Solving The Inequality |x-9| ≤ 2 Algebraically

When we talk about solving inequalities, especially those involving absolute values, it can sometimes feel like we're navigating a maze of possibilities. But don't worry, it's a systematic process, and with a little practice, you'll be breezing through them. Today, we're going to tackle the inequality $|x-9| \leq 2$ and break down exactly how to solve it algebraically. This means we'll be using mathematical rules and steps to arrive at our solution, rather than just guessing or plotting points. The core concept behind absolute value is distance. The expression $|a|$ represents the distance of the number 'a' from zero on the number line. So, when we see an inequality like $|x-9| \leq 2$, we're essentially asking: "For what values of 'x' is the distance between 'x' and 9 less than or equal to 2?" This geometric interpretation is super helpful because it gives us a visual anchor. Imagine yourself standing at the number 9 on the number line. The inequality is telling you to find all the points 'x' that are 2 units or closer to 9. This means you can move 2 units to the left of 9, and 2 units to the right of 9, and everything in between (including the endpoints) will satisfy the condition. This immediate insight helps us anticipate the kind of solution we're looking for – a range of numbers, not just a single value. The algebraic approach allows us to formalize this understanding and express it with precision. We'll see how the definition of absolute value directly leads to two separate, simpler inequalities that we can then solve independently. Understanding the properties of absolute value is fundamental, and this problem is a perfect showcase for those properties. So, let's get ready to dive deep into the mechanics of solving this specific inequality, making sure every step is clear and easy to follow. We'll also touch upon why this method is robust and applicable to a wide range of similar problems you might encounter.

Understanding the Absolute Value Property

The key to unlocking inequalities with absolute values lies in understanding their fundamental definition and properties. For any real number 'a', the absolute value $|a|$ is defined as:

• $|a| = a$, if $a \geq 0$
• $|a| = -a$, if $a < 0$

This definition tells us that the absolute value of a number is always non-negative. It strips away the sign, leaving only the magnitude. Now, consider an inequality of the form $|E| \leq k$, where 'E' is some expression (in our case, $x-9$) and 'k' is a positive constant (in our case, 2). This inequality $|E| \leq k$ is equivalent to the compound inequality $-k \leq E \leq k$. This is the crucial property we'll use. It effectively breaks down the absolute value problem into two standard linear inequalities that can be solved simultaneously. Why does this work? Think back to the distance interpretation. $|x-9| \leq 2$ means that the expression $x-9$ must be a value whose distance from zero is less than or equal to 2. The numbers whose distance from zero is less than or equal to 2 are all the numbers between -2 and 2, inclusive. So, $x-9$ must fall within this range. That is, $x-9$ must be greater than or equal to -2 AND less than or equal to 2. This directly translates to $-2 \leq x-9 \leq 2$. This property is a cornerstone for solving absolute value inequalities, and recognizing it is the first major step towards a successful solution. We can also consider the case where the inequality is $|E| \geq k$. This is equivalent to $E \geq k$ OR $E \leq -k$. Notice the 'OR' here, which leads to two separate solution sets that are then combined. For our specific problem, $|x-9| \leq 2$, we are firmly in the first scenario, the 'less than or equal to' case, which simplifies nicely into a single, continuous interval.

Algebraic Steps to Solve $|x-9| \leq 2$

Now, let's apply the property we just discussed to our specific inequality, $|x-9| \leq 2$. Our goal is to isolate 'x' and find the range of values that satisfy the condition.

Step 1: Rewrite the absolute value inequality as a compound inequality.

Using the property that $|E| \leq k$ is equivalent to $-k \leq E \leq k$, we can rewrite our inequality as:

$-2 \leq x-9 \leq 2$

This single statement encompasses all the possibilities for $x-9$ to be within 2 units of zero.

Step 2: Isolate 'x' in the compound inequality.

Our objective now is to get 'x' by itself in the middle. To do this, we need to undo the operations being performed on 'x'. Currently, 9 is being subtracted from 'x'. To reverse subtraction, we add. We must perform the same operation on all three parts of the compound inequality to maintain the balance and the validity of the statement.

Add 9 to all three parts of the inequality:

$-2 + 9 \leq x-9 + 9 \leq 2 + 9$

Step 3: Simplify the inequality.

Now, perform the addition in each section:

$7 \leq x \leq 11$

And there you have it! We've successfully isolated 'x'. This result tells us that any value of 'x' that is greater than or equal to 7 AND less than or equal to 11 will satisfy the original inequality $|x-9| \leq 2$. This range includes the numbers 7, 8, 9, 10, and 11, as well as all the decimal and fractional values in between.

Verification (Optional but Recommended):

To ensure our solution is correct, we can test values from within the interval and outside the interval.

• Test a value within the interval (e.g., $x=9$): $|9-9| = |0| = 0$. Since $0 \leq 2$, this value works.
• Test an endpoint (e.g., $x=7$): $|7-9| = |-2| = 2$. Since $2 \leq 2$, this endpoint works.
• Test another endpoint (e.g., $x=11$): $|11-9| = |2| = 2$. Since $2 \leq 2$, this endpoint works.
• Test a value outside the interval (e.g., $x=6$): $|6-9| = |-3| = 3$. Since $3 \not\leq 2$, this value does not work.
• Test a value outside the interval (e.g., $x=12$): $|12-9| = |3| = 3$. Since $3 \not\leq 2$, this value does not work.

These tests confirm that our solution $7 \leq x \leq 11$ is indeed correct. This algebraic method is clean, efficient, and guarantees that we haven't missed any solutions.

Understanding the Solution Set

The solution we arrived at, $7 \leq x \leq 11$, represents a closed interval on the number line. This means that both the lower bound (7) and the upper bound (11) are included in the set of solutions. In interval notation, this is written as $[7, 11]$. The square brackets indicate that the endpoints are part of the interval. Graphically, if you were to draw this on a number line, you would shade the region between 7 and 11, and you would place solid dots at 7 and 11 to signify that these values are included. This interval represents all the possible values of 'x' for which the distance between 'x' and 9 is less than or equal to 2. Think about it: if 'x' is 7, the distance from 9 is $|7-9| = |-2| = 2$. If 'x' is 11, the distance from 9 is $|11-9| = |2| = 2$. If 'x' is any number between 7 and 11, say 9.5, the distance from 9 is $|9.5-9| = |0.5| = 0.5$, which is certainly less than 2. The algebraic manipulation we performed directly leads to this continuous range. It's not just a few isolated numbers; it's an entire segment of the number line. This comprehensive range is what makes the absolute value inequality so powerful. It's a concise way to express a set of values that satisfy a specific condition related to distance or magnitude. Understanding that the solution to $|E| \leq k$ is a single, continuous interval is key to visualizing and interpreting these results correctly. This is in contrast to inequalities of the form $|E| \geq k$, which typically result in two separate intervals (e.g., $x \geq a$ or $x \leq b$), representing values that are far from a certain point rather than close to it. The clarity of the solution $7 \leq x \leq 11$ means we have found all the numbers that meet the criteria, and no others. This is the hallmark of a complete algebraic solution.

Conclusion

We've successfully navigated the process of solving the inequality $|x-9| \leq 2$ algebraically. By understanding the fundamental properties of absolute value, specifically how $|E| \leq k$ translates to $-k \leq E \leq k$, we were able to transform the absolute value inequality into a straightforward compound inequality. The subsequent steps involved basic algebraic manipulations – adding 9 to all parts of the inequality – to isolate 'x'. This led us directly to the solution $7 \leq x \leq 11$, which represents all values of 'x' that are within a distance of 2 units from 9, inclusive. This solution can be expressed as the closed interval $[7, 11]$. Remember, the absolute value fundamentally deals with distance, and inequalities involving it often describe a range of numbers. Whether you're dealing with distances, errors, or other applications, mastering these algebraic techniques will provide you with a robust toolset for problem-solving. Keep practicing with different variations of absolute value inequalities, and you'll find your confidence grow. For further exploration into inequalities and algebraic concepts, you might find resources from Khan Academy or Math is Fun to be incredibly helpful.