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Solving Absolute-Value Equations (page 1 of 2)

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 simple:

• Solve | x | = 3

I've pretty much already solved this: | 3 | = 3 and | –3 | = 3, so x must be 3 or –3. But how are you supposed to solve this if you don't already know the answer? You use the positive/negative property of the absolute value to split the equation into two cases, and you use the fact that the minus sign " – " indicates "the opposite sign", not necessarily a negative number.

(For example, if you 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 you are changing the sign on x.)

Whatever x might be, the absolute value makes it positive. Since x might have been positive and might have been negative, you have to acknowledge this fact when you take the absolute-value bars off, by splitting the equation into two cases. If x was positive to start with, then you can bring it out without changing its sign, so x = 3. But x might also have been negative, in which case you would have to change the sign on x for the absolute value to come out positive, so you also have –x = 3, which solves as x = –3.

Then the solution is x = –3, 3.

These equations always work this way: to be able to remove the bars, you have to isolate the absolute value onto one side, and then split the equation into two cases.

You can verify the solution graphically. When you have | x | = 3, you are, in effect, setting two line equations equal to each other, and finding where they cross. In this case, as you can see below:

...the two lines, y₁ = | x | and y₂ = 3, crossed at two x-values:  x = –3 and x = 3.

Let's do some more: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

• Solve | x + 2 | = 7

To clear the absolute-value bars, split the equation into two cases, one for each sign:

(x + 2) = 7     or     –(x + 2) = 7
x + 2 = 7       or     –x – 2 = 7
x = 5             or     –9 = x

Then the solution is x = –9, 5.

To confirm this graphically, look for the intersections of  y₁ = | x + 2 | and  y₂ = 7:

• Solve | 2x – 3 | – 4 = 3

First, isolate the absolute-value part:

| 2x – 3 | – 4 = 3
| 2x – 3 | = 7

Now clear the bars by splitting the equation into two cases, one for each sign:

(2x – 3) = 7     or      –(2x – 3) = 7
2x – 3 = 7       or      –2x + 3 = 7
2x = 10           or       –2x = 4
x = 5               or      x = –2

So the solution is  x = –2, 5.

To confirm this graphically, look for the intersections of  y₁ = | 2x – 3 | – 4 and  y₂ = 3:

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