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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.