# Solving Compound Inequalities

In this video, we will be learning how to solve compound inequalities.

“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                                   $x\geq10$

## 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.

Notice that the lines we drew are opposite of each other.

Because the inequality we have is an “or”. So the numbers between the two are not the answer.

Our answer satisfies one or the other, not both.