# Absolute-value problems

Consider how the absolute value of a real number x relates to a fixed positive number, say 4:
On the inside is one region where |x| < 4, on the outside are two regions where |x| > 4, and these regions are separated by two points where |x| = 4. Using this picture it's easy to solve these equation and inequalities:
This statement …has this solution … with this standard abbreviation … and this solution set.
|x| < 4; x < 4 and x > 4; −4 < x < 4; x ∈ (−4, 4).
|x| > 4; x < −4 or x > 4; [none]; x ∈ (−∞, −4) ∪ (4, ∞).
|x| = 4; x = −4 or x = 4; x = ±4; x ∈ {−4, 4}.
All of this works in much greater generality, except for the abbreviation ±4 (so I won't mention that again).

## Inequalities with absolute values

The general principles for solving inequalities with absolute vales are actually a bit simpler than the general principle for solving equations, so I'll start with inequalities.

Here are the relevant rules:

This statement … is equivalent to this statment.
|a| < b; b < a < b.
|a| ≤ b; b ≤ a ≤ b.
|a| > b; a < −b or a > b.
|a| ≥ b; a ≤ −b or a ≥ b.

These pairs of statements are equivalent for any real numbers a and b, whether positive, negative, or zero. This means that you can substitute any expressions, however complicated, for a and b above, and the equivalence will be valid. For the examples here, I'll use a linear expression for a and a constant for b; then the statement on the right will be one that you already know how to solve. However, the rule applies to any expressions whatsoever.

For example, here's an inequality where the absolute value of a linear expression is less than a constant:

• |t − 5| < 12 — original statement;
• −12 < t − 5 < 12 — freed of absolute values;
• −7 < t < 17 — add 5 to all three sides;
• t ∈ (−7, 17) — solution set (optional).

And here's one where the absolute value of a linear expression is greater than a constant:

• |2y + 3| > 6 — original statement;
• 2y + 3 < −6 or 2y + 3 > 6 — freed of absolute values;
• 2y < −9 or 2y > 3 — subtract 3 from both sides of each inequality;
• y < −9/2 or y > 3/2 — divide both sides of each inequality by 2;
• y ∈ (−∞, −9/2) ∪ (3/2, ∞) — solution set (optional).

Here's an example with a weak inequality; notice that it uses the same rule as for a strict inequality:

• |7 − x| ≤ 2 — original statement;
• −2 ≤ 7 − x ≤ 2 — freed of absolute values;
• −9 ≤ −x ≤ −5 — subtract 7 from all sides;
• 9 ≥ x ≥ 5 — take the opposite of all sides and reverse the inequalities;
• 5 ≤ x ≤ 9 — swap the order to keep things increasing;
• x ∈ [5, 9] — solution set (optional).

There are also some degenerate problems along this line. Here's an example:

• |n + 3| < −4 — original statement;
• 4 < n + 3 < −4 — freed of absolute values;
• 1 < n < −7 — subtract 3 from all sides;
• False — since 1 > −7;
• n ∈ ∅ — solution set (optional).
You can solve this problem very quickly if you remember that an absolute value is never negative, so it couldn't possibly be less than −4. Even if you don't notice this, however, and go through the steps anyway, still they won't send you astray as long as you do them correctly. But when you get to the end, you should notice that the compound inequality is never true; a graph can really help here.

## Equations with absolute values

The general rule for an equation with an absolute value is a bit complicated:
This statement … is equivalent to this statment.
|a| = b; a = −b or a = b, and b ≥ 0.
Fortunately, when you replace b by a constant, it's obvious whether the last condition is true or not. So normally you just have to look at two possibilities.

Here's an example to show what I mean:

• |2r + 5| = 7 — original statement;
• 2r + 5 = −7 or 2r + 5 = 7, and 7 ≥ 0 — freed of absolute values;
• 2r + 5 = −7 or 2r + 5 = 7 — since in fact 7 > 0;
• 2r = −12 or 2r = 2 — subtract 5 from both sides of each equation;
• r = −6 or r = 1 — divide both sides of each equation by 2;
• r ∈ {−6, 1} — solution set (optional).

Normally, you wouldn't even bother to write down the bit about 7 ≥ 0; since you can see right away that this is true, you go on directly to the next step. Here's an example where I do just that:

• |4c − 8| = 6 — original statement;
• 4c − 8 = −6 or 4c − 8 = 6 — freed of absolute values, since 6 ≥ 0;
• 4c = 2 or 4c = 14 — add 8 to both sides of each equation;
• c = 1/2 or c = 7/2 — divide both sides of each equation by 4;
• c ∈ {1/2, 7/2} — solution set (optional).

Still, you do have to think about that bit; compare this example:

• |2x − 3| = −5 — original statement;
• 2x − 3 = 5 or 2x − 3 = −5, and −5 ≥ 0 — freed of absolute values;
• False — since in fact −5 < 0;
• x ∈ ∅ — solution set (optional).
If I'd forgotten to check whether −5 ≥ 0 and just pressed on with the two equations, then I'd have gotten −1 and 4 as solutions, but these are wrong. (If you put them in for x in |2x − 3|, you'll get 5, not −5.) Again, if you remember that an absolute value can never be negative, then you won't fall into this trap!

## Isolating the absolute value

The rules above only apply directly if the absolute value is alone on one side of the equation or inequality. If this is not so, then you can still solve it, but you first have to isolate the absolute value.

To do this, pretend that the absolute value is itself a single thing (don't pay any attention for now to what's inside it), and this thing is the variable that you're solving for.

For example:

• |y + 2| − 5 = 7 — original problem;
• |y + 2| = 12 — add 5 to both sides to isolate the absolute value;
• y + 2 = −12 or y + 2 = 12 — now freed of absolute values, since 12 ≥ 0;
• y = −14 or y = 10 — subtract 2 from both sides of each equation;
• y ∈ {−14, 10} — solution set (optional).
Here, I had to add 5 to both sides to get the absolute value alone, just like I would have done if I'd wanted to solve u − 5 = 7.

Here's another example:

• 3|k − 9| ≤ 12 — original problem;
• |k − 9| ≤ 4 — divide both sides by 3 to isolate the absolute value;
• −4 ≤ k − 9 ≤ 4 — now freed of absolute values;
• 5 ≤ k ≤ 13 — add 9 to all sides;
• k ∈ [5, 13] — solution set (optional).
This time the first step —to divide by 3— is the same as it would be to solve 3x ≤ 12.

## Equating two absolute values

You might also see an equation with two (or even more) absolute values in it! In general, this can be a quite complicated affair; first, you have to isolate one absolute value and get rid of it, then (in every statement that results) you have to isolate the other absolute value and get rid of it. In principle, you can always do it, but it can be a big mess!

However, if you simply want to say that the two absolute values are equal, then everything becomes much easier. The reason is that two real numbers have the same absolute value exactly when they are either equal or opposite:

This statement … is equivalent to this statment.
|a| = |b|; a = b or a + b = 0.

For example:

• |2x + 4| = |x − 9| — original problem;
• 2x + 4 = x − 9 or (2x + 4) + (x − 9) = 0 — freed of absolute values;
• x + 4 = −9 or 3x − 5 = 0 — subtract x from both sides of the first equation, and simplify the left side of the second equation;
• x = −13 or 3x = 5 — subtract 4 from both sides of the first equation, and add 5 to both sides of the second equation;
• x = −13 or x = 5/3 — divide both sides of the second equation by 3;
• x ∈ {−13, 5/3} — solution set (optional).
This is a little more complicated than the other examples on this page, because the same steps won't solve both equations; you really have to treat them separately. But you can do it.

Occasionally you can get a degenerate one of these too. For example:

• |3t + 4| = |3t + 2| — original problem;
• 3t + 4 = 3t + 2 or (3t + 4) + (3t + 2) = 0 — freed of absolute values;
• 4 = 2 or 6t + 6 = 0 — subtract 3t from both sides of the first equation, and simplify the left side of the second equation;
• 6t + 6 = 0 — in fact 4 > 2;
• 6t = −6 — subtract 6 from both sides of the remaining equation;
• t = −1 — divide both sides of the equation by 6;
• t ∈ {−1} — solution set (optional).
So usually there are two solutions, one for each equation; but sometimes there may be only one (if one equation is always false), or everything may be a solution (if one equation is always true, for example |5 − y| = |y − 5|).
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