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Textbook by: Kenny M. Felder. E-mail the author

# Absolute Value Equations

Module by: Kenny M. Felder. E-mail the author

Summary: This module introduces absolute value equations.

Absolute value is one of the simplest functions—and paradoxically, one of the most problematic.

On the face of it, nothing could be simpler: it just means “whatever comes in, a positive number comes out.”

5 = 5 5 = 5 alignl { stack { size 12{ \lline 5 \rline =5} {} # {} } } {}
(1)
5 = 5 5 = 5 size 12{ \lline - 5 \rline =5} {}
(2)

Absolute values seem to give us permission to ignore the whole nasty world of negative numbers and return to the second grade when all numbers were positive.

But consider these three equations. They look very similar—only the number changes—but the solutions are completely different.

 ∣ x ∣ = 10 ∣ x ∣ = 10 size 12{ \lline x \rline ="10"} {} ∣ x ∣ = − 10 ∣ x ∣ = − 10 size 12{ \lline x \rline = - "10"} {} ∣ x ∣ = 0 ∣ x ∣ = 0 size 12{ \lline x \rline =0} {} x=10x=10 size 12{x="10"} {} works. x=10x=10 size 12{x="10"} {} doesn’t work. x=0x=0 size 12{x=0} {} is the only solution. Hey, so does x=−10x=−10 size 12{x= - "10"} {}! Neither does x=−10x=−10 size 12{x= - "10"} {}. Concisely, x=±10x=±10 size 12{x= +- "10"} {}. Hey...absolute values are never negative!

We see that the first problem has two solutions, the second problem has no solutions, and the third problem has one solution. This gives you an example of how things can get confusing with absolute values—and how you can solve things if you think more easily than with memorized rules.

For more complicated problems, follow a three-step approach.

1. Do the algebra to isolate the absolute value.
2. Then, think it through like the simpler problems above.
3. Finally, do more algebra to isolate xx.

In my experience, most problems with this type of equation do not occur in the first and third step. And they do not occur because students try to think it through (second step) and don’t think it through correctly. They occur because students try to take “shortcuts” to avoid the second step entirely.

## Example 1: Absolute Value Equation (No Variable on the Other Side)

3 | 2 x + 1 | - 7 = 5 3 | 2 x + 1 | - 7 = 5 size 12{3 \lline 2x+1 \rline - 7=5} {}

• Step 1:: Algebraically isolate the absolute value
• 3 | 2 x + 1 | = 12 3 | 2 x + 1 | = 12 size 12{3 \lline 2x+1 \rline ="12"} {}
• | 2 x + 1 | = 4 | 2 x + 1 | = 4 size 12{ \lline 2x+1 \rline =4} {}
• Step 2:: Think!
• For the moment, forget about the quantity 2 x + 1 2 x + 1 size 12{2x+1} {}; just think of it as something. The absolute value of “something” is 4. So, in analogy to what we did before, the “something” can either be 4, or –4. So that gives us two possibilities...
• 2x + 1 = 4 2x + 1 = 4
• 2x + 1 = 4 2x + 1 = 4
• Step 3:: Algebraically solve (both equations) for x
• 2 x =3 2 x =3 size 12{2x=3} {} or 2 x =5 2 x =5 size 12{2x= - 5} {}
• x = 3 2 x = 3 2 size 12{x= { { size 8{3} } over { size 8{2} } } } {} or x = 5 2 x = 5 2 size 12{x= - { { size 8{5} } over { size 8{2} } } } {}

So this problem has two answers: x = 3 2 x = 3 2 size 12{x= { { size 8{3} } over { size 8{2} } } } {} and x = 5 2 x = 5 2 size 12{x= - { { size 8{5} } over { size 8{2} } } } {}

## Example 2: Absolute Value Equation (No Variable on the Other Side)

6x25+7=46x25+7=4 size 12{ { {6 \lline x - 2 \lline } over {5} } +7=4} {}

• Step 1: Algebraically isolate the absolute value:
6 | x - 2 | 5 = -3 6 | x - 2 | 5 =-3
(3)
6 | x - 2 | = -15 6|x-2|=-15
(4)
| x - 2 | = -15 6 = -5 2 |x-2|= -15 6 = -5 2
(5)
• Step 2: Think!: The absolute value of “something” is 2 1 2 2 1 2 . But wait—absolute values are never negative! It can’t happen! So we don’t even need a third step in this case: the equation is impossible.
• No solution.

The “think” step in the above examples was relatively straightforward, because there were no variables on the right side of the equation. When there are variables on the right side, you temporarily “pretend” that the right side of the equation is a positive number, and break the equation up accordingly. However, there is a price to be paid for this slight of hand: you have to check your answers, because they may not work even if you do your math correctly.

## Example 3: Absolute Value Equation with Variables on Both Sides

2x + 3 = 11 x + 42 2x + 3 = 11 x + 42 size 12{ \lline 2x+3 \rline = - "11"x+"42"} {}

We begin by approaching this in analogy to the first problem above, x=10x=10 size 12{ \lline x \rline ="10"} {}. We saw that x x size 12{x} {} could be either 10, or –10. So we will assume in this case that 2x + 3 2x + 3 size 12{2x+3} {} can be either 11 x + 42 11 x + 42 size 12{ - "11"x+"42"} {} , or the negative of that, and solve both equations.

### Problem One

• 2 x + 3 = -11 x + 42 2x+3=-11x+42
(6)
• 13 x + 3 = 42 13x+3=42
(7)
• 13 x = 39 13x=39
(8)
• x = 3 x=3
(9)

### Problem Two

• 2 x + 3 = - ( - 11 x + 42 ) 2x+3=-(-11x+42)
(10)
• 2 x + 3 = 11 x - 42 2x+3=11x-42
(11)
• -9 x + 3 = -42 -9x+3=-42
(12)
• -9 x = -45 -9x=-45
(13)
• x = 5 x=5
(14)

So we have two solutions: x=3x=3 size 12{x=3} {} and x=5x=5 size 12{x=5} {}. Do they both work? Let’s try them both.

### Problem One

1. 2 ( 3 ) + 3 = 11 ( 3 ) + 42 2 ( 3 ) + 3 = 11 ( 3 ) + 42 size 12{ \lline 2 $$3$$ +3 \rline = - "11" $$3$$ +"42"} {} .
• 9=99=9 size 12{ \lline 9 \rline =9} {}.

### Problem Two

1. 2 ( 5 ) + 3 = 11 ( 5 ) + 42 2 ( 5 ) + 3 = 11 ( 5 ) + 42 size 12{ \lline 2 $$5$$ +3 \rline = - "11" $$5$$ +"42"} {} .
• 13=1313=13 size 12{ \lline "13" \rline = - "13"} {}.

We see in this case that the first solution, x=3x=3 size 12{x=3} {}, worked; the second, x=5x=5 size 12{x=5} {}, did not. So the only solution to this problem is x=3x=3 size 12{x=3} {}.

However, there was no way of knowing that in advance. For such problems, the only approach is to solve them twice, and then test both answers. In some cases, both will work; in some cases, neither will work. In some cases, as in this one, one will work and the other will not.

## Hard and fast rule:

Whenever an absolute value equation has variables on both sides, you have to check your answer(s). Even if you do all the math perfectly, your answer(s) may not work.

OK, why is that? Why can you do all the math right and still get a wrong answer?

Remember that the problem x=10x=10 size 12{ \lline x \rline ="10"} {} has two solutions, and x=10x=10 size 12{ \lline x \rline = - "10"} {} has none. We started with the problem 2x+3=11x+422x+3=11x+42 size 12{ \lline 2x+3 \rline = - "11"x+"42"} {}. OK, which is that like? Is the right side of the equation like 10 or –10? If you think about it, you can convince yourself that it depends on what xx size 12{x} {} is. After you solve, you may wind up with an xx size 12{x} {}-value that makes the right side positive; that will work. Or, you may wind up with an xx size 12{x} {}-value that makes the right side negative; that won’t work. But you can’t know until you get there.

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