Use absolute values when you are interested in the magnitude of an expression but not whether it's positive or negative. For example, if your location is at the origin of a graph where east is a positive direction, west is a negative direction, and you are only interested in gasoline consumption, you can ignore the positive or negative direction associated with a trip. Absolute value statements are sometimes hard to visualize compared to interval notation statements.
Combine two expressions to create one interval-related statement. The two statements may set the limits of an interval that includes both statements (inclusionary). For example, if z is a positive, single-digit number, you can write 0 < z < 10. The two statements may also define two parts of the real numbers outside of an interval (exclusionary). For example, saying that k has more than two digits is really two statements: "k < -99" and "k > 99." It can be combined into a single interval notation statement by using "&" like this: k < -99 & k > 99.
Convert absolute value expressions to interval notation by writing two separate statements that account for both positive and negative values. In the first statement, replace the absolute value symbols with parentheses and precede them with a negative sign. The second statement is the same except the parentheses are preceded by a plus sign.
Remember these two rules.
1. Any "less-than" absolute-value inequalities follow this pattern: If it's | x | < Z, it can always be re-expressed in the form --Z < x < Z.
2. Any "greater than" absolute-value inequalities follow this pattern: If it's | x | > Z, it can be re-expressed in the form x < --Z or x > a.
Solve |3x + 7| < 12. First rewrite it using the pattern for "less than" (See Section 1, Step 4 for patterns): -12< 3x + 7 < 12
Now subtract 7 from all sides and divide by 3 to get "x" by itself:
-19 < 3x < 5
-19/3 < x < 5/3
So, the solution to |3x + 7| < 12 is -19/3 < x < 5/3.
Express the interval notation --1 < x < 5 as an absolute value. Start by looking at the notation's endpoints -1 and 5. These integers are six whole units apart; half of six is three. So restate the equation to get -3 and +3 on either side. To do this, subtract 2 from both sides.
You now have -3 < x-2 < 3. Check the "less than" pattern in Section 1, Step 4. You will see that it is | x | < Z = --Z < x < Z. According to the pattern, therefore, you can write this as an absolute value like this: | x-2 | < 3.