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: |2x  3|  x  1 Solution: Treat this as an equality, and solve for the boundary-values: |2x  3|  x  1 Since | y |  y if y is positive or 0, and | y |  y if y is negative, we must solve two equations in order to get the boundary-values:

2x  3  x  1 and 2x  3  (x  1)

2 Solving each of these gives the boundary-values x  4 and x  . Next we plot 3 these on a number line, and then pick three numbers to test, one in each zone:

test-numbers

0 2 5 2 4 3

boundary-values

Test x  0 : |2  0  3|  0  1: | 3 |  1, TRUE

Test x  2 : |2  2  3|  2  1, |1|  3 , FALSE

Test x  5: |2  5  3|  5  1, |7 |  6 , TRUE

of 4 2 Therefore, the solution is: x  or x  4 3

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, “  ” or “  ”) then a boundary-value that causes a fraction to be undefined can not be included in the solution. 3 x  3 Example:  x  2 2 Solution: One of the boundary-values is x  2 because this makes the fraction 0. Treating it

3 x  3 as an equality,  , we cross-multiply to get: x2  x  6  6. To x  2 2

solve this equation, we subtract 6 from both sides, then factor: x2  x  12  0 , (x  3)(x  4)  0 . The solutions of this equation are x  3 and x  4 . These two boundary-values are also solutions of the original inequality since it includes an equality, but the boundary-value x  2 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-numbers

–5 0 2.5 4 –4 2 3

boundary-values

3 5  3 3 Test –5:  ,   1, TRUE 5  2 2 8 3 0  3 3 3 Test 0:  ,   , FALSE 0  2 2 2 2 3 2.5  3 Test 2.5:  , 6  2.75, TRUE 2.5  2 2 3 4  3 3 7 Test 4:  ,  , FALSE 4  2 2 2 2

Therefore, the solution is: x  4 or 2  x  3 .

of 4 Exercises: Solve. Show all work, including test-values.

1. | x  2 |  3 2. | x  3|  2

3. |3  2x |  5 4. |2x  1|  3

5. |3x  2 |  x  2 6. | x |  x

7. |3  x |  x  3 8. | x  2 |  x  2

9. 2x  | x  2 | 10. | x  5|  x  5

of 4 1 1 1 1 11.  12.   x 5 x 5

2 1 4 x  5 13.  14.  x  3 x  2 x  2 2

x 2 3 x  5 15.  16.  2 x  3 x 2