Solving |x+10|=3x-2 A Step-by-Step Guide
Navigating the world of absolute value equations can sometimes feel like traversing a maze. These equations, distinguished by the absolute value notation, require a nuanced approach to ensure accurate solutions. In this comprehensive guide, we will dissect the equation |x+10|=3x-2, providing a step-by-step solution and exploring the underlying concepts. This will not only help you solve this specific problem but also equip you with the skills to tackle similar challenges with confidence.
Understanding Absolute Value
Before diving into the solution, it's crucial to grasp the fundamental concept of absolute value. The absolute value of a number is its distance from zero on the number line, irrespective of direction. This means that the absolute value of a number is always non-negative. For example, |5| = 5 and |-5| = 5. This property is key to understanding how we approach solving absolute value equations.
When dealing with an equation like |x+10|=3x-2, the absolute value introduces two possible scenarios. The expression inside the absolute value, (x+10), can be either positive or negative, but its absolute value will always be non-negative. Therefore, we need to consider both possibilities to find all potential solutions.
Setting Up the Two Cases
To solve an absolute value equation, we essentially split it into two separate equations, each representing one of the possible scenarios:
- Case 1: The expression inside the absolute value is positive or zero. This means (x+10) ≥ 0. In this case, the absolute value does not change the expression, so we can write the equation as: x + 10 = 3x - 2
- Case 2: The expression inside the absolute value is negative. This means (x+10) < 0. In this case, the absolute value makes the expression positive, which is equivalent to multiplying the expression by -1. So, we write the equation as: -(x + 10) = 3x - 2
By addressing both cases, we ensure that we capture all potential solutions to the original absolute value equation. The next step involves solving each of these equations independently.
Solving Case 1: x + 10 = 3x - 2
Let's begin by solving the first case, where we assume (x+10) is positive or zero. The equation we need to solve is:
x + 10 = 3x - 2
The goal here is to isolate x on one side of the equation. We can achieve this through a series of algebraic manipulations. First, let's gather the x terms on one side. We can subtract x from both sides of the equation:
10 = 2x - 2
Next, we want to isolate the term with x. We can do this by adding 2 to both sides of the equation:
12 = 2x
Finally, to solve for x, we divide both sides of the equation by 2:
x = 6
So, we have found a potential solution: x = 6. However, it's crucial to remember that this is just a potential solution. We need to verify that it satisfies the original absolute value equation and the condition for this case, (x+10) ≥ 0.
Verification for Case 1
To verify the solution, we substitute x = 6 back into the original equation |x+10|=3x-2:
|6 + 10| = 3(6) - 2
|16| = 18 - 2
16 = 16
The equation holds true. Now, let's check the condition (x+10) ≥ 0:
6 + 10 ≥ 0
16 ≥ 0
This condition is also satisfied. Therefore, x = 6 is a valid solution to the absolute value equation.
Solving Case 2: -(x + 10) = 3x - 2
Now, let's move on to the second case, where we assume (x+10) is negative. The equation we need to solve is:
-(x + 10) = 3x - 2
First, we need to distribute the negative sign on the left side of the equation:
-x - 10 = 3x - 2
Similar to Case 1, our goal is to isolate x. Let's gather the x terms on one side by adding x to both sides:
-10 = 4x - 2
Next, we isolate the term with x by adding 2 to both sides:
-8 = 4x
Finally, we solve for x by dividing both sides by 4:
x = -2
Again, we have found a potential solution: x = -2. We need to verify if this solution satisfies the original absolute value equation and the condition for this case, (x+10) < 0.
Verification for Case 2
To verify the solution, we substitute x = -2 back into the original equation |x+10|=3x-2:
|-2 + 10| = 3(-2) - 2
|8| = -6 - 2
8 = -8
This equation does not hold true. Therefore, x = -2 is not a solution to the absolute value equation. It's an extraneous solution that arose from the process of solving the equation.
We also need to check the condition (x+10) < 0:
-2 + 10 < 0
8 < 0
This condition is also not satisfied, further confirming that x = -2 is not a valid solution.
Identifying Extraneous Solutions
As we saw in Case 2, it's possible to arrive at solutions that do not actually satisfy the original equation. These are called extraneous solutions. Extraneous solutions often arise when dealing with absolute value equations, radical equations, or rational equations, where the process of solving involves squaring both sides or multiplying by expressions that could be zero.
The key takeaway here is that verification is crucial. After solving each case, always substitute the potential solutions back into the original equation and check if they hold true. Also, ensure that the solutions satisfy the initial conditions for each case (e.g., (x+10) ≥ 0 for Case 1 and (x+10) < 0 for Case 2).
The Solution
After carefully solving both cases and verifying the potential solutions, we have found that only one solution satisfies the original equation |x+10|=3x-2:
x = 6
Therefore, the correct answer is:
C. x = 6
Conclusion
Solving absolute value equations requires a systematic approach. By breaking down the equation into cases based on the sign of the expression inside the absolute value, we can solve each case separately. However, it's crucial to remember to verify the solutions to avoid extraneous results. This detailed guide has not only provided the solution to |x+10|=3x-2 but also equipped you with the knowledge and skills to confidently tackle similar absolute value problems in the future. Remember to always check your answers and understand the underlying principles, and you'll master the art of solving absolute value equations.
