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1. Equations & Inequalities in One Variable
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United States of AmericaTennessee
Integrated Math 1

1.05 Absolute value equations

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Lesson

Concept summary

The absolute value of a number is the distance of that number from 0 on a number line. Distance is expressed as a positive value, and so the absolute value of a number is always positive. Absolute value is sometimes called magnitude.

To denote the absolute value of a quantity, we use a single vertical bar on either side. The absolute value of x can be thought of as:

  • If x \geq 0, then \left|x\right|=x
  • If x < 0, then \left|x\right|=-x
Absolute value equation

An equation that contains a variable expression inside absolute value bars.

Example:

\left|x\right|=4

Extraneous solution

A solution of an equation that emerges from the process of solving the problem, but is not a valid solution to the problem.

An absolute value equation can be solved by applying arithmetic operations as usual, but sometimes one or more solutions to an absolute value equation will be extraneous solutions depending on the context.

Worked examples

Example 1

Solve each absolute value equation.

a

\left|x\right|=6

Solution

The solutions to this equation will be any values of x that have an absolute value of 6.

The solutions are x=6 and x=-6.

Reflection

Notice that this equation has different solutions from the equation \left|x-6\right|=0 which has only a single solution of x=6. This shows that adding or subtracting terms in or out of the absolute value bars changes the equation, similar to parentheses.

b

\left|2.5x\right|=5

Solution

The solutions to this equation will be any values of x that make 2.5x have an absolute value of 5.

The solutions are x=2 and x=-2.

Reflection

Notice that this equation has the same solutions as the equation 2.5\left|x\right|=5. This demonstrates that multiplying or dividing a constant term in or out of the absolute value bars does not change the equation.

c

\left|3x\right|+4=2x+9

Approach

This equation is harder to solve because it involves x-terms both inside and outside absolute value bars. One way to work around this is to split up the problem into two cases: when 3x\geq0, and when 3x<0.

Solution

When 3x\geq0, we know that \left|3x\right|=3x, so we can write and solve the equation:

\displaystyle \left|3x\right|+4\displaystyle =\displaystyle 2x+9
\displaystyle 3x+4\displaystyle =\displaystyle 2x+9Take the case where 3x\geq0
\displaystyle x+4\displaystyle =\displaystyle 9Subtract 2x from both sides
\displaystyle x\displaystyle =\displaystyle 5Subtract 4 from both sides

When 3x<0, we know that \left|3x\right|=-3x, we can write and solve the equation:

\displaystyle \left|3x\right|+4\displaystyle =\displaystyle 2x+9
\displaystyle -3x+4\displaystyle =\displaystyle 2x+9Take the case where 3x<0
\displaystyle 4\displaystyle =\displaystyle 5x+9Add 3x to both sides
\displaystyle -5\displaystyle =\displaystyle 5xSubtract 9 from both sides
\displaystyle -1\displaystyle =\displaystyle xDivide both sides by 5

Using two cases, we have found two solutions: x=5 and x=-1.

We can substitute each solution into the starting equation to check that they are valid.

\displaystyle \left|3(5)\right|+4\displaystyle =\displaystyle 2(5)+9
\displaystyle 15+4\displaystyle =\displaystyle 10+9
\displaystyle 19\displaystyle =\displaystyle 19

and

\displaystyle \left|3(-1)\right|+4\displaystyle =\displaystyle 2(-1)+9
\displaystyle 3+4\displaystyle =\displaystyle -2+9
\displaystyle 7\displaystyle =\displaystyle 7

Reflection

If substituting either solution into the starting equation resulted in an untrue equation, then that solution would be extraneous.

Example 2

Clay wants to find the point in a 100 yard dash where his distance in yards from the middle of the race is equal to 40 less than twice the number of yards he has run so far.

Find the number of yards that Clay can have run to make this equality true.

Approach

The varying factor in this context is the number of yards that Clay has run. We can represent this with the variable y.

Clay's distance from the middle of the race can be described as the difference between the number of yards he has run, y, and 50 yards. We can write this as the absolute value expression \left|y-50\right|.

Since y represents how far Clay has run in the race, we can express "40 less than twice the yards he has run so far" as the expression 2y-40.

Putting these two expressions together, we get the equation that Clay wants to solve:\left|y-50\right|=2y-40

Solution

To solve this equation, we can use the two cases method.

Taking the case where y-50\geq0, we have:

\displaystyle y-50\displaystyle =\displaystyle 2y-40
\displaystyle -50+40\displaystyle =\displaystyle 2y-y
\displaystyle -10\displaystyle =\displaystyle y

Taking the case where y-50<0, we have:

\displaystyle -(y-50)\displaystyle =\displaystyle 2y-40
\displaystyle 50-y\displaystyle =\displaystyle 2y-40
\displaystyle 50+40\displaystyle =\displaystyle 2y+y
\displaystyle 90\displaystyle =\displaystyle 3y
\displaystyle 30\displaystyle =\displaystyle y

Solving using the two cases method gives us the possible solutions y=-10 and y=30.

We can see that if we substitute y=-10 into our starting equation we will get \left|-10-50\right|=2(-10)-40 which is not true. This tells us that y=-10 is an extraneous solution.

We can also see that if we substitute y=30 into our starting equation we will get \left|30-50\right|=2(30)-40 which is true.

So Clay must run exactly 30 yards so that his distance from the middle of the race is equal to 40 less than twice the number of yards he has run.

Reflection

Notice also that the solution y=-10 is made non-valid by the context, since Clay cannot run a negative number of yards.

Example 3

A freediver holds her breath and dives below the surface of the water. Her depth, d in meters, after t seconds, is given by:

d=\dfrac{2}{3}\left| t-75\right| -50

a

Write an equation that could be used to solve for t.

Approach

We need to rearrange the equation to isolate t. To rearange this formula, we can consider the two cases involved and use the properties of equality and inverse operations to get the solution.

Solution

Taking the case where t-75\geq0, we have:

\displaystyle d\displaystyle =\displaystyle \dfrac{2}{3}\left( t-75\right) -50
\displaystyle d+50\displaystyle =\displaystyle \dfrac{2}{3}\left( t-75\right)Addition property of equality
\displaystyle \dfrac{3}{2}\left(d+50\right)\displaystyle =\displaystyle t-75Multiplication property of equality
\displaystyle \dfrac{3}{2}\left(d+50\right)+75\displaystyle =\displaystyle tAddition property of equality
\displaystyle t\displaystyle =\displaystyle \dfrac{3}{2}\left(d+50\right)+75Symmetric property of equality

Taking the case where t-75\lt 0, we have:

\displaystyle d\displaystyle =\displaystyle -\dfrac{2}{3}\left( t-75\right) -50
\displaystyle d+50\displaystyle =\displaystyle -\dfrac{2}{3}\left( t-75\right)Addition property of equality
\displaystyle -\dfrac{3}{2}\left(d+50\right)\displaystyle =\displaystyle t-75Multiplication property of equality
\displaystyle -\dfrac{3}{2}\left(d+50\right)+75\displaystyle =\displaystyle tAddition property of equality
\displaystyle t\displaystyle =\displaystyle -\dfrac{3}{2}\left(d+50\right)+75Symmetric property of equality

Reflection

Notice the two equations that result only differ by the sign in front of \dfrac{3}{2}. This means we could efficiently write the two formulas together as follows: t=\pm \dfrac{3}{2}\left(d+50\right)+75

b

Find the time(s) when the diver is 40 meters below the surface. (d=-40)

Approach

Substitute d=-40 into the formula for each case found in part (a) and check each solution is valid in the context.

Solution

Taking the case where d-75\geq0, we have:

\displaystyle t\displaystyle =\displaystyle \dfrac{3}{2}\left(d+50\right)+75
\displaystyle t\displaystyle =\displaystyle \dfrac{3}{2}\left(-40+50\right)+75Substitute d=-40 into the equation
\displaystyle t\displaystyle =\displaystyle \dfrac{3}{2}\left(10\right)+75Evaluate the addition
\displaystyle t\displaystyle =\displaystyle 15+75Evaluate the multiplication
\displaystyle t\displaystyle =\displaystyle 90Evaluate the addition

Taking the case where d-75\geq0, we have:

\displaystyle t\displaystyle =\displaystyle -\dfrac{3}{2}\left(d+50\right)+75
\displaystyle t\displaystyle =\displaystyle -\dfrac{3}{2}\left(-40+50\right)+75Substitute d=-40 into the equation
\displaystyle t\displaystyle =\displaystyle -\dfrac{3}{2}\left(10\right)+75Evaluate the addition
\displaystyle t\displaystyle =\displaystyle -15+75Evaluate the multiplication
\displaystyle t\displaystyle =\displaystyle 60Evaluate the addition

The diver will reach 40 meters below the surface at 60 and 90 seconds into the dive.

Reflection

Here both solutions are valid in the context, with the diver at 40 meters below the surface at 60 seconds on her descent and again at 90 seconds on her ascent.

The substitution process can be made more efficient by noting which part of the formula is different in the two cases. In this example, only the final step differs and the second substitution could jump to this step.

Outcomes

M1.N.Q.A.1

Use units as a way to understand real-world problems.*

M1.N.Q.A.1.C

Define and justify appropriate quantities within a context for the purpose of modeling.*

M1.A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems in a real-world context.*

M1.A.REI.A.1

Understand solving equations as a process of reasoning and explain the reasoning. Construct a viable argument to justify a solution method.

M1.A.REI.B.2

Solve linear and absolute value equations and inequalities in one variable.

M1.A.REI.B.2.B

Solve absolute value equations and inequalities in one variable. Represent solutions algebraically and graphically.

M1.MP1

Make sense of problems and persevere in solving them.

M1.MP2

Reason abstractly and quantitatively.

M1.MP3

Construct viable arguments and critique the reasoning of others.

M1.MP4

Model with mathematics.

M1.MP6

Attend to precision.

M1.MP7

Look for and make use of structure.

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