Solving Compound Inequalities: A Step-by-Step Guide

Hey math enthusiasts! Ready to dive into the world of compound inequalities? Don't worry, it's not as scary as it sounds. In this guide, we'll break down how to solve each compound inequality, graph its solution, and express it using interval notation. Let's get started, shall we?

Understanding Compound Inequalities

Before we jump into the problems, let's quickly recap what compound inequalities are all about. Basically, a compound inequality is just two or more inequalities joined together by the words "and" or "or." When we see "and", it means we're looking for solutions that satisfy both inequalities. Think of it as a stricter requirement. On the other hand, "or" means we're looking for solutions that satisfy at least one of the inequalities. This gives us more flexibility. The goal is to find the range of values that make the inequality true. The values are the possible solutions to the inequality and in order to completely describe the set of solutions, you can show the solution using interval notation.

Graphing these solutions on a number line can make the problem easier to solve. When graphing compound inequalities, you will often deal with the inequalities in the form of "less than", "less than or equal to", "greater than", and "greater than or equal to". Each of these different forms has a different way of showing up on the number line. For "less than" and "greater than" inequalities, the number line will have an open circle and will not include the number. For "less than or equal to" and "greater than or equal to" inequalities, the number line will have a closed circle and will include the number.

Now, let's get into the main course: solving some problems!

Problem 1: Solving and Graphing "And" Inequalities

Let's kick things off with the first problem: $-4 + a \leq -4$ and $\frac{a}{3} \geq -1$. This is a classic example of an "and" compound inequality, so we need to find the values of a that satisfy both inequalities simultaneously.

First, let's tackle the first inequality, $-4 + a \leq -4$. To isolate a, we'll add 4 to both sides: $a \leq 0$. This tells us that a must be less than or equal to 0.

Next, let's solve the second inequality, $\frac{a}{3} \geq -1$. To isolate a, we'll multiply both sides by 3: $a \geq -3$. This means a must be greater than or equal to -3.

Now, we need to find the values of a that satisfy both conditions: $a \leq 0$ and $a \geq -3$. This means a must be between -3 and 0, including -3 and 0. On the number line, we'll have a closed circle at -3 and a closed circle at 0, with a line connecting them.

Interval Notation

To write this solution in interval notation, we use square brackets to indicate that the endpoints are included: $[-3, 0]$. This interval represents all the numbers from -3 to 0, including -3 and 0.

Problem 2: Solving Compound Inequalities with a Chain

Alright, let's move on to the second problem: $-44 \leq 8k - 4 \leq -20$. This one looks a little different, but it's still a compound inequality. Notice that the variable k is in the middle, between two inequality symbols. This is often called a "chain" inequality. The good news is, we can solve this in a straightforward way.

Our goal is to isolate k in the middle. To do this, we'll perform the same operations on all three parts of the inequality. First, we'll add 4 to all three parts: $-44 + 4 \leq 8k - 4 + 4 \leq -20 + 4$. This simplifies to $-40 \leq 8k \leq -16$.

Next, we'll divide all three parts by 8: $\frac{-40}{8} \leq \frac{8k}{8} \leq \frac{-16}{8}$. This gives us $-5 \leq k \leq -2$. This means that k must be greater than or equal to -5 and less than or equal to -2.

Graphing and Interval Notation

On the number line, we'll have a closed circle at -5 and a closed circle at -2, with a line connecting them. In interval notation, the solution is $[-5, -2]$. This means the values of k are between -5 and -2, inclusive.

Problem 3: Another Chain Inequality Challenge

Let's tackle another chain inequality: $7 \leq -5 - 3x \leq 16$. This problem is similar to the last one, but we have a bit more work to do to isolate x.

First, we'll add 5 to all three parts of the inequality: $7 + 5 \leq -5 - 3x + 5 \leq 16 + 5$. This simplifies to $12 \leq -3x \leq 21$.

Now, here's a crucial step: we need to divide all three parts by -3. Remember that when you multiply or divide an inequality by a negative number, you must flip the direction of the inequality symbols. So, we get $\frac{12}{-3} \geq \frac{-3x}{-3} \geq \frac{21}{-3}$, which simplifies to $-4 \geq x \geq -7$. It's a bit easier to read if we rewrite this as $-7 \leq x \leq -4$.

Graphing and Interval Notation

On the number line, we'll have a closed circle at -7 and a closed circle at -4, with a line connecting them. In interval notation, the solution is $[-7, -4]$. Notice that the smaller number comes first in interval notation.

Problem 4: Solving "And" Inequalities with Variables on Both Sides

Let's get into something slightly more complex: $-x + 8 > 8x - 10$ and $3 - 7x > -9x - 7$. This problem has two separate inequalities, and each one has variables on both sides. We'll solve each inequality individually, and then combine the solutions.

First, let's solve $-x + 8 > 8x - 10$. To isolate x, let's add x to both sides and add 10 to both sides: $8 + 10 > 8x + x$. This simplifies to $18 > 9x$. Dividing both sides by 9 gives us $2 > x$, which is the same as $x < 2$.

Next, let's solve $3 - 7x > -9x - 7$. Add $9x$ to both sides and subtract 3 from both sides: $-7x + 9x > -7 - 3$, simplifying to $2x > -10$. Dividing both sides by 2 gives us $x > -5$.

Now, we need to find the values of x that satisfy both conditions: $x < 2$ and $x > -5$. This means x must be between -5 and 2, but not including -5 or 2 (because the inequality symbols are "greater than" and "less than").

Graphing and Interval Notation

On the number line, we'll have an open circle at -5 and an open circle at 2, with a line connecting them. In interval notation, the solution is $(-5, 2)$.

Problem 5: Solving a Triple Inequality

Let's tackle the final problem: $10n + 4 < 3 + 9n < 8 + 10n$. This is another compound inequality where the variable n is in the middle. We'll split this into two separate inequalities: $10n + 4 < 3 + 9n$ and $3 + 9n < 8 + 10n$.

First, let's solve $10n + 4 < 3 + 9n$. Subtracting $9n$ from both sides and subtracting 4 from both sides gives us $n < -1$.

Next, let's solve $3 + 9n < 8 + 10n$. Subtracting $9n$ from both sides and subtracting 8 from both sides gives us $-5 < n$, or $n > -5$.

Now, we need to find the values of n that satisfy both conditions: $n < -1$ and $n > -5$. This means n must be between -5 and -1, not including -5 or -1.

Graphing and Interval Notation

On the number line, we'll have an open circle at -5 and an open circle at -1, with a line connecting them. In interval notation, the solution is $(-5, -1)$.

Conclusion: Mastering Compound Inequalities

And there you have it, guys! We've covered a variety of compound inequality problems. Remember to take it step by step, pay close attention to the "and" and "or" conditions, and don't forget to flip those inequality signs when multiplying or dividing by a negative number. Keep practicing, and you'll become a compound inequality pro in no time! Good luck!