Solving Absolute Value Equations Step By Step Guide To |6x+3|=21

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In the realm of mathematics, absolute value equations often present a unique challenge, demanding a careful and systematic approach to unravel their solutions. These equations, characterized by the presence of absolute value symbols, require a deep understanding of the concept of absolute value and its implications for solving equations. In this comprehensive guide, we will delve into the intricacies of solving absolute value equations, using the specific example of ∣6x+3∣=21|6x+3|=21 to illustrate the key steps and strategies involved.

At its core, the absolute value of a number represents its distance from zero on the number line. This distance is always non-negative, regardless of whether the original number is positive or negative. For instance, the absolute value of 5, denoted as ∣5∣|5|, is simply 5, as 5 is 5 units away from zero. Similarly, the absolute value of -5, denoted as βˆ£βˆ’5∣|-5|, is also 5, as -5 is also 5 units away from zero. This fundamental property of absolute value lies at the heart of solving absolute value equations.

When confronted with an absolute value equation like ∣6x+3∣=21|6x+3|=21, we must recognize that the expression inside the absolute value symbols, in this case, 6x+36x+3, can potentially be either 21 or -21. This is because both 21 and -21 have an absolute value of 21. Therefore, to solve the equation, we must consider both possibilities and solve the resulting equations separately. This splitting into cases is the cornerstone of solving absolute value equations.

Step-by-Step Solution of ∣6x+3∣=21|6x+3|=21

To effectively solve the absolute value equation ∣6x+3∣=21|6x+3|=21, we will follow a structured, step-by-step approach that ensures we capture all possible solutions. This systematic method involves splitting the equation into two separate cases, solving each case independently, and finally, verifying the solutions to ensure their validity. Let's embark on this journey of solving absolute value equations, unraveling the intricacies of each step.

Step 1: Split the Equation into Two Cases

The first crucial step in solving an absolute value equation is to recognize the inherent duality introduced by the absolute value. Since the absolute value of a number represents its distance from zero, the expression inside the absolute value symbols can be either positive or negative while still yielding the same absolute value. In our case, ∣6x+3∣=21|6x+3|=21 implies that the expression 6x+36x+3 can be either 21 or -21. This insight leads us to split the original equation into two distinct cases:

  • Case 1: 6x+3=216x+3 = 21
  • Case 2: 6x+3=βˆ’216x+3 = -21

By considering both possibilities, we ensure that we account for all potential solutions of the absolute value equation. This splitting into cases is the cornerstone of solving absolute value equations, allowing us to transform a single equation with absolute values into two simpler linear equations.

Step 2: Solve Each Case Independently

Now that we have successfully split the absolute value equation into two separate cases, the next step is to solve each case independently. This involves applying standard algebraic techniques to isolate the variable xx in each equation. Let's delve into the solution process for each case:

Case 1: 6x+3=216x+3 = 21

To solve this linear equation, we begin by isolating the term containing xx. This is achieved by subtracting 3 from both sides of the equation:

6x+3βˆ’3=21βˆ’36x + 3 - 3 = 21 - 3

This simplifies to:

6x=186x = 18

Next, to isolate xx, we divide both sides of the equation by 6:

rac{6x}{6} = rac{18}{6}

This yields the solution for Case 1:

x=3x = 3

Case 2: 6x+3=βˆ’216x+3 = -21

Following a similar approach as in Case 1, we begin by isolating the term containing xx. Subtracting 3 from both sides of the equation, we get:

6x+3βˆ’3=βˆ’21βˆ’36x + 3 - 3 = -21 - 3

This simplifies to:

6x=βˆ’246x = -24

Now, to isolate xx, we divide both sides of the equation by 6:

rac{6x}{6} = rac{-24}{6}

This gives us the solution for Case 2:

x=βˆ’4x = -4

Therefore, by solving each case independently, we have obtained two potential solutions for the absolute value equation: x=3x = 3 and x=βˆ’4x = -4.

Step 3: Verify the Solutions

Obtaining potential solutions is not the end of the journey. It is crucial to verify these solutions by substituting them back into the original absolute value equation. This step ensures that the solutions are valid and do not introduce any contradictions. Let's verify the solutions we obtained in the previous step:

Verifying x=3x = 3

Substituting x=3x = 3 into the original equation ∣6x+3∣=21|6x+3|=21, we get:

∣6(3)+3∣=21|6(3) + 3| = 21

Simplifying the expression inside the absolute value symbols:

∣18+3∣=21|18 + 3| = 21

∣21∣=21|21| = 21

Since the absolute value of 21 is indeed 21, the equation holds true, confirming that x=3x = 3 is a valid solution.

Verifying x=βˆ’4x = -4

Now, let's substitute x=βˆ’4x = -4 into the original equation:

∣6(βˆ’4)+3∣=21|6(-4) + 3| = 21

Simplifying the expression inside the absolute value symbols:

βˆ£βˆ’24+3∣=21|-24 + 3| = 21

βˆ£βˆ’21∣=21|-21| = 21

Since the absolute value of -21 is also 21, the equation holds true, confirming that x=βˆ’4x = -4 is also a valid solution.

Therefore, after verifying both potential solutions, we can confidently conclude that the solutions to the absolute value equation ∣6x+3∣=21|6x+3|=21 are x=3x = 3 and x=βˆ’4x = -4.

Conclusion

In this comprehensive guide, we have embarked on a detailed exploration of solving absolute value equations, using the specific example of ∣6x+3∣=21|6x+3|=21 to illustrate the key steps and strategies involved. We have demonstrated the importance of splitting the equation into two separate cases, solving each case independently, and verifying the solutions to ensure their validity. By mastering these techniques, you will be well-equipped to tackle a wide range of absolute value equations, unraveling their solutions with confidence and precision. Remember, the key to success lies in understanding the fundamental properties of absolute value and applying a systematic approach to problem-solving. With practice and perseverance, you can conquer the challenges posed by absolute value equations and expand your mathematical prowess.

Therefore, the correct answer is A. x=3x=3 and x=βˆ’4x=-4