4. Solving Inequalities with Absolute Values
For inequalities involving absolute values ie. |x|, we use the following relationships, for some number n:
If `|f(x)| > n`, then this means:
`f(x) < -n` or `f(x) > n`
If `|f(x)| < n`, then this means:
`-n < f(x) < n`
Example 1
Solve the inequality |x − 3| < 2.
Answer
Applying the relationships discussed above:
`- 2 < x - 3 < 2`
Adding `3` to all sides, we get:
`-2 + 3 < x - 3 + 3 < 2 + 3`
`1 < x < 5`
Here is the graph of our solution:
Convince yourself that this answer is correct by checking. Try `x = 0` (should fail, because it is outside the range of our answer), `x = 3` (should work) and `x = 10` (should fail). Every time you check like this, it becomes clearer why we solve it this way.
Example 2
Solve the inequality |2x − 1| > 5.
Answer
Applying the relationships discussed earlier:
`2x - 1 < -5 \ or \ 2x - 1 > 5`
Solving both inequalities, we get:
`2x < -5 + 1` or `2x > 5 + 1` `2x < -4` or `2x > 6` `x < -2 ` or `x > 3 `
Here's the graph of our solution:
Example 3
Solve the inequality `2|(2x)/3 + 1|>=4`
Answer
`2|(2x)/3 + 1|>=4`
This gives
`|(2x)/3 + 1|>=2`
So either...
`(2x)/3+1<=-2`
OR
`(2x)/3+1>=2`
Multiply both sides by 3
`2x + 3 ≤ -6 ``2x ≤ -9`
`x<=-9/2`
OR
`2x + 3 ≥ 6 ``2x ≥ 3`
`x>=3/2`
Solution: `x<=-9/2 = -4.5,\ "or"\ x>=3/2=1.5`
Here's the graph of our solution:
Example 4
Solve the inequality `|3 − 2x| < 3`
Answer
`|3 − 2x| < 3`
`-3 < 3 - 2x < 3`
Subtract `3` from all sides:
`-6 < -2x < 0`
Divide all sides by `-2`
`3 > x > 0`
(note the change in sense due to dividing by a negative number)
A better way to write this is: `0 < x < 3`, since we usually have smaller numbers on the left hand side.
Here's the solution graph:
Example 5
A technician measures an electric current which is `0.036\ "A"` with a possible error of `±0.002\ "A"`. Write this current, i, as an inequality with absolute values.
Answer
The possible error of `± 0.002\ "A"` means that the difference between the actual current and the value of `0.036\ "A"` cannot be more than `0.002\ "A"`.
So the values of i we have can be expressed as:
`0.034 ≤ i ≤ 0.038`
We can simply write this as:
`|i − 0.036 | ≤ 0.002`
Here's the solution graph:
Exercises
1. Solve ` |5 − x| ≤ 2`
Answer
`-2 ≤ 5 - x ≤ 2`
`-7 ≤ - x ≤ -3`
`7 ≥ x ≥ 3`
It is better to write this as: `3 ≤ x ≤ 7`.
Here's the graph of the solution:
2. Solve `|(2x-9)/4|<1`
Answer
3. Solve `|(4x)/3-5|>=7`
Answer
`|(4x)/3-5|>=7`
| `(4x)/3-5<=-7` | OR | `(4x)/3-5>=7` |
`4x-15<=-21` `4x<=-6` `x<=-1.5` |
`4x-15>=21` `4x>=36` `x>=9` |
So the solution is: `x ≤ -1.5` or `x ≥ 9`.
Here's the graph of the solution:
4. Solve `|x^2+3x-1|<3`
Answer
`-3<x^2+3x-1<3`
Now we need to break this up into 2 parts:
Left part
`-3<x^2+3x-1`
`0<x^2+3x+2`
`x^2+3x+2>0`
`(x+2)(x+1)>0`
Critical values for the left side are:
`x = -2 and x = -1`.
This gives:
`x < -2 and x > -1`.
Right part
`x^2+3x-1<3`
`x^2+3x-4<0`
`(x+4)(x-1)<0`
Critical values:
`x = -4` and `x = 1`
This gives:
`-4 < x < 1`
It's easier to graph these together on one axis to decide the final result:
These 3 regions intersect in the following 2 places:
This gives us our final solution: −4 < x < −2 and −1 < x < 1.
Remember to check numbers inside and outside the given solution region to make sure it works.