How to Solve Compound Inequalities

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.