# 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:

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### 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:

Easy to understand math videos:

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### Example 3

Solve 2*x* ≤ 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:

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### Example 4

Solve the inequality 3 − 2*x* ≥ 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.

Easy to understand math videos:

MathTutorDVD.com

## Inequalities with Three Members

### Example 6

Solve −1 < 2*x* + 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:

Easy to understand math videos:

MathTutorDVD.com

### Example 7

Solve 2*x *<* x* − 4 ≤ 3*x* + 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:

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### Exercises

Solve the following inequalities for *x*:

1. Solve 3 − 3*x* < − 1

Answer

`3 - 3x < -1`

`- 3x < -4 ``x> 4/3 ~~ 1.333...`

Here is the solution graph:

2. Solve −2(*x* + 4) > 1 − 5*x*

Answer

`-2(x + 4) > 1 - 5x`

`-2x - 8 > 1 - 5x ``3x - 8 > 1`

`3x > 9`

`x > 3`

The solution graph:

Easy to understand math videos:

MathTutorDVD.com

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:

Easy to understand math videos:

MathTutorDVD.com

4. Solve *x* − 1 < 2*x* + 2 < 3*x* +
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:

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