Solving an Equation or Inequality Containing an Absolute Value

The absolute value of a real number |x,| denoted |\vert x \vert| , is the following: ||\vert x \vert = \begin{cases} \ \ \ x & \text{if}\ x \geq 0\\ -x & \text{if}\ x < 0\end{cases}||

When an absolute value is isolated on one side of an equality, the algebraic expression or number on the other side must be positive. ||\vert x\vert=\underbrace{\dots}_{\ge\ 0}||

• If it is a strictly negative number, we stop solving the inequality since there are no solutions.

• If it is an algebraic expression, then it’s possible to impose a restriction: the value of the algebraic expression must be greater than or equal to |0.|

1. When the absolute value is equal to a positive number ||\vert x+3 \vert = 5||Since |5| is a positive number, this equation has two solutions.

2. When the absolute value is equal to a negative number ||\vert x-4 \vert = -25||Since |-25| is a negative number, this equation has no solution.

3. When the absolute value is equal to an algebraic expression ||\vert 2x-5 \vert = -3x+12||Since the absolute value of |2x-5| is equal to an algebraic expression, this can be positive or negative depending on the value of |x.| Thus, the solutions of this equation are valid if and only if the inequality |-3x+12 \geq 0| is respected. This is a restriction. Thus, |-3x+12 \geq 0| if and only if |x \leq 4.| ||\begin{align} -3x+12 &\geq 0 \\ &\Updownarrow \\ -3x &\geq -12 \\ &\Updownarrow \\ x &\leq 4 \end{align}||In other words, after solving the equation, verify that all the solutions found are less than or equal to |4.|

1. Isolate the absolute value.

2. Check if the absolute value is equal to a positive number and set the restriction, if necessary.

3. Apply the definition of absolute value to form two equations.

4. Solve the two obtained equations.

5. Verify the solutions.

6. Write the solution set.

Solve the equation |\vert 2x + 6 \vert =3.|

1. Isolate the absolute value
   The absolute value is already isolated.

2. Check if the absolute value is equal to a positive number
   Since |3\geq 0,| continue solving the equation.

3. Apply the definition of the absolute value to form two equations

||\vert 2x + 6 \vert = 3\ \Leftrightarrow \begin{cases}\! 2x+6=3\\ \quad\text{or} \\ \!2x+6=-3 \end{cases}||

4. Solve the two equations obtained

5. Verify the solutions

||\begin{align} x_1\rightarrow \left\vert 2\! \left(-\dfrac{3}{2}\right) + 6 \right\vert &= \vert 3 \vert = 3 \\\\ x_2\rightarrow \left\vert 2\! \left(- \dfrac{9}{2}\right) + 6 \right\vert &= \vert{-3} \vert = 3 \end{align}||Both solutions are valid.

6. Give the solution set
   The solution set is: |x\in \left\lbrace -\dfrac{9}{2},-\dfrac{3}{2} \right\rbrace.|

In step 3, there are two ways to form the two equations.

For example, with the equation |\vert a \vert = b,| it’s possible to form the two equations to solve…

When the absolute value is equal to a number and not an algebraic expression, as in the previous example, it is much easier to choose option B – which is the reason for using this strategy. If option A were chosen, here is what would have been obtained: ||\vert 2x + 6 \vert = 3\\ \Updownarrow\\ \overbrace{\begin{align} \begin{aligned} 2x+6&=3 \\ 2x&=-3 \\ x_1 &= -\dfrac{3}{2} \\ \phantom{x} \end{aligned} \qquad \begin{aligned} -(2x+6)&=3 \\ -2x-6&=3\\-2x&=9 \\ x_2 &= -\dfrac{9}{2} \end{aligned} \end{align}}||The same two solutions are found, regardless of the chosen strategy.

Solve the equation |\dfrac{1}{2} \vert x+5 \vert = x-2.|

1. Isolate the absolute value ||\begin{align} \dfrac{1}{2} \vert x + 5 \vert \color{#ec0000}{\times 2} &= (x-2) \color{#ec0000}{\times 2} \\ \vert x + 5 \vert &= 2x - 4 \end{align}||

2. Check if the absolute value is equal to a positive number and set the restriction
   Since the absolute value is equal to an algebraic expression, a restriction must be imposed on |x.| ||\begin{align} 2x-4 \geq 0 \\ 2x \geq 4 \\ x \geq 2 \end{align}||

3. Apply the definition of absolute value to form two equations ||\vert x+5 \vert = 2x-4\\ \Updownarrow \\ \begin{cases} x+5 = 2x-4 \\ \quad\text{or} \\ -(x+5) = 2x-4 \end{cases}||

4. Solve the two equations obtained

5. Verify the solutions
   |x_1=9| respects the restriction since |9\geq 2.| Therefore, it is a solution of the equation.
   |x_2=-\dfrac{1}{3}| does not respect the restriction since |-\dfrac{1}{3} \not\geq 2.| Therefore, |-\dfrac{1}{3}| is not part of the solution set.

6. Write the solution set
   The only solution of the equation is |x=9.|

There are two ways to verify the solutions found.

1. By making sure that the restrictions are respected, as in the previous example.

2. By substituting the two values of |x| obtained in step 4 in the original equation.

   In step 5 of the previous example, the following verification could be done. ||\begin{align} \dfrac{1}{2} \vert \color{#3a9a38}x + 5 \vert &= x-2 \\ x_1 \rightarrow\ \dfrac{1}{2} \vert \color{#3a9a38}{9} + 5 \vert &\overset{?}{=} \color{#3a9a38}{9}-2 \\ 7 &= 7 \\\\ x_2 \rightarrow\ \dfrac{1}{2} \left\vert \color{#3a9a38}{-\dfrac{1}{3}} + 5 \right\vert &\overset{?}{=} \color{#3a9a38}{-\dfrac{1}{3}} -2 \\ \dfrac{7}{3} &\color{#ec0000}{\not=} -\dfrac{7}{3} \end{align}||The solution |-\dfrac{1}{3}| must therefore be rejected.

Solve the equation |3 \vert x - 1 \vert + 6 = 0.|

1. Isolate the absolute value ||\begin{align} 3 \vert x - 1 \vert + 6 &= 0 \\ 3\vert x - 1 \vert &= -6 \\ \vert x - 1 \vert &= -2 \end{align}||

2. Check if the absolute value is equal to a positive number
   Stop solving the equation since an absolute value cannot be equal to a negative number. Give the answer directly.

Answer: There is no solution. The solution set is: |x \in \varnothing.|

Note: In the image below, notice how the absolute value function with an equation of |y=3\vert x - 1 \vert + 6| never intersects the horizontal line with the equation |\color{#333fb1}{y=0}.| This explains why the equation |3 \vert x - 1 \vert + 6 = \color{#333fb1}{0}| has no solution.

Solve the equation |2\vert x - 4 \vert +1 = 3x+2.|

1. Isolate the absolute value ||\begin{align} 2 \vert x - 4 \vert +1 &= 3x+2 \\ 2\vert x -4 \vert &= 3x+1 \\ \vert x -4 \vert &= \dfrac{3x+1}{2} \end{align}||

2. Check if the absolute value is equal to a positive number and set the restriction
   Since the absolute value is equal to an algebraic expression, impose a restriction on |x.| ||\begin{align} \dfrac{3x+1}{2} &\geq 0 \\ 3x+1 &\geq 0 \\ 3x &\geq -1 \\ x &\geq \dfrac{-1}{3} \end{align}||

3. Apply the definition of absolute value to form two equations  ||\vert x-4 \vert = \dfrac{3x+1}{2}\\ \ \\ \Updownarrow \\ \ \\ \begin{cases} \ \ \ \ \, x-4\,\, =\dfrac{3x+1}{2} \\ \qquad\qquad \text{ou}\\ -(x-4) =\dfrac{3x+1}{2}\ \end{cases}||

4. Solve the two equations obtained

5. Verify the solutions
   |x_1=-9| does not respect the restriction since |-9 \not\geq \dfrac{-1}{3}.| Therefore, |-9| is not part of the solution set.
   However, |x_2=\dfrac{7}{5}| respects the restriction since |\dfrac{7}{5} \geq \dfrac{-1}{3}.|

6. Write the solution set
   The sole solution of this equation is |x=\dfrac{7}{5}.|

1. Replace the inequality symbol with the equal symbol.

2. Isolate the absolute value.

3. Check if the absolute value is equal to a positive number and set the restriction, if necessary.

4. Apply the definition of absolute value.

5. Solve the equations.

6. Verify the solutions of the equation formed in step 1.

7. Determine the solution set of the inequality using a graph or a number line.

Solve the inequality |\vert 3 - 2x \vert > 9.|

1. Replace the inequality symbol with an equal symbol ||\vert 3 - 2x \vert = 9||

2. Isolate the absolute value
   The absolute value is already isolated.

3. Check if the absolute value is equal to a positive number
   Since |9\geq 0,| continue to solve the equation.

4. Apply the definition of absolute value ||\vert 3-2x \vert = 9\\ \ \\ \Updownarrow \\ \ \\ \begin{cases} 3-2x =9\\ \qquad \text{or} \\ 3-2x =-9 \end{cases}||

5. Solve the equations

6. Verify the solutions of the equation formed in step 1
   Since there are no restrictions, both solutions are valid and can be verified. ||\begin{align} x_1 &\rightarrow \vert 3-2(\color{#3a9a38}{-3})\vert = \vert 9 \vert = 9 \\ x_2 &\rightarrow \vert 3-2(\color{#3a9a38}{6})\vert = \vert{-9}\vert = 9 \end{align}||The two solutions of the equation are indeed valid.

7. Determine the solution set of the inequality using a graph

Answer Using the graph, it can be concluded that the solution set is |x\in\ ]-\infty,-3[\, \cup\, ]6,+\infty[.|

Note: Since the inequality sign is |>,| the interval boundaries are not included.

To determine the solution set (step 7), the following visual aids can be used.

1. A graph of the situation
   This option was chosen in the previous example. It is not rare, in word problems, for the graph of the situation to have already been sketched at this stage. If that is the case, the solution set can be quickly determined.

2. A number line
   Here's what this visual aid would have looked like for the previous example:

   On a number line, place the solutions found in step 5.
   Place empty points if the inequality symbol is |>| or |<.> Place solid points if the inequality symbol is |\geq| or |\leq.|

The dots separate the line into three sections: |]-\infty, -3[,| |]-3,6[,| and |]6,\infty[.|

It is necessary to test a random value in each of these sections by verifying it with the initial inequality.

• For |]-\infty, -3[,| use |x=-5.| ||\begin{align} \vert 3 - 2x \vert &> 9 \\ \vert3-2(-5)| &\overset{?}{>} 9 \\ 13 &> 9 \end{align}||Since the inequality is true, the interval |]-\infty, -3[| is part of the solution set.

• For |]-3,6[,| use |x=0.| This obtains |3 \not> 9,| so the interval |]-3,6[| is not part of the solution set.

• For |]6,\infty[,| use |x=10.| This obtains |17>9.| Since the inequality is true, the interval |]6,\infty[| is part of the solution set.

The solution set indeed corresponds to that determined using the graph, which is |x\in\ ]-\infty,-3[\, \cup\, ]6,+\infty[.|

On the number line, colour the sections that correspond to the solution set.

Solve the inequality |\vert 5 - x \vert \leq 3x+1.|

1. Replace the inequality symbol with an equal symbol ||\vert 5 - x \vert = 3x +1||

2. Isolate the absolute value
   The absolute value is already isolated.

3. Check if the absolute value is equal to a positive number and set the restriction ||\begin{align} 3x+1 &\geq 0 \\ 3x &\geq -1 \\ x &\geq -\dfrac{1}{3} \end{align}||

4. Apply the definition of absolute value ||\vert 5-x \vert = 3x+1\\ \ \\ \Updownarrow \\ \ \\ \begin{cases} \quad 5-x\ \,= 3x+1 \\ \qquad \text{or} \\ -(5-x) =3x+1 \end{cases}||

5. Solve the two equations

6. Verify the solutions of the equation formed in step 1
   |x_1=1| respects the restriction because |1\geq -\dfrac{1}{3}.|
   |x_2=-3| violates the restriction because |-3 \not\geq -\dfrac{1}{3}.|
   Thus, |1| is part of the solution set, but |-3| is not part of it.

7. Determine the solution set of the inequality using a graph

Answer: The solution set is |[1,+\infty[.|

Note: The lower boundary |(1)| of the interval is included since the inequality sign is |\leq.|

Two special cases can arise when solving an absolute value inequality.

1. If the absolute value is greater than a negative number, the solution set is |\mathbb{R}.|

2. If the absolute value is less than a negative number, there is no solution.

1. Solve the inequality |2\vert x+4 \vert +1 \geq -3.| ||\begin{align} 2\vert x+4 \vert +1 &\geq -3 \\ 2\vert x+4 \vert &\geq -4 \\ \vert x+4 \vert &\geq -2 \end{align}||This last inequality is always true because an absolute value always gives a number greater than |0.|

   Answer: The solution set is |x\in \mathbb{R}.|

   In the graph illustrating the initial inequality, just as in that of the reduced inequality, observe how the sketch of the absolute value function is always above the horizontal line for all possible values of |x.|

2. Solve the inequality |-\dfrac{1}{2}\vert x+4 \vert +2> 3.| ||\begin{align} -\dfrac{1}{2}\vert x+4 \vert +2 &> 3 \\ \color{#ec0000}{\dfrac{\color{black}{-\frac{1}{2}\vert x+4 \vert}}{-\frac{1}{2}}} &>\ \color{#ec0000}{\dfrac{\color{black}{1}}{-\frac{1}{2}}}\end{align} \\ \qquad\ \underbrace{\vert x+4 \vert < -2}\\ \qquad\ \text{Impossible}||When dividing or multiplying by a negative number, the inequality sign must be reversed. This is why the inequality changed from |>| at |<.|

   The resulting inequality is always false because an absolute value is always a number greater than |0.|

   Answer: This inequality has no solution. The solution set is |x\in \varnothing.|

   Once again, the graph of the starting situation and the graph illustrating the reduced inequality can be sketched to validate the answer.

Sometimes, graphing the situation reveals that there is no point of intersection. In that case, the solution set can be given without doing any algebraic calculations.