Lesson Explainer: Absolute Value Equations Mathematics

In this explainer, we will learn how to solve equations involving the absolute value.

Recall that the absolute value of a real number is its distance from 0 on the number line. For example, in the expression |5| (which can be read as the absolute value of 5), the number 5 is given using absolute value notation, the two vertical bars. Since 5 is located 5 units from 0 on the number line, the value of the expression is 5. The value of the expression |5| (which can be read as the absolute value of 5) is also 5, because 5 is also located 5 units from 0 on the number line.

Definition: The Absolute Value

The absolute value of a number is the magnitude of the number without regard to its sign: |𝑥|=𝑥𝑥0,𝑥𝑥<0.ifif

Let us also recap how to plot the graph of an absolute value function. To do this, we can complete a table of values for 𝑦=|𝑥|:

𝑥3210123
𝑦3210123

Then, we can plot the coordinates on the Cartesian plane to draw the graph:

Being able to apply the definition of the absolute value and plot absolute value graphs is very useful when solving absolute value equations, so practicing these skills is very important.

Let us now introduce the concept of absolute value equations.

Consider the equation |𝑥+1|=3.

Before we look at this algebraically, it can often be useful to think about the problem graphically. So, on the same set of axes, let us plot 𝑦=|𝑥+1| and 𝑦=3. To do this, we can complete a table of values for 𝑦=|𝑥+1|:

𝑥3210123
𝑦2101234

From this graph, we can see that the two lines intersect at two distinct points, when 𝑥=4 and when 𝑥=2.

So, the solution to the equation |𝑥+1|=3 is 𝑥=4 or 𝑥=2. This can be written as a solution set in the form {4,2}.

Now, we can also see from the graph how we might approach solving this equation algebraically. Recalling our definition of the absolute value function, we can represent this using piecewise notation: |𝑥+1|=𝑥+1𝑥1,(𝑥+1)𝑥<1.ifif

As |𝑥+1|=3, we will solve the two equations 𝑥+1=3 and (𝑥+1)=3.

We start with 𝑥+1=3(1)𝑥=2.subtractingfrombothsides

Then, we look at (𝑥+1)=3()𝑥1=3(1)𝑥=4(1)𝑥=4.distributingtheparenthesesaddingtobothsidesdividingbothsidesby

Combining these two equations, we can see that the solutions are 𝑥=4 or 𝑥=2, which agrees with our initial findings from inspecting the graphs.

An alternative algebraic method, as opposed to using piecewise notation, is to set the absolute value expression equal to the positive and the negative of the other quantity.

In this example, |𝑥+1|=3, we need to solve two equations: 𝑥+1=3 and 𝑥+1=3.

Subtracting 1 from both sides of both equations, we once again obtain the solutions 𝑥=2 and 𝑥=4.

This method can be summarized as follows.

How To: Solving Absolute Value Equations

  1. Isolate the absolute value expression.
  2. Set the quantity inside the absolute value notation equal to the positive and the negative of the quantity on the other side of the equation.
  3. Solve for the unknown in both equations.
  4. Check the answer analytically or graphically.

The two methods are equivalent, and while we can use either to solve absolute value equations, one or the other of the methods can make the process more efficient.

We will now solve a variety of absolute value equations both algebraically and graphically.

Example 1: Solving Absolute Value Equations

What is the solution set of the equation 3|𝑥|66=0?

Answer

We will solve this question first by using an algebraic approach and then by using a graphical approach.

To solve this equation algebraically, we will follow the following four-step process:

  1. Isolate the absolute value expression.
  2. Set the quantity inside the absolute value notation equal to the positive and the negative of the quantity on the other side of the equation.
  3. Solve for the unknown in both equations.
  4. Check the answer analytically or graphically.

In order to isolate the absolute value expression, we begin by rearranging the equation to make |𝑥| the subject: 3|𝑥|66=0(66)3|𝑥|=66(3)|𝑥|=22.addingtobothsidesdividingbothsidesby

Recalling that the absolute value of a real number is its distance from zero, we have two possible solutions:

Either 𝑥=22 or 𝑥=22

The solution set of the equation 3|𝑥|66=0 contains the values 22 and 22.

We can check these solutions by substituting the values into the original equation, 3|𝑥|66=0. In both cases, the left-hand side of the equation equals zero.

We will now consider how we can demonstrate this solution graphically. Consider the equation 𝑦=|𝑥|. The graph of this equation can be drawn on a pair of coordinate axes as follows:

As |𝑥|=22, we can draw the horizontal line 𝑦=22 onto the same graph.

Our two lines intersect at two distinct points 𝑥=22 and 𝑥=22, which are the two solutions to the equation 3|𝑥|66=0.

Therefore, the solution set is {22,22}.

Example 2: Solving Absolute Value Equations

Find algebraically the solution set of the equation |𝑥+4|=𝑥+4.

Answer

Recalling our definition of the absolute value function, we can solve this equation using piecewise notation: |𝑥+4|=𝑥+4𝑥4,(𝑥+4)𝑥<4.ifif

As |𝑥+4|=𝑥+4, we can consider solving the two equations 𝑥+4=𝑥+4 and (𝑥+4)=𝑥+4.

Firstly, when 𝑥4, we solve the equation 𝑥+4=𝑥+4.

As the left-hand side of our equation is exactly the same as the right-hand side, this equation is true for all values of 𝑥. However, we were restricting the solutions to values of 𝑥 greater than or equal to 4, so the solution to this part of the equation is 𝑥4.

Secondly, when 𝑥<4, we need to solve the equation (𝑥+4)=𝑥+4()𝑥4=𝑥+4(𝑥)4=2𝑥+4(4)8=2𝑥(2)4=𝑥.distributingtheparenthesesaddingtobothsidessubtractingfrombothsidesdividingbothsidesby

In this case, as we restricted this to the interval of 𝑥<4, this is not a valid solution to this part of the equation.

However, this solution is already covered in the solution to our first equation. The solution of the equation |𝑥+4|=𝑥+4 is 𝑥4, which in interval notation can be written [4,[.

We can verify this solution graphically. Let us begin by drawing the equation 𝑦=𝑥+4 on the cartesian plane.

By drawing the equation 𝑦=|𝑥+4| on the same axes, we see that both graphs are identical for values of 𝑥 such that 𝑥4.

This confirms our earlier solution, |𝑥+4|=𝑥+4, when 𝑥4.

Therefore, the solution set is [4,[.

The remaining examples in this explainer will involve solving absolute value equations that result in quadratic expressions.

Example 3: Solving Absolute Value Equations

Find algebraically the solution set of the equation 11𝑥+44|𝑥+4|=𝑥.

Answer

In this question, we begin by noticing that since the denominator of a fraction cannot equal zero, the left-hand side of our equation is undefined when 𝑥=4. This means that we have two different situations to consider: firstly, when 𝑥+4>0 and, secondly, when 𝑥+4<0.

Recalling our definition of the absolute value function, we can then represent this part of the equation using piecewise notation: |𝑥+4|=𝑥+4𝑥>4,(𝑥+4)𝑥<4.ifif

When 𝑥>4,we solve the equation 11𝑥+44𝑥+4=𝑥((𝑥+4))11𝑥+44=𝑥(𝑥+4)()11𝑥+44=𝑥+4𝑥(11𝑥44)0=𝑥7𝑥44()0=(𝑥11)(𝑥+4).multiplyingbothsidesbydistributingtheparenthesessubtractingandfrombothsidesfactoringtheright-handside

So, eitheror𝑥=11𝑥=4.

As we have already established that 𝑥 cannot equal 4, this equation has one valid solution.

When 𝑥<4, we need to solve the equation 11𝑥+44(𝑥+4)=𝑥((𝑥+4))11𝑥+44=𝑥(𝑥+4)()11𝑥+44=𝑥4𝑥(𝑥4𝑥)𝑥+15𝑥+44=0()(𝑥+11)(𝑥+4)=0.multiplyingbothsidesbydistributingtheparenthesesaddingandtobothsidesfactoringtheleft-handside

So, eitheror𝑥=11𝑥=4.

Once again, we established that 𝑥 cannot equal 4, so this equation also has one valid solution.

The solutions of the equation 11𝑥+44|𝑥+4| are 𝑥=11 and 𝑥=11, which can be written in set notation as {11,11}.

Example 4: Solving Absolute Value Linear Equations

Find algebraically the solution set of the equation |𝑥3||𝑥+1|=4.

Answer

Recalling our definition of the absolute value function, we can represent the |𝑥3| part of the equation using piecewise notation: |𝑥3|=𝑥3𝑥>3,(𝑥3)𝑥3.ifif

We can also represent the |𝑥+1| part of the equation using piecewise notation: |𝑥+1|=𝑥+1𝑥1,(𝑥+1)𝑥<1.ifif

This means that we need to consider three scenarios.

Firstly, when 𝑥<1, we know that 𝑥3<0 and 𝑥+1<0. We need to solve the equation (𝑥3)(𝑥+1)=4𝑥+3+𝑥+1=44=4.

So, |𝑥3||𝑥+1|=4 is true for all values of 𝑥<1.

Secondly, when 1𝑥3, we know that 𝑥30 and 𝑥+10. We need to solve the equation (𝑥3)(𝑥+1)=4𝑥+3𝑥1=42𝑥+2=42𝑥=2𝑥=1.

So, |𝑥3||𝑥+1|=4 is also true when 𝑥=1.

Finally, when 𝑥>3, we know that 𝑥3>0 and 𝑥+1>0. We need to solve the equation (𝑥3)(𝑥+1)=4𝑥3𝑥1=44=4.

So, there are no solutions of |𝑥3||𝑥+1|=4 when 𝑥>3.

We can therefore conclude that the equation |𝑥3||𝑥+1|=4 is true for all values of 𝑥 less than or equal to 1. This can be written in set notation as ],1].

Example 5: Finding the Solution Set of Quadratic Equations Involving Absolute Value

Find algebraically the solution set of the equation 4𝑥|𝑥|4𝑥=0.

Answer

To solve this equation, we will follow the following four-step process:

  1. Isolate the absolute value expression.
  2. Set the quantity inside the absolute value notation equal to the positive and the negative of the quantity on the other side of the equation.
  3. Solve for the unknown in both equations.
  4. Check your answer analytically or graphically.

We have 4𝑥|𝑥|4𝑥=0()4𝑥(|𝑥|1)=0.factoringtheleft-handside

Either 4𝑥=0(4)𝑥=0dividingbothsidesby or |𝑥|1=0(1)|𝑥|=1.addingtobothsides

Recalling that the absolute value of a real number is its distance from zero, there are two possible solutions to this equation:

Either 𝑥=1 or 𝑥=1.

Combining our two results, we have three possible solutions, 𝑥=1, 𝑥=0, and 𝑥=1. This can be written as the solution set {1,0,1}.

Example 6: Finding the Solution Set of Root Equations Involving Absolute Value

Find the solution set of the equation 4𝑥28𝑥+49=|𝑥+4|.

Answer

We begin by recalling that the absolute value of any function must be positive. This means that we can square both sides of our equation without adding any extraneous solutions: 4𝑥28𝑥+49=(|𝑥+4|)4𝑥28𝑥+49=(𝑥+4)()4𝑥28𝑥+49=𝑥+8𝑥+16(𝑥+8𝑥+16)3𝑥36𝑥+33=0(3)𝑥12𝑥+11=0()(𝑥11)(𝑥1)=0.distributingtheparenthesessubtractingfrombothsidesdividingbothsidesbyfactoringtheleft-handside

So, eitheror𝑥=11𝑥=1.

Checking that these solutions satisfy the original equation, we have

LHS: 4(11)28(11)+49=225=15 RHS: |(11)+4|=|15|=15 and

LHS: 4(1)28(1)+49=25=5 RHS: |(1)+4|=|5|=5.

We can therefore conclude that there are two solutions to the equation 4𝑥28𝑥+49=|𝑥+4|. They are 𝑥=11 and 𝑥=1. This can be written as the solution set {1,11}.

This can be represented graphically as shown below.

The points of intersection of 𝑦=4𝑥28𝑥+49 and 𝑦=|𝑥+4| occur when 𝑥=11 and 𝑥=1.

The solution set is, therefore, {1,11}.

Example 7: Finding the Solution Set of Quadratic Equations Involving Absolute Value by Factorization

Find algebraically the solution set of the equation |𝑥+13𝑥+21|=21.

Answer

Recalling our definition of the absolute value function, we have two different situations to consider: firstly, when 𝑥+13𝑥+210 and, secondly, when 𝑥+13𝑥+21<0.

In the first scenario, we solve the equation 𝑥+13𝑥+21=21(21)𝑥+13𝑥=0()𝑥(𝑥+13)=0.subtractingfrombothsidesfactoringtheleft-handside

So, eitheror𝑥=0𝑥=13.

We check that these two solutions are valid by substituting them into the equation |𝑥+13𝑥+21|=21.

When 𝑥=0, |(0)+13(0)+21|=21|21|=2121=21.

When 𝑥=13, |(13)+13(13)+21|=21|169169+21|=21|21|=2121=21.

Therefore, the solutions 𝑥=0 and 𝑥=13 are both valid.

When 𝑥+13𝑥+21<0, we solve the equation 𝑥+13𝑥+21=21()𝑥13𝑥21=21(21)𝑥13𝑥42=0(1)𝑥+13𝑥+42=0()(𝑥+6)(𝑥+7)=0.distributingtheparenthesessubtractingfrombothsidesdividingbothsidesbyfactoringtheleft-handside

So, eitheror𝑥=6𝑥=7.

We check that these two solutions are valid by substituting them into the equation |𝑥+13𝑥+21|=21.

When 𝑥=6, |(6)+13(6)+21|=21|3678+21|=21|21|=2121=21.

When 𝑥=7, |(7)+13(7)+21|=21|4991+21|=21|21|=2121=21.

Therefore, the solutions 𝑥=6 and 𝑥=7 are both valid.

Combining our two results, there are four possible solutions to the equation |𝑥+13𝑥+21|=21. They are 𝑥=13, 𝑥=0, 𝑥=6, and 𝑥=7. This can be written as the solution set {13,7,6,0}.

We can also demonstrate this graphically. Let us begin by sketching the equation 𝑦=𝑥+13𝑥+21 on the Cartesian plane.

In order to sketch the graph of 𝑦=|𝑥+13𝑥+21| we reflect the portion of the graph below the 𝑥-axis in the line 𝑦=0, as shown in the figure below.

Drawing the horizontal line 𝑦=21, we see that this intersects 𝑦=|𝑥+13𝑥+21| at four distinct points.

These correspond to the values 𝑥=13, 𝑥=0, 𝑥=6, and 𝑥=7 as identified in the algebraic solution above.

Therefore, the solution set is {13,7,6,0}.

We will finish this explainer by recapping some of the important concepts.

Key Points

  • The absolute value is the magnitude of a number without regard to its sign.
  • We solve absolute value equations by considering the set of values for which the absolute value can be positive and negative.
  • We then work out the solution sets separately and check if each solution matches the above criteria.
  • Absolute value equations can be solved both graphically and algebraically.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.