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Absolute Value Equations

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• Pranshu Gaba
• Alan Enrique Ontiveros Salazar
• Kai Hsien Boo
• Agnishom Chattopadhyay
• Aditya Virani
• Prince Loomba
• Keshav Ramesh

Absolute value equations are equations involving expressions with the absolute value functions . This wiki intends to demonstrate and discuss problem solving techniques that let us solve such equations.

A very basic example would be as follows:

Find all values of \(x\) satisfying \(|x-2| + |x-4| = 4.\)

Usually, the basic approach is to analyze the behavior of the function before and after the point where they reach 0. For example, for \(|x-a|\) one could analyze the cases where \(x > a\) or \(x < a\), or even \(x = a\) if required. However, these problems are often simplified with a more sophisticated approach like being able to eliminate some of the cases, or graphing the functions. In this wiki, we intend to discuss this techniques along with strategies on when to use which.

Methodology

Technique - squaring both sides, technique - casework, technique - sketching graph, problem solving - miscellaneous.

Introduction to absolute value equations

Methodology to solve absolute value equations: the techniques used to solve absolute value equations and when to use which one

Take an example to describe the following methodology:

1) Understanding absolute value--positive, negative case (or graph approach) 2) Determining possible solutions 3) Verifying solutions

Explain - How do we use this technique to solve absolute value equations?

Remember to verify the possible solutions - why and how?

2-3 examples in increasing order of difficulty - explaining how we squared both sides to solve more difficult problems

Followed by 1-2 TIY problems - relevant to be solved by case work technique

Suppose we have an equation of the form \(\lvert a \rvert = \lvert b \rvert\). Since both sides are positive, we can square them without adding extraneous solutions: \[a^2=b^2.\] Then solve it as an ordinary equation: \[\begin{align} a^2-b^2&=0 \\ (a+b)(a-b)&=0. \end{align}\] So we see that \(a=-b\) or \(a=b\).

Solve the equation \(\lvert 3x+4 \rvert = \lvert 2x-7 \rvert\) for real \(x\). We square both sides to obtain \[(3x+4)^2=(2x-7)^2.\] Here we don't need to expand both sides; just apply the difference of two squares to find the factors: \[\begin{align} (3x+4+2x-7)(3x+4-2x+7)&=0 \\ (5x-3)(x+11)&=0. \end{align} \] The solutions are \(\left\{\frac{3}{5}, -11\right\}. \ _ \square\)

Because absolute value can be defined as piecewise functions, depending on where the value of \(x \) is with respect to the number line, you have to work with a different "piece" of the piecewise function.

General steps:

Using the definition of absolute value as a piecewise function, "undo" the absolute value sign(s) and write cases. For example, we know that the expression in the absolute value sign can either be positive or negative.

Solve each case for \( x\).

Verify the solutions.

Find all real values of \(x\) such that \( | 3x – 4 | – 2 = 3.\) We first isolate the absolute value onto one side: \[\begin{align} | 3x – 4 | – 2 & = 3\\ | 3x – 4 | &= 5. \end{align}\] Now, we "undo" the absolute value signs and split the equation into its two cases, the positive case and the negative case: \[\begin{array}{rlcccrl} (3x – 4) &= 5 &&\text{ or } && –(3x – 4) &= 5\\ 3x – 4 &= 5 &&\text{ or } && –3x + 4 &= 5\\ 3x &= 9 &&\text{ or } && –3x &= 1\\ x &= 3 &&\text{ or } && x &= –\frac{1}{3}. \ _\square \end{array}\]

Find all real values of \(x \) such that \( |x+1| + |2x+3| = 5 \). There are four possible cases, but one will be eliminated due to impossibility: Case 1. If \(x+1 \) and \( 2x+3 \) are both positive, then \[\begin{align} x+1 + 2x+3 &= 5 \\ 3x + 4 &= 5 \\ 3x &= 1 \\ x &= \dfrac{1}{3}. \end{align} \] Case 2. If \( x+1 \) is negative and \( 2x + 3 \) is positive, then \[\begin{align} - x - 1 + 2x + 3 &= 5 \\ x + 2 &= 5 \\ x &= 3. \end{align} \] However, when \( x = 3 \), \( x+1 \) and \( 2x+3 \) are both positive, so this is not a valid solution to the equation. Case 3. If \( x+1 \) and \( 2x+3 \) are both negative, then \[\begin{align} -x - 1 - 2x - 3 &= 5 \\ -3x - 4 &= 5 \\ -3x &= 9 \\ x &= -3. \end{align} \] Case 4. If \( x +1 \) is positive and \( 2x + 3 \) is negative, it is an impossible case. Graph the two lines if you are not convinced. Therefore, the solution set is \(\left \{ -3, \frac{1}{3} \right \}.\ _\square\)

Find all real values of \(x\) such that \[|x+2|+|2x+6|+|3x-3|=12.\] In this problem we are dealing with 3 terms of absolute values. Their turning points (the values of \(x\) such that they change sign) of the three terms are \(x=-2, x=-3, x=1,\) respectively. Hence, we need to check the cases \(-\infty < x \leq -3\), \(-3<x\leq -2\), \(-2 < x \leq 1\), \(1<x<\infty\). Case 1. \(\, -\infty < x \leq -3\) In this case, the three terms will always be negative. Hence, \[\begin{aligned} -(x+2)-(2x+6)-(3x-3)&=12 \\ x &= -\frac{17}{6}. \end{aligned}\] However, \(x=-\frac{17}{6} >-3\) is not within the domain \(-\infty < x \leq -3\). Thus this solution is not valid. Case 2. \(\, -3<x\leq -2\) In this case, the three terms will be negative, positive, and negative, respectively. Hence, \[\begin{aligned} -(x+2)+(2x+6)-(3x-3)&=12 \\ x &= -\frac{5}{2}. \end{aligned}\] \(x=-\frac{5}{2}\) lies between \(-3\) and \(-2\). Thus \(\boxed{x=-\frac{5}{2}}\) is one of the solutions. Case 3. \(\, -2 < x \leq 1\) In this case, the three terms will be positive, positive, and negative, respectively. However, \[\begin{aligned} (x+2)+(2x+6)-(3x-3)=11 \neq 12. \end{aligned}\] Thus there is no solution within this domain. Case 4. \(\, 1<x<\infty\) In this case, the three terms are always positive. Hence, \[\begin{aligned} (x+2)+(2x+6)+(3x-3)&=12 \\ x &= \frac{7}{6}, \end{aligned}\] which lies between \(1\) and \(\infty\). Thus \(\boxed{x=\frac{7}{6}}\) is another solution. In conclusion, \(x=-\frac{5}{2}\) and \(x=\frac{7}{6}\) are the solutions for the given equation. \(_\square\)

Find all real values of \( x \) such that \( |x||x+1| = 2 \). Case 1. \(\, x, x+1 \) both positive \[\begin{align} x(x+1)-2 &= 0 \\ x^2 +x - 2 &= 0 \\ x &= 1, x = -2. \end{align} \] Reject \( x = -2 \) because it does not make both \(x \) and \( x +1 \) positive. Case 2. \(\, x\) negative, \( x + 1 \) positive \[\begin{align} -x(x+1)-2 &= 0 \\ -x^2 - x - 2 &= 0 \\ x^2 + x + 2 &= 0 \\ x &= \dfrac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot 2 }}{2}. \end{align} \] We only asked for real solutions, so at this point we ignore this case because we're going to get imaginary results. Case 3. \(\, x\) positive, \( x +1 \) negative This is an impossible case (graph the lines and you'll see why), so we can ignore it. Case 4. \(\, x, x+1 \) both negative Because they're both negative, the negatives end up "canceling" and become positive, which was what we had in Case 1. However, the restriction is different from Case 1 (here, both \( x \) and \( x +1 \) have to be negative, not positive ), so instead of rejecting \( x = -2 \), we reject \( x = 1 \) from this case. Basically, in this specific case 4, \( x = 1 \) is not a possible solution, but it does not mean it's not a possible solution for Case 1 because we're simply going piece by piece in this piecewise function--in the end we will take the union of all possible solutions. Thus, the solutions are \( \left \{ -2, 1 \right \} \). \(_\square\)

Sometimes absolute value equations have a ridiculous number of cases and it would take too long to go through every single case. Therefore, we can instead graph the absolute value equations using the definition of absolute value as a piecewise function. To get each piece, you must figure out the domain of each piece. This method is highly beneficial when the question writer asks for the number of solutions instead of the actual solutions. Let's work through some examples to see how this is done.

Find all real solutions to \( |3x-4| = 5 \). To graph this, there are two possible cases: when \( 3x - 4\) is positive, and when \( 3x-4 \) is negative. When is \( 3x-4 \) positive? \[\begin{align} 3x - 4 &> 0 \\ 3x &> 4 \\ x &> \dfrac{4}{3}. \end{align} \] (Also, when \( x < \frac{4}{3} \), \( 3x- 4 \) will be negative.) We know that there will be a "turning point" at \( x = \frac{4}{3} \) for the graph of \( y = |3x-4| \). Finally, using the definition of absolute value, we know that when \( x > \frac{4}{3} \), \( y = 3x - 4 \), and when \( x \leqslant \frac{4}{3} \), \( y = -3x + 4 \). We now just need to graph \( y = 5 \) and look for the intersections. You can see that the solutions are \(\left \{ -\frac{1}{3} , 3 \right \}.\ _\square\). Another benefit of this graphing technique is that you do not need to verify any of the solutions--since we are only graphing the pieces that are actually mathematically possible, we get all the solutions we are looking for, no less and no more. If you could not discern the solutions from the picture, you can simply solve the equation for each case.

Find all real solutions to \( |x+1| + |2x+3| = 5 \). The possible cases are that \(\hspace{0.5cm}\) 1. \( \, x+1, 2x+3 \) are both positive; \(\hspace{0.5cm}\) 2. \( \, x+1 \) is negative and \( 2x+3 \) is positive; \(\hspace{0.5cm}\) 3. \( \, x+1 , 2x + 3 \) are both negative. We need to figure out the domains for which each of these holds. Case 1 holds when \( x > -1 \). Case 2 holds when \( -\frac{3}{2} < x< -1 \). Case 3 holds when \( x < -\frac{3}{2} \). Now, let's write our piecewise function. When \( x > -1 \), we have \( y = x+1 + 2x + 3 = 3x + 4 \). When \( -\dfrac{3}{2} < x< -1 \), we have\( y = -x -1 + 2x + 3 = x + 2 \). When \( x < - \dfrac{3}{2} \), we have \( y = -x - 1 -2x - 3 = -3x -4 \). As you can see in the graph, the solutions for the given equation are \(\left \{ -3, \frac{1}{3} \right \}.\ _\square\) .

Find all real solutions to \( |x||x+1| = 2 \). To graph this, we again only look at the possible cases and when they would occur: \(\hspace{0.5cm}\) 1. \(\, x, x+1 \) both positive \(\hspace{0.5cm}\) 2. \(\, x \) negative, \( x+1 \) positive \(\hspace{0.5cm}\) 3. \(\, x, x+1 \) both negative. Case 1 is true when \( x>0 \). Case 2 is true when \( -1 < x < 0 \). Case 3 is true when \( x<-1 \). When \( x>0 \) and when \( x< -1 \), we have \( y = x(x+1) = x^2 + x \). When \( -1 < x < 0 \), we have \( y = -x(x+1) = -x^2 - x \). It is evident that the solutions are \(\{-2, 1\}.\ _\square\)

Any other technique (fact, definition) you can use to solve the problems? Otherwise move on to the followings.

3-4 examples solved by using a mix of more than one of above techniques

Add guiding text in between. Guiding text means phrasing the section in a way that it keeps on telling the reader what's going on in this section.

3-4 TIY problems - using multiple techniques to solve

What is the sum of all real numbers \(x\) satisfying \[ x^2-\sqrt{x^2} = \lvert x-1 \rvert +5?\] Observe that \(\sqrt{x^2}=\lvert x \rvert.\) Then the given equation becomes \[ x^2-\lvert x \rvert= \lvert x-1 \rvert +5.\] If \(x<0,\) then we rewrite the equation to obtain \[\begin{align} x^2-(-x)&=-(x-1)+5\\ x^2+2x-6&=0\\ x&=-1\pm \sqrt{7}\\ x&=-1-\sqrt{7}. \qquad (\text{since } x<0) \end{align}\] If \(0\le x<1,\) then we rewrite the equation to obtain \[\begin{align} x^2-x&=-(x-1)+5\\ x^2&=6\\ x&=\pm \sqrt{6}, \end{align}\] which do not satisfy the assumption \(0\leq x<1.\) Thus there are no solutions in this interval. If \(x\ge 1,\) then we rewrite the equation to obtain \[\begin{align} x^2-x&=x-1+5\\ x^2-2x-4&=0\\ x&=1\pm \sqrt{5}\\ x&=1+\sqrt{5}. \qquad (\text{since } x\ge 1) \end{align}\] Therefore, the above three cases give two solutions \(x=-1-\sqrt{7}\) and \(x=1+\sqrt{5},\) the sum of which is \(\sqrt{5}-\sqrt{7}.\) \(_\square\)

[IMO 1959/2] Solve the equation \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A\) for real \(x\) (where the square roots are only defined for non-negative values), when \(A=\sqrt{2}\); \(A=1\); \(A=2\). Here we don't see any absolute value involved with the equation. Before doing anything, note that our first restriction for \(x\) is \(x \geq \frac{1}{2}\) and for \(A\) is \(A>0\). Intuitively, we could square both sides to get rid of some square roots: \[\begin{align} x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{x^2-(2x-1)}&=A^2 \\ 2x+2\sqrt{(x-1)^2}&=A^2. \end{align}\] Great! We find a perfect square inside the square root, so an absolute value will appear: \[2x+2|x-1|=A^2.\] Now we are going to find the possible cases for \(A\): When \(x-1 > 0\), we have \[\begin{align} 2x+2(x-1)&=A^2 \\ x&=\dfrac{A^2+2}{4}. \end{align}\] Then, by our assumption of \(x-1 > 0\), we get that this solution only works when \(A^2 > 2\). When \(x-1 \leq 0\), something interesting happens: \[\begin{align} 2x-2(x-1)&=A^2 \\ 2&=A^2. \end{align}\] So, when \(A^2=2 (\text{or }A=\sqrt{2})\), the equation becomes independent of \(x\), implying that any value of the interval \(x \in \left[\frac{1}{2},1\right]\) will be a solution for the first point. When \(A=1\), there are no solutions by our restriction of \(A^2 \geq 2\). Finally, when \(A=2\) we have \[x=\dfrac{2^2+2}{4}=\frac{3}{2}.\ _\square\] What happens if we allow the square roots to admit negative values?

Sometimes, in minimization problems, it often helps us to see that the value of an expression inside the absolute value is at least 0.

\[ y = \Big| \big| \small| x - 5 \small| + 5 \big| - 5 \Big| + 5 \]

What is the smallest possible value of \(y\)?

\[\] Notation : \( | \cdot | \) denotes the absolute value function.

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Solving Equations with Absolute Values

Absolute values on one side of the equation.

When solving an equation with absolute values, it is necessary to split the equation into two equations, one resulting in a positive value and the other resulting in a negative value. We can then solve the two equations to obtain two possible solutions.

x + 2 = 3 or x + 2 = –3

x = 1 or x = –5 (subtract 2 from both sides)

2 x – 6 = 8 or 2 x – 6 = –8

2 x = 14 or 2 x = –2 (add 6 to both sides)

x = 7 or x = –1 (divide 2 to both sides)

The following video shows how to solve multi-step equations with absolute values.

Absolute Values On Both Side Of The Equation

The same method can be applied when there are absolute values on both side of the equation.

3 x + 3 = 2 x + 5 or 3 x + 3 = –(2 x + 5)

Solving the first equation:

3 x + 3 = 2 x + 5

3 x – 2 x = 5 – 3

Solving the second equation:

3 x + 3 = –(2 x + 5)

3 x + 3 = –2 x – 5

3 x + 2 x = –5 –3

5 x = –8

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What is Absolute Value?

Practice questions, solving absolute value equations – methods & examples.

Solving equations containing an absolute value is as simple as working with regular linear equations . Before we can embark on solving absolute value equations, let’s take a review of what the word absolute value means.

In mathematics, the absolute value of a number refers to the distance of a number from zero, regardless of direction. The absolute value of a number x is generally represented as | x | = a, which implies that, x = + a and -a.

We say that the absolute value of a given number is the positive version of that number . For example, the absolute value of negative 5 is positive 5, and this can be written as: | − 5 | = 5.

Other examples of absolute values of numbers include:  |− 9| = 9, |0| = 0, − |−12| = −12 etc. From these examples of absolute values, we simply define absolute value equations as equations containing expressions with absolute value functions.

How to Solve Absolute Value Equations?

The following are the general steps for solving equations containing absolute value functions:

• Isolate the expression containing the absolute value function.
• Get rid of the absolute value notation by setting up the two equations so that in the first equation, the quantity inside absolute notation is positive. In the second equation, it is negative. You will remove the absolute notation and write the quantity with its suitable sign.
• Calculate the unknown value for the positive version of the equation.
• Solve for the negative version of the equation, in which you will first multiply the value on the other side of the equal sign by -1, and then solve.

In addition to the above steps, there are other important rules you should keep in mind when solving absolute value equations.

• The ∣x∣is always positive: ∣x∣ → +x.
• In | x| = a, if the  a on the right is a positive number or zero, then there is a solution.

Solve the equation for x: |3 + x| − 5 = 4.

• Isolate the absolute value expression by applying the Law of equations. This means, we add 5 to both sides of the equation to obtain;

| 3 + x | − 5 + 5 = 4 + 5

| 3 + x |= 9

• Calculate for the positive version of the equation. Solve the equation by assuming the absolute value symbols.

| 3 +  x  | = 9 → 3 +  x  = 9

Subtract 3 from both sides of the equation.

3 – 3 + x = 9 -3

• Now calculate for the negative version of the equation by multiplying 9 by -1.

3 +  x  | = 9 → 3 +  x  = 9 × ( −1)

Also subtract 3 from both side to isolate x.

3 -3 + x = – 9 -3

Therefore 6 and -12 are the solutions.

Solve for all real values of x such that | 3x – 4 | – 2 = 3.

• Isolate the equation with absolute function by add 2 to both sides.

= | 3x – 4 | – 2 + 2 = 3 + 2

= | 3x – 4 |= 5

Assume the absolute signs and solve for the positive version of the equation.

| 3x – 4 |= 5→3x – 4 = 5

Add 4 to both sides of the equation.

3x – 4 + 4 = 5 + 4

Divide: 3x/3 =9/3

Now solve for the negative version by multiplying 5 by -1.

3x – 4 = 5→3x – 4 = -1(5)

3x – 4 = -5

3x – 4 + 4 = – 5 + 4

Divide by 3 on both sides.

Therefore, 3 and 1/3 are the solutions.

Solve for all real values of x: Solve | 2 x  – 3 | – 4 = 3

Add 4 to both sides.

| 2 x  – 3 | -4 = 3 →| 2 x  – 3 | = 7

Assume the absolute symbols and solve for the positive version of x.

2 x  – 3 = 7

2x – 3 + 3 = 7 + 3

Now solve for the negative version of x by multiplying 7 by -1

2 x  – 3 = 7→2 x  – 3 = -1(7)

Add 3 to both sides.

2x – 3 + 3 = – 7 + 3

x = – 2

Therefore, x  = –2, 5

Solve for all real numbers of x: | x + 2 | = 7

Already the absolute value expression is isolated, therefore assume the absolute symbols and solve.

| x + 2 | = 7 → x + 2 = 7

Subtract 2 from both sides.

x + 2 – 2 = 7 -2

Multiply 7 by -1 to solve for the negative version of the equation.

x + 2 = -1(7) → x + 2 = -7

Subtract by 2 on both sides.

x + 2 – 2 = – 7 – 2

Therefore, x = -9, 5

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Absolute value equations.

Solving absolute value equations is based on the idea that absolute value represents the distance between a point on the number line and zero. In this lesson, we will look at a few examples to understand how to solve these equations and also take a bit of a look at this idea of distance as it relates to solving absolute value equations. Table of Contents

Steps for solving absolute value equations

• Step-by-step examples
• Absolute value equations with no solutions or one solution
• Why does our method work?

Another perspective

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When given an absolute value equation, you will first need to isolate the absolute value part of the equation. Then, you will write two equations based on the definition of absolute value (though sometimes, there will end up only being one equation). This sounds complicated, but it is only a step or two more than solving the typical linear equation.

Let’s try these steps out with some examples!

In this first example, the absolute value part of the equation is already isolated, so only step two will apply. Whether or not this first step applies or not, you will always have zero, one, or two solutions to any absolute value equation.

Solve the equation: \(|5x – 2| = 13\)

As mentioned, the absolute value part is already isolated. Therefore, we will solve two equations without the absolute value: one where the 13 is positive and one where 13 is negative.

Equation 1:

\(\begin{align}5x – 2 &= 13\\ 5x &= 15\\ x &= 3\end{align}\)

Equation 2:

\(\begin{align}5x – 2 &= -13\\ 5x &= -11 \\ x &= -\dfrac{11}{5}\end{align}\)

So, there are two solutions to this equation: \(x = \bbox[border: 1px solid black; padding: 2px]{3, -\dfrac{11}{5}}\)

In this next example, there will be a little more work since the absolute value part of the equation is not isolated. In this situation, you will always need to isolate this term before you write your two equations, or you will end up with incorrect answers.

Solve the equation: \(4 + 3|x – 5| = 16\)

Your first step here is to use algebra to isolate the absolute value part of the equation.

\(4 + 3|x – 5| = 16\)

Subtract 4 from both sides.

\(3|x – 5| = 12\)

Divide both sides by 3.

\(|x – 5| = 4\)

Now you can write and solve two equations, one where the 4 is negative and one where the 4 is positive. Remember to drop the absolute value symbol at this step.

\(\begin{align}x – 5 &= 4\\ x &= 9\end{align}\)

\(\begin{align}x – 5 &= -4\\ x &= 1\end{align}\)

Once again, there are two solutions to the equation: \(x = \bbox[border: 1px solid black; padding: 2px]{1, 9}\)

Absolute value equations with one solution or no solutions

In both of our examples above, there were two solutions so you may think that this is always the case. While this is often right, there are cases where there is only one solution and even when there are none. The next two examples will show when this happens.

Example – one solution

Solve the equation: \(6|x – 2| – 1 = -1\)

As usual, we will first isolate the absolute value equation.

\(6|x – 2| – 1 = -1\)

Add 1 to both sides.

\(6|x – 2| = 0\)

Divide both sides by 6.

\(|x – 2| = 0\)

Normally at this stage, we would write two equations without the absolute value bars, but writing 0 with a positive or negative is the same thing. So we only have one equation:

\(x – 2 = 0\)

Adding 2 to both sides then gives the only solution.

\(x = \bbox[border: 1px solid black; padding: 2px]{2}\)

We will look more closely at why this happens, but first let’s look at how you might end up with no solutions.

Example – no solutions

Solve the equation: \(|10x – 1| + 3 = -8\)

To isolate the absolute value, subtract 3 from both sides.

\(|10x – 1| = -11\)

At this step, it can be determined that there are no solutions to the equation. Why? The absolute value of any number is positive. Here, we have the absolute value of something is negative. This is not possible so there are no possible x-values that make this equation true. Therefore, you can write:

Answer : No solutions

The absolute value of any number is always positive. Use this to determine when there are no solutions to an absolute value equation.

Notice that in both examples, the steps were the same as before. You will always follow those two steps when solving any absolute value equation.

Why does this work?

You can think of the absolute value of any number as representing how far it is from zero on the number line. Consider \(|3|\) and \(|–3|\) below.

This is why the absolute value is always positive – it is representing a distance. Now think of an equation where the absolute value part is isolated, such as \(|5x + 1| = 2\). Is the absolute value is 2, then all you know is that \(5x + 1\) is 2 units from zero on the number line. This gives two possibilities:

So this is why we end up with two different equations. In the case of only one solution, you end up with an absolute value expression equal to zero. Since this means that the distance from zero on the number line is zero, you end up with only one equation.

When you study the graphs of absolute value equations, you can see the three cases of one solution, no solution, and two solutions graphically. This is due to the shape of the graph of the absolute value function. This is a bit beyond the scope of this lesson, but in the graph below, you can see the graph of \(y = |x – 1|\) and \(y = 2\). Notice that the two graphs intersect at two points. These represent the two solutions to the equation \(|x – 1| = 2\).

You can probably see how a horizontal line might cross the graph at exactly one point (one solution) or at no points (no solutions). This would be just changing the number on the right hand side of the equation.

Absolute value equations are always solved with the same steps: isolate the absolute value term and then write equations based on the definition of the absolute value. There may end up being two solutions, one solution, or no solutions. To catch when there is no solution, always remember that absolute values must be positive, but remember to apply this idea only after the absolute value term has been isolated.

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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:

Content Continues Below

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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

URL: https://www.purplemath.com/modules/solveabs.htm

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6.2: Solving Absolute Value Equations

• Last updated
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• Page ID 45192

• Victoria Dominguez, Cristian Martinez, & Sanaa Saykali
• Citrus College via ASCCC Open Educational Resources Initiative

To solve absolute value equations, first consider the following two properties of absolute value :

Definition: Properties of Absolute Value

Property 1: For \(b > 0\), \(|a| = b\) if and only if \(a = b\) or \(a = −b\)

Property 2: For any real numbers \(a\) and \(b\), \(|a| = |b|\) if and only if \(a = b\) or \(a = −b\)

• Before Property 1 is applied, isolate the absolute value expression to either side of the equation.
• Check the solutions by substituting them back into the original equation.
• Solutions are presented as a solution set of the form \(\{p, q\}\), where \(p\) and \(q\) are any real numbers.
• The solution set of an absolute value equation is graphed as points on a number line.

Example 6.2.1

Solve each equation and graph the solution set.

• \(|x| = 7\)
• \(|5x – 3| = 2\)
• \(|20 – x| = −80\)
• To solve \(|x| = 7\), apply Property 1 with \(a = x\) and \(b = 7\).

Therefore, the solutions are, \(x = −7\) and \(x = 7\), and the solution set is \(\{-7,7\}\). The graph of the solution set is as shown in the figure below.

• The equation-solving method used in part a can be extended to the given equation in this part with \(a = 5x – 3\) and \(b = 2\).

Thus, the absolute value equation \(|5x – 3| = 2\) is equivalent to:

\(\begin{array} &&5x − 3 = 2 &\text{ or } &5x − 3 = −2 &\text{Property 1} \\ &5x = 5 &\text{ or } &5x = 1 &\text{Add \(3\) to both sides of the equations} \\ &x = 1 &\text{ or } &x = \dfrac{1}{5} &\text{Divide by \(5\) both sides of the equations} \end{array}\)

Now, check if \(x = 1\) and \(x = \dfrac{1}{5}\) are solutions to the given absolute value equation.

\(\begin{array} &&\text{For } x = 1 &\text{For } x = \dfrac{1}{5} &\\ &|5x − 3| = 2 &|5x − 3| = 2 &\text{Given} \\ &|5(1) − 3| \stackrel{?}{=} 2 &|5 \left( \dfrac{1}{5} \right) − 3| \stackrel{?}{=} 2 &\text{Substitute the \(x\)-values} \\ &|5 − 3| \stackrel{?}{=} 2 &|1 − 3| \stackrel{?}{=} 2 &\text{Simplify} \\ &|2| \stackrel{?}{=} 2 &|− 2| \stackrel{?}{=} 2 &\text{Apply the absolute value definition} \\ &2 = 2\; \checkmark &2 = 2\; \checkmark \end{array}\)

Since the above equations are true, then, \(x = 1\) and \(x = \dfrac{1}{5}\) are solutions to the given absolute value equation. The solution set is \(\left\{\dfrac{1}{5} , 1\right\}\). The graph of the solution set is as shown in the figure below.

• Since an absolute value can never be negative, there are no real numbers \(x\) that makes \(|20 – x| = −80\) true. The equation has no solution and the solution set is \(∅\).

Example 6.2.2

Solve and graph the solution set.

• \(\left| \dfrac{4}{3} x + 3 \right| + 8 = 18\)
• \(4 \left| \dfrac{1}{3}x − 6 \right| − 5 = −5\)
• \(|4x – 3| = |x + 6|\)
• Notice that the absolute value expression is not isolated which means the properties cannot be applied. First, isolate \(\left| \dfrac{4}{3}x + 3 \right|\) on the left side of the equation, then, apply Property 1.

\(\begin{array} &&\left| \dfrac{4}{3} x + 3 \right| + 8 = 18 &\text{Given equation} \\ & \left| \dfrac{4}{3} + 3 \right| = 10 &\text{Subtract \(8\) from both sides of the equation} \end{array}\)

With the absolute value now isolated, solve\(\left| \dfrac{4}{3} + 3 \right| = 10\) using Property 1, with \(a = \dfrac{4}{3} x + 3\) and \(b = 10\) as follows,

\(\begin{array} && &\left| \dfrac{4}{3} + 3 \right| = 10 & & \\ &\dfrac{4}{3} + 3 = 10 &\text{ or } & \dfrac{4}{3} + 3 = -10 &\text{Property 1} \\ &\dfrac{4}{3} x = 7 &\text{ or } &\dfrac{4}{3}x = −13 &\text{Subtract \(3\) from both sides} \\ &x = \dfrac{21}{4} &\text{ or } &x = −\dfrac{39}{4} &\text{Multiply both sides by \(\dfrac{3}{4}\)} \end{array}\)

Check the solutions \(x = −\dfrac{39}{4}\) and \(x = \dfrac{21}{4}\) by substituting them into the original absolute value equation. The solution set is \(\left\{ −\dfrac{39}{4}, \dfrac{21}{4} \right\}\) and the graph of the solution set is as shown in the figure below.

• Similar to part a, isolate the absolute value expression. So, first isolate \(\left| \dfrac{1}{3} x − 6 \right|\) on the left side of the equation and apply Property 1.

\(\begin{array} &&4 \left| \dfrac{1}{3}x − 6 \right| − 5 = −5 &\text{Given equation} \\ &4 \left| \dfrac{1}{3}x − 6 \right| = 0 &\text{Add \(5\) to both sides of the equation} \\ &\left| \dfrac{1}{3}x − 6 \right| = 0 &\text{Divide by \(4\) both sides of the equation} \end{array}\)

The absolute value is isolated. Since \(0\) is the only number whose absolute value is \(0\), the expression \(\dfrac{1}{3}x − 6\) must be equal to \(0\). So,

\(\begin{array} &&\dfrac{1}{3}x − 6 = 0 & \\ &\dfrac{1}{3}x − 6 &\text{Add \(6\) to both sides of the equation} \\ &x = 18 &\text{Multiply both sides by \(3\)}\end{array}\)

The solution is \(18\) and the solution set is \(\{18\}\). Verify that it satisfies the original equation. The graph of the solution set is as shown in the figure below.

• \(|4x − 7| = |x + 14|\) Notice that to solve \(|4x − 7| = |x + 14|\), use Property 2 with \(a = 4x − 7\) and \(b = x + 14\).

\(\begin{array} && &|4x − 7| = |x + 14| & &\text{Given} \\ &4x−7 = x+14 &\text{ or } &4x − 7 = −(x + 14) &\text{Property 2} \\ &4x−7 = x+14 &\text{ or } &4x − 7 = −x − 14 &\text{Distribute \(−1\) to simplify the right equation} \\ &4x = x + 21 &\text{ or } &4x = −x − 7 &\text{Add \(7\) to both sides of each equality} \\ &3x = 21 &\text{ or } &5x = −7 &\text{Simplify} \\ &x = 7 &\text{ or } &x = −\dfrac{7}{5} &\text{Divide each equation by the \(x\)-coefficient} \end{array}\)

Check the solutions \(x = −\dfrac{7}{5}\) and \(x = 7\) by substituting them into the original absolute value equation. The solution set is \(\left\{ −\dfrac{7}{5}, 7\right\}\). The graph of the solution is as shown in the figure below.

Exercise 6.2.1

Solve each equation, check the solution and graph the solution set.

• \(|x| = 19\)
• \(|x − 4| = 10\)
• \(|2x − 5| = 12\)
• \(\left|\dfrac{x}{11} \right| = 2.5\)
• \(|x − 3.8| = −2.7\)
• \(|3x − 4.5| = 9.3\)
• \(\dfrac{8}{3} |x − 6| = 14\)
• \(|x + 15| − 19 = 7\)
• \(|11x + 3| + 28 = 16\)
• \( \left| \dfrac{8}{7} x + 9 \right| − 2 = 8\)
• \( −3|2x − 7| + 13 = 13\)
• \( 8 − 5|10x + 6| = 5\)
• \( |5x − 14| = |3x − 9|\)
• \( |15x| = |x − 21|\)
• \( |4x − 7| = |5(2x + 3)|\)
• \( \dfrac{7}{8} = \dfrac{3x}{2} + \dfrac{2x}{5}\)