with the absolute value in the given inequality and solve \(x + 2 \lt -(x^2 - 4)\) which may be rewritten as \(x^2 + x - 2 \lt 0\) Solution set to the above inequality is given by the interval: (-2 , 1) We need to select only the solutions within the interval \((-2 , 2)\). The solution set to above inequality is given by the intersection of the intervals \((-2 , 1)\) and \((-2 , 2)\) which gives (-2 , 1) Conclusion: The solution set to the given inequality is \((-\infty, -2) \cup (-2 , 1) \cup (3 , +\infty)\) Check the above answer to the inequality graphically using the graphs of the two sides of the given inequality shown below.
## References and links to absolute value equations and Inequalities.Graph absolute value functionsSolve Equations with Absolute Value. Absolute Value Functions. Definition of the Absolute Value Function. Maths Problems, Questions and Online Self Tests. |