Inequalities and Absolute Value
PreCalculus Name: __________________________
Worksheet: Inequalities and Absolute Value
The rules for solving linear inequalities are simple, and correspond to the same rules for solving equalities except when you multiply or divide both sides by a negative, in which case you must “flip” the inequality. However, when solving more complicated inequalities the situation is not always as obvious. In particular, absolute-value inequalities cannot be done in the “obvious way.”
A safer approach is to treat an inequality as an equality, solve that, use the solutions as “boundary-values” to partition the number-line into “zones of possible solutions,” and finally test numbers in each “zone.” The following example illustrates this procedure applied to an absolute-value inequality:
Example: [pic]
Solution: Treat this as an equality, and solve for the boundary-values: [pic]
Since [pic] if y is positive or 0, and [pic] if y is negative, we must solve two equations in order to get the boundary-values:
[pic] and [pic]
Solving each of these gives the boundary-values [pic] and [pic]. Next we plot these on a number line, and then pick three numbers to test, one in each zone:
Test [pic]: [pic]: [pic], TRUE
Test [pic]: [pic], [pic], FALSE
Test [pic]: [pic], [pic], TRUE
Therefore, the solution is: [pic] or [pic]
Inequalities which contain algebraic fractions must be treated even more carefully. The boundary-values not only contain values of the unknown that make the equality true, but they may also contain values that make a fraction undefined (that is, its denominator is 0). Furthermore, if the inequality includes an equality (that is, “[pic] ” or “[pic]”) then a boundary-value that causes a fraction to be undefined can not be included in the solution.
Example: [pic]
Solution: One of the boundary-values is [pic] because this makes the fraction 0. Treating it as an equality, [pic], we cross-multiply to get: [pic]. To solve this equation, we subtract 6 from both sides, then factor: [pic],
[pic]. The solutions of this equation are [pic] and [pic]. These two boundary-values are also solutions of the original inequality since it includes an equality, but the boundary-value [pic] is not a solution since it makes a fraction undefined. We must now test numbers in 4 zones determined by these three boundary-values:
Test –5: [pic], [pic], TRUE
Test 0: [pic], [pic], FALSE
Test 2.5: [pic], [pic], TRUE
Test 4: [pic], [pic], FALSE
Therefore, the solution is: [pic] or [pic].
Exercises: Solve. Show all work, including test-values.
1. [pic] 2. [pic]
3. [pic] 4. [pic]
5. [pic] 6. [pic]
7. [pic] 8. [pic]
9. [pic] 10. [pic]
11. [pic] 12. [pic]
13. [pic] 14. [pic]
15. [pic] 16. [pic]
-----------------------
4
[pic]
0
2
5
boundary-values
test-numbers
test-numbers
boundary-values
4
2
3
2.5
0
–4
–5
................
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