# Inequalities of the Linear Kind

This next session is about Linear Inequalities. An inequality is similar to an equation. The difference is that it will not have an equal sign. It will be important to have a good understanding of inequalities because they pop up everywhere.

**Introduction**

An example of an inequality is: \( x^2 -1 < 10 \). In this way they are similar to linear functions. They can have any number of variables or exponents. Solving an inequality is finding the values for one or more variables that make the statement true. Inequalities can also be linear or nonlinear.

A solution to any particular linear inequality will often be written as an interval. These will be real numbers and not imaginary. Intervals can be open or closed. This depends on the values. A closed interval looks like \( [2,5] \). An open interval take this form: \( (3,9) \).

Inequalities can be solved in a few different ways. The main way is just like linear equations since they are similar. However, do realize that they can be solved using tables if they are in the correct form.

**Example 1**

\( 2x-3 < \frac{x+2}{-3} \)

The first step you do is multiply by -3 and then reverse the inequality sign.

Anytime you multiply or divide a negative number through an equation you will reverse the inequality sign.

\( -6x +9 > x + 2 \)

Now add 6x to both sides.

\(9 > 7x + 2\)

Subtract 2 from both sides now. Our goal is to have the same type of values on either side.

\(7 > 7x \)

Divide both sides by 7.

\(1 > x \)

**Example 2**

Solve: \(-3(4z-4) \geq 4 - (z - 1) \)

Distribute everything to make it easier.

\(-12z + 12 \geq 4 -z + 1 \)

Group everything as much as possible. This makes it easier to deal with and find a solution.

\( -12z +12 \geq -z + 5 \)

We should now group the terms together so we can better deal with them.

This means combining the <z> terms.

\( -11z + 12 \geq + 5 \)

Now we need to isolate the variable.

To do that, we subtract 12 from both sides.

\( -11z \geq -7 \)

the last thing to do is get rid of the coefficient in front of our variable.

\( z \leq \frac{7}{11} \)

**Example 3**

Solve: \( 1-x \geq \frac{1}{2}x -2 \)

Here we have an inequality with a fraction in it. It follows the same rules as other equations.

Let us first subtract one from each side, so we can isolate the variable.

\( -x \geq \frac{1}{2} x - 3 \)

Since we have <x> variables on each side of the equation, we need to get them all on one side.

We do this by subtracting the <x> that also has a constant on its side.

\( \frac{-3}{2}x \geq -3 \)

Our next step is to get rid of the coefficient in front of the <x>.

We do this by multiplying \( \frac{-3}{2} \) to each side of the equation.

Remember, any time we multiply or divide an inequality by a negative we also reverse the inequality sign.

\( x \leq 2 \)

**Example 4**

Solve: \( -4 \leq 5x + 1 < 21 \)

This is a different type of inequality.

They are compound inequalities.

It means there are multiple inequalities in the same equation.

Let us see how to work with these.

\( -4 \leq 5x + 1 < 21 \)

Our first goal is to isolate the variable in the middle.

This means no constants with the variable.

Subtract 1 from each portion.

If there are two inequalities then you have three portions.

\( -5 \leq 5x < 20 \)

Now we isolate the variable so there is no longer a coefficient.

To do that, we divide by 5 because that is the coefficient of x.

\( -1 \leq x < 4 \)

**Example 5**

Solve: \( 32 \leq 70 - 29x \leq 50 \)

This is another compound inequality.

We need to isolate the variable and the first step is to get rid of the constants.

Let’s start with subtracting 70 from each portion of the inequality.

You can subtract from each portion like this when you have a constant in each portion.

\( -38 \leq -29x \leq -20 \)

The next step is to get rid of the coefficient in front of our variable.

To do this, we divide all portions by <-29>.

Since we are multiplying or dividing by a negative, we will reverse each inequality.

\( \frac{-38}{-29} \geq x \geq \frac{-20}{-29} \)

We can simplify these terms now.

A negative divided by a negative is a positive, so let us rewrite these terms.

\( \frac{38}{29} \geq x \geq \frac{20}{29} \)

That looks nicer now that we have done that.

We can do more though.

Let’s also write it with the smallest number on the left.

We will reverse the signs again when we do this.

\( \frac{20}{29} \leq x \leq \frac{38}{29} \)

**Example 6**

Solve: \( \frac{-x}{2} + 1 \leq 3 \)

We want to isolate the variable.

To do this, we get rid of the constants first.

In this example we subtract <-1> from each term.

\( \frac{-x}{2} \leq 2 \)

Our next step is to get rid of the coefficient in front of the variable.

It may not be obvious, but it really \(\frac{-1}{2} * x \).

Since this is the case, we will multiply both terms by -2.

We are multiplying each term by a negative, this means we reverse the inequality.

\( x \geq -4 \)

**Example 7**

\( -8 < \frac{3x-1}{2} \leq 5 \)

Here we have a more complicated looking example.

The middle term is a fraction and we will need to deal with it first.

To start, multiply each term by 2 since that is the denominator of the fraction with our variable.

\( -16 < 3x - 1 \leq 10 \)

That will be much easier to deal with.

Now we want to get rid of the constant in the term with our variable.

Add +1 to each term.

\( -15 < 3x \leq 11 \)

Now we want to get rid of the coefficient in front of our variable which is 3.

Divide all terms by 3.

\( -5 < x \leq \frac{11}{3} \)

**Example 8**

Solve: \( 5(x-6) < 2x - 2(1-x) \)

We need to get this problem in the correct form, so we need to distribute these terms first.

\( 5x - 30 < 2x -2 + 2x \)

Next, let us simplify a bit and combine terms so that it makes more sense.

In the second term, let us combine both variables together.

\( 5x - 30 < 4x - 2 \)

Let us simplify some more.

We should subtract 4x in the second term since it is the smaller of the two variables.

\( x - 30 < -2 \)

Now we should isolate the variable and add 30 to both sides.

\( x < 28 \)