On the inside is one region where |

This statement … | has this solution. |
---|---|

|x| < 4; |
−4 < x < 4; |

|x| > 4; |
x < −4 or x > 4; |

|x| = 4; |
x = −4 or x = 4. |

This works in much greater generality.

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 in this class,
we'll typically 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.

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 sides.

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

- |2
*y*+ 3| > 6 — original statement; - 2
*y*+ 3 < −6 or 2*y*+ 3 > 6 — freed of absolute values; - 2
*y*< −9 or 2*y*> 3 — subtract 3 from both sides of each inequality; *y*< −9/2 or*y*> 3/2 — divide both sides of each inequality by 2.

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 (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.

|Fortunately, when you replacea| =b⇔a= −bora=b, andb≥ 0.

Here's an example to show what I mean:

- |2
*r*+ 5| = 7 — original statement; - 2
*r*+ 5 = −7 or 2*r*+ 5 = 7, and 7 ≥ 0 — freed of absolute values; - 2
*r*+ 5 = −7 or 2*r*+ 5 = 7 — since in fact 7 > 0; - 2
*r*= −12 or 2*r*= 2 — subtract 5 from both sides of each equation; *r*= −6 or*r*= 1 — divide both sides of each equation by 2.

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:

- |4
*c*− 8| = 6 — original statement; - 4
*c*− 8 = −6 or 4*c*− 8 = 6 — freed of absolute values, since 6 ≥ 0; - 4
*c*= 2 or 4*c*= 14 — add 8 to both sides of each equation; *c*= 1/2 or*c*= 7/2 — divide both sides of each equation by 4.

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

- |2
*x*− 3| = −5 — original statement; - 2
*x*− 3 = 5 or 2*x*− 3 = −5, and −5 ≥ 0 — freed of absolute values; - False — since in fact −5 < 0.

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;*y*= −14 or*y*= 10 — subtract 2 from both sides of each equation.

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.

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. In symbols,

|a| = |b| ⇔a=bora+b= 0.

For example:

- |2
*x*+ 4| = |*x*− 9| — original problem; - 2
*x*+ 4 =*x*− 9 or (2*x*+ 4) + (*x*− 9) = 0 — freed of absolute values; *x*+ 4 = −9 or 3*x*− 5 = 0 — subtract*x*from both sides of the first equation, and simplify the left side of the second equation;*x*= −13 or 3*x*= 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.

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

- |3
*t*+ 4| = |3*t*+ 2| — original problem; - 3
*t*+ 4 = 3*t*+ 2 or (3*t*+ 4) + (3*t*+ 2) = 0 — freed of absolute values; - 4 = 2 or 6
*t*+ 6 = 0 — subtract 3*t*from both sides of the first equation, and simplify the left side of the second equation; - 6
*t*+ 6 = 0 — in fact 4 > 2; - 6
*t*= −6 — subtract 6 from both sides of the remaining equation; *t*= −1 — divide both sides of the equation by 6.

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