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]

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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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