In this video, we will be learning how to solve linear inequalities. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

For Example:

$1\textless3x-2\leq -5$

First we split the inequalities:

$1\textless3x-2$                                               $3x-2\leq -5$

$1+2\textless3x-2 +2$                                $3x-2+2\leq -5 +2$

$\dfrac{3}{3}\textless\frac{3x}{3}$                                                         $\dfrac{3x}{3}\textless\frac{-3}{3}$

$1\textless x$                                                          $x\leq-1$

### Example 1

$-10\textless2x-4\textless 2$

First we split the inequalities:

$-10\textless2x-4$                                             $2x-4\textless 8$

$-10+4\textless2x-4+4$                                $2x-4+4\textless 2+4$

$-6\textless2x$                                                      $2x\textless 6$

$\dfrac{-6}{2}\textless\dfrac{2x}{2}$                                                     $\dfrac{2x}{6}\textless\dfrac{6}{2}$

$-3\textless x$                                                        $3\textless x$

### Example 2

5x+3\leq18
First, subtract 3 on both sides
5x+3-3\leq18-3
5x\leq15
Then, divide 5 on both sides to isolate x
\dfrac{5x}{5}\leq \dfrac{15}{5}
x\leq 3

## Video-Lesson Transcript

In this lesson, we’ll go over solving linear inequalities.

This is very similar to solving linear equations except for one thing:

If we multiply or divide by a negative number, we must flip the inequality sign.

So a sign like this $\textless$ could be flipped the other way and become this $>$.

Let’s look at an example:

$2x - 5 \textless 3$

So let’s just treat the inequality sign as a regular equal sign as we solve.

First, let’s add $5$ on both sides.

$2x - 5 + 5 \textless 3 + 5$ $2x \textless 8$

From here we have to divide by $2$ to isolate the $x$.

$\dfrac{2x}{2} \textless \dfrac{8}{2}$

And we end up with

$x \textless 4$

So if we need to graph it, let’s draw a number line and draw an open circle at $4$. Open circle because $x$ is not equal to $4$. Then make an arrow going to the left.

Next, we have

$-3x + 7 \textless 1$

Again, we’re going to treat it as a regular equation when solving $x$.

First thing we have to do is to get rid of $7$, so we subtract $7$ on both sides.

$-3x + 7 - 7 \textless 1 - 7$ $-3x \textless -6$

Then let’s solve for $x$ by dividing by $-3$.

$\dfrac{-3x}{-3} \textless \dfrac{-6}{-3}$

And because we’re dividing by $-3$, we have to flip the inequality sign.

$x > 2$

Remember, when we divide by a negative number, we always have to flip the sign.

It doesn’t matter if the dividend is positive or negative. All we care about is when the divisor is negative, that’s the time we flip the sign.

If we graph the answer, let’s draw a number line.

Draw an open circle at number $2$. It’s not a filled circle because it is not equal to.

Since $x$ is greater, draw a line going to the right.

Again, solving inequalities is very similar to solving regular equations except if we multiply or divide by a negative number we have to flip the sign.

Another difference is that we’re not going to have an explicit answer for $x$ or an explicit solution for $x$. It’s going to be a range of numbers.

We won’t have $x$ is equal to a number.

Our answer is $x$ is any number less than or greater than a number. All the way up to infinity.

Here we have a more complicated inequality.

$1 \textless 3x - 2 \leq 16$

We have two different conditions.

$1 \textless 3x - 2$ and $3x - 2 \leq 16$

What we should do is separate this into two different inequalities.

$1 \textless 3x - 2$ and $3x - 2 \leq 16$

Let’s solve the inequality on the left first.

$1 \textless 3x - 2$

Let’s add $2$ on both sides.

$1 + 2 \textless 3x - 2 + 2$ $3 \textless 3x$

Then solve $x$ by dividing by $3$.

$\dfrac{3}{3} \textless \dfrac{3x}{3}$ $1 \textless x$

Now, let’s solve the other inequality.

$3x - 2 \leq 16$

Let’s start by adding $2$ on both sides.

$3x - 2 + 2 \leq 16 + 2$ $3x \leq 18$

Then solve for $x$ by dividing both sides by $3$.

$\dfrac{3x}{3} \leq \dfrac{18}{3}$ $x \leq 6$

Let’s draw a number line to graph these two inequalities starting with $0$ and ending in $7$.

For $1 \textless x$, we have to draw an open circle at number $1$. Open circle because it is not equal to. And since it’s greater than, draw a line going to the right.

Now for $x \leq 6$, so let’s draw a shaded circle at $6$ since it’s also equal to it. Then draw a line going to the left.

Further, draw a line to the other circle.

Now this line segment represents our solution.

Our answer can also be written as:

$1 \textless x \leq 6$

Notice that the two endpoints are the end numbers as well – $1$ and $6$.

And $x$ is somewhere in between these two numbers but can also be equal to $6$.

Let’s have another complex inequality.

$1 \textless 3x - 2 \leq -5$

Let’s break this down into two simple inequalities.

$1 \textless 3x - 2$ and $3x - 2 \leq -5$

Let’s work on the first inequality by adding $2$ on both sides.

$1 + 2 \textless 3x - 2 + 2$ $3 \textless 3x$

To solve $x$, divide both sides by$3$.

$\dfrac{3}{3} \textless \dfrac{3x}{3}$ $1 \textless x$

Now, let’s solve the other inequality.

$3x - 2 \leq -5$

Let’s start off by adding $2$ on both sides.

$3x - 2 + 2 \leq {-5} + 2$ $3x \leq {-3}$

To solve for $x$, we’ll divide both sides by $3$.

$\dfrac{3x}{3} \leq \dfrac{-3}{3}$ $x \leq {-1}$

So, now we graph this by drawing a number line.

Draw an open circle at $1$ since it’s not equal to. Then draw a line going to the right since $x$ is greater than $1$.

Next, draw a shaded circle at $-1$ because $x$ could equal to it. Then draw a line going to the left since $x$ is less than $-1$.

The two arrows are pointing in different directions.

Now, we can write our solution as

$x \leq -1$ or $x > 1$

$x$ won’t be able to satisfy both, so we write “or”.