2. Solving Linear Inequalities
The procedure for solving linear inequalities in one variable is similar to solving basic equations. (See Solving Equations.)
We need to be careful about the sense of the equality when multiplying or dividing by negative numbers.
Following are several examples of solving equations involving inequalities.
Example 1
Solve x + 2 < 4
Answer
We need to subtract `2` from both sides of the inequality.
`x+2<4`
`x<4-2`
`x<2`
The graph of this solution is as follows:
Example 2
Solve `x/2>4`
Answer
We need to multiply both sides of the inequality by `2`.
`x/2>4`
`x>4xx2`
`x>8`
Here's the graph of this solution:
Example 3
Solve 2x ≤ 4
Answer
We need to divide both sides of the inequality by `2`.
`2x<=4`
`x<=4/2`
`x<=2`
Here's the graph of this solution:
Example 4
Solve the inequality 3 − 2x ≥ 15
Answer
In this example, we need to subtract 3 from both sides; then divide both sides by `-2` (remembering to change the direction of the inequality).
`3-2x>=15`
`-2x>=15-3`
`-2x>=12`
`x<=12/(-2)`
`x<=-6`
Here's the graph of this solution:
(Note the change in sense due to dividing by a negative number)
Check: Always check your solution and you can be sure your answer is correct.
In this case, any number less than `-6` should "work" in the original equation, and any number bigger than `-6` should fail.
Let's take `x = -10` (a convenient number less than `-6`)
LHS `= 3 − 2(-10) = 3 + 20 = 23`. This is more than `15` so it is true.
Now let's take `x = 0` (a convenient number greater than `-6`)
LHS `= 3 − 2(0) = 3`. This is NOT more than `15`, which is what we hoped for.
So we can be sure our answer is correct.
Example 5
Solve the inequality `3/2(1-x)>1/4-x`
Answer
`3/2(1-x)>1/4-x`
Multiplying both sides by 4 gives us:
`6(1-x)>1-4x`
` 6-6x>1-4x`
`-6x+4x>1-6`
`-2x> -5`
`x<5/2`
(Note the change in sense in the last line, due to dividing by a negative number).
Here's the graph of this solution:
Check: Taking x = 0 (which should work):
`"LHS" = 3/2(1 − 0) = 3/2`
`"RHS" = 1/4`
It is TRUE that `3/2 > 1/4`, so that is good.
Now we take x = 3 (a convenient number bigger than 5/2, which should not work):
`"LHS" = 3/2(1 − 3) = -3`
`"RHS" = 1/4 − 3 = -2 3/4`
It is NOT true that `-3 > -2 3/4` and so `x=3` fails, as we hoped.
We can be sure our answer (`x<5/2`) is correct.
Inequalities with Three Members
Example 6
Solve −1 < 2x + 3 < 6
Answer
`- 1 < 2x + 3 < 6`
Subtract `3` from all 3 sides
`- 1 - 3 < 2x + 3 - 3 < 6 - 3`
`- 4 < 2x < 3`
Divide all sides by `2`
`-2 < x < 3/2`
The solution graph:
Example 7
Solve 2x < x − 4 ≤ 3x + 8
Answer
Another way of solving more complicated inequalities with 3 members (sides) is to rewrite the inequality as
`2x < x - 4` and `x - 4 ≤ 3x + 8`
Then solving each of these inequalities, we obtain:
LHS inequality:
`x < - 4`
RHS inequality:
`x - 4 ≤ 3x + 8`
`- 4 ≤ 2x + 8`
`- 12 ≤ 2x`
`x ≥ -6 `
Bearing in mind that the final solution has to satisfy both inequalities, we obtain:
`x < -4` and `x ≥ -6`
These two portions appear as:
On considering the region where the two portions intersect, we get
`-6 ≤ x < -4`
This is the final solution graph:
Exercises
Solve the following inequalities for x:
1. Solve 3 − 3x < − 1
Answer
`3 - 3x < -1`
`- 3x < -4 ``x> 4/3 ~~ 1.333...`
Here is the solution graph:
2. Solve −2(x + 4) > 1 − 5x
Answer
`-2(x + 4) > 1 - 5x`
`-2x - 8 > 1 - 5x ``3x - 8 > 1`
`3x > 9`
`x > 3`
The solution graph:
3. Solve `x/5-2>2/3(x+3)`
Answer
`x/5-2>2/3(x+3)`
Multiply throughout by 5:
`x-10>10/3(x+3)`
Multiply throughout by 3:
`3x-30>10(x+3)`
`3x-30>10x+30`
`3x>10x+60`
`-7x>60`
`x<(-60)/7~~-8.57`
The solution graph:
4. Solve x − 1 < 2x + 2 < 3x + 1
Answer
We need to find the intersection of the "true" values.
`x - 1 < 2x + 2` and `2x + 2 < 3x + 1`
`x < 2x + 3` and `2x < 3x - 1`
`x > -3` and `x > 1`
These 2 inequalities on the one axis appear as follows:
The intersection of these 2 regions is `x > 1`.
Always check your answers!
Here's the final solution graph:
