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

“And” inequalities are drawn toward one another

“Or” inequalities are drawn away from one another

For Example:

$3x+5\textless1$ or $2x-8\geq12$

Solve the inequalities individually first

$3x+5\textless1$                                                          $2x-8\geq12$

$3x+5-5\textless1-5$                                       $2x-8+8\geq12+8$

$\frac{3x}{3}\textless\frac{-4}{3}$                                                                 $\frac{2x}{2}\geq\frac{20}{2}$

$x\textless\frac{-4}{3}$                            or                                   $x\geq10$

## Examples of Solving Compound Inequalities

### Example 1

$8x+8\geq -64$ and $-7-8x\geq79$

Solve the inequalities individually first

$8x+8\geq -64$                                                          $-7-8x\geq-79$

$8x+8\geq -64+8$                                                   $-7-8x+7\geq-79+7$

$\frac{8x}{8}\geq\frac{-72}{8}$                                                                   $\frac{-8x}{-8}\geq\frac{-72}{-8}$

$x\geq-9$                                and                              $x\geq9$

### Example 2

$2x\textless10$ or $\dfrac{x}{2}\geq3$

Solve the inequalities individually first

$2x\textless10$                                                        $\dfrac{x}{2}\geq3$

$\dfrac {2x}{2}\textless\dfrac{10}{2}$                                                       $x\geq(3)(2)$

$x\textless5$                               or                         $x\geq6$

## Video-Lesson Transcript

Let’s go over solving compound inequalities.

Compound inequality means we have two inequalities that are associated.

For example:

$2x + 5 \textless 7$ and $3x + 8 \geq {-21}$

Here we have two constraints for the value of $x$.

Let’s solve the first one first.

$2x + 5 \textless 7$

Let’s subtract $5$ from both sides.

$2x + 5 - 5 \textless 7 - 5$
$2x \textless 2$

To solve for $x$, let’s divide both sides by $2$.

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

Now we have

$x \textless 1$

But $x$ is not just less than $1$. It should also satisfy the other inequality.

$3x + 8 \geq {-21}$

So let’s solve this by first subtracting $8$ on both sides.

$3x + 8 - 8 \geq {-21} - 8$
$3x \geq {-29}$

Now let’s solve for $x$ by dividing both sides by $3$.

$\dfrac{3x}{3} \geq \dfrac{-29}{3}$
$x \geq \dfrac{-21}{3}$

which we can also write as an improper fraction

$x \geq -9\dfrac{2}{3}$

If we graph this, draw a number line with the lowest number at $-9\dfrac{2}{3}$ and the highest number is $1$.

Since $x$ is less than $1$, let’s draw an open circle on number $1$. We also know that $x \geq -9\dfrac{2}{3}$, so let’s draw a solid circle on $-9\dfrac{2}{3}$ then draw a line between the two points.

So our solution can be written as:

$-9\dfrac{2}{3} \leq x \textless 1$

Our next example is written as one long inequality.

$2 \leq 3x + 8 \textless 14$

So we’ll solve this as a whole at once.

Let’s start off with subtracting $8$ from all sides of the inequality.

$2 - 8 \leq 3x + 8 - 8 \textless 14 - 8$
$-6 \leq 3x \textless 6$

Now, let’s solve for $x$ by dividing the inequality by $3$.

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

Now we have

$-2 \leq x \textless 2$

Let’s graph the solution by drawing a number line with the lowest number as $-2$ and the highest number is $2$ because $x$ is everything between the two.

Let’s draw a solid circle at number $-2$ to represent the equal sign and an open circle at $2$. Then draw a line in between.

Now, it’s not always the same number.

Let’s have another example

$2 \leq 3x + 8 \textless 20$

Let’s solve it the same way as above.

$2 - 8 \leq 3x + 8 - 8 \textless 20 - 8$
$-6 \leq 3x \textless 12$

Then

$\dfrac{-6}{3} \leq \dfrac{3x}{3} \textless \dfrac{12}{3}$

$-2 \leq x \textless 4$

This time our number line starts at $-2$ and ends with $4$.

The circle on $-2$ is solid because it has an equal sign and an open circle at $4$.

Let’s move on to another type of compound inequality.

Here we have “or”.

For example:

$3x + 5 \textless 1$ or $2x - 8 \geq 12$

We have to solve this individually.

$3x + 5 \textless 1$

Subtract $5$ from both sides

$3x + 5 - 5 \textless 1 - 5$
$3x \textless -4$

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

$\dfrac{3x}{3} \textless \dfrac{-4}{3}$
$x \textless -1\dfrac{1}{3}$

Now let’s solve the other one.

$2x - 8 \geq 12$

Let’s add $8$ on both sides

$2x - 8 + 8 \geq 12 + 8$
$2x \geq 20$

Let’s solve for $x$ by dividing both sides by $2$.

$\dfrac{2x}{2} \geq \dfrac{20}{2}$
$x \geq 10$

$x \textless -1\dfrac{1}{3}$ or $x \geq 10$
Let’s draw line number with $-1\dfrac{1}{3}$ as the lowest number and the highest number is $10$.
We draw an open circle at $-1\dfrac{1}{3}$ and draw a line going to the left. Then draw a solid circle at $10$ and a line going to the right.