Solve The Following Inequality:$\[ |6x - 3| \leq 1 \\]Choose The Correct Answer Below And Fill In The Values. Round To Three Decimal Places If Needed.A. \[$\square \leq X \leq \square\$\] B. \[$x \leq \square \quad \text{OR} \quad

Introduction

In this article, we will focus on solving the given inequality $|6x - 3| \leq 1$. This type of inequality involves absolute values, which can be challenging to solve. However, with the correct approach and steps, we can find the solution to this inequality. We will break down the solution into manageable parts and provide a clear explanation of each step.

Understanding Absolute Value Inequalities

Before we dive into solving the inequality, let's briefly discuss what absolute value inequalities are. An absolute value inequality is an inequality that involves the absolute value of an expression. The absolute value of a number is its distance from zero on the number line. For example, the absolute value of 5 is 5, and the absolute value of -5 is also 5.

Solving the Inequality

To solve the inequality $|6x - 3| \leq 1$, we need to consider two cases:

Case 1: $6x - 3 \geq 0$

In this case, the absolute value inequality becomes $6x - 3 \leq 1$. We can solve this inequality by adding 3 to both sides, which gives us $6x \leq 4$. Dividing both sides by 6, we get $x \leq \frac{2}{3}$.

Case 2: $6x - 3 < 0$

In this case, the absolute value inequality becomes $-(6x - 3) \leq 1$. We can simplify this inequality by distributing the negative sign, which gives us $-6x + 3 \leq 1$. Subtracting 3 from both sides, we get $-6x \leq -2$. Dividing both sides by -6, we get $x \geq \frac{1}{3}$.

Combining the Solutions

Now that we have solved the inequality for both cases, we need to combine the solutions. The solution to the inequality $|6x - 3| \leq 1$ is the intersection of the two solutions we found in the previous steps. Therefore, the solution is $x \leq \frac{2}{3}$ and $x \geq \frac{1}{3}$.

Conclusion

In this article, we solved the inequality $|6x - 3| \leq 1$. We broke down the solution into manageable parts and provided a clear explanation of each step. We found that the solution to the inequality is $x \leq \frac{2}{3}$ and $x \geq \frac{1}{3}$. This solution can be written in interval notation as $\left[\frac{1}{3}, \frac{2}{3}\right]$.

Final Answer

The final answer is $\boxed{\left[\frac{1}{3}, \frac{2}{3}\right]}$.

Discussion

The solution to the inequality $|6x - 3| \leq 1$ is a closed interval, which means that the endpoints are included in the solution. This is because the inequality is of the form $|x - a| \leq b$, where $a$ and $b$ are positive numbers. In this case, $a = \frac{3}{6} = \frac{1}{2}$ and $b = 1$. Therefore, the solution is a closed interval centered at $\frac{1}{2}$ with a radius of 1.

Related Topics

• Solving linear inequalities
• Solving absolute value inequalities
• Interval notation

References

• [1] "Algebra and Trigonometry" by Michael Sullivan
• [2] "College Algebra" by James Stewart
• [3] "Precalculus" by Michael Sullivan
  

Introduction

In our previous article, we solved the inequality $|6x - 3| \leq 1$ and found that the solution is $x \leq \frac{2}{3}$ and $x \geq \frac{1}{3}$. In this article, we will provide a Q&A section to help clarify any doubts or questions that readers may have.

Q&A

Q: What is the meaning of the absolute value inequality $|6x - 3| \leq 1$?

A: The absolute value inequality $|6x - 3| \leq 1$ means that the distance between $6x - 3$ and 0 is less than or equal to 1.

Q: How do I solve an absolute value inequality?

A: To solve an absolute value inequality, you need to consider two cases: when the expression inside the absolute value is positive and when it is negative. In the case of $|6x - 3| \leq 1$, we considered two cases: $6x - 3 \geq 0$ and $6x - 3 < 0$.

Q: What is the difference between a closed interval and an open interval?

A: A closed interval is an interval that includes the endpoints, while an open interval is an interval that does not include the endpoints. In the case of the solution to the inequality $|6x - 3| \leq 1$, the solution is a closed interval $\left[\frac{1}{3}, \frac{2}{3}\right]$.

Q: How do I write the solution to an inequality in interval notation?

A: To write the solution to an inequality in interval notation, you need to determine whether the solution is a closed interval, an open interval, or a combination of both. In the case of the solution to the inequality $|6x - 3| \leq 1$, the solution is a closed interval $\left[\frac{1}{3}, \frac{2}{3}\right]$.

Q: Can I use a calculator to solve an absolute value inequality?

A: Yes, you can use a calculator to solve an absolute value inequality. However, it's always a good idea to check your work by hand to make sure that the solution is correct.

Q: What are some common mistakes to avoid when solving absolute value inequalities?

A: Some common mistakes to avoid when solving absolute value inequalities include:

• Not considering both cases when the expression inside the absolute value is positive and when it is negative.
• Not including the endpoints in the solution.
• Not writing the solution in interval notation.

Conclusion

In this article, we provided a Q&A section to help clarify any doubts or questions that readers may have about solving the inequality $|6x - 3| \leq 1$. We hope that this article has been helpful in providing a clear understanding of how to solve absolute value inequalities.

Final Answer

The final answer is $\boxed{\left[\frac{1}{3}, \frac{2}{3}\right]}$.

Discussion

The solution to the inequality $|6x - 3| \leq 1$ is a closed interval, which means that the endpoints are included in the solution. This is because the inequality is of the form $|x - a| \leq b$, where $a$ and $b$ are positive numbers. In this case, $a = \frac{3}{6} = \frac{1}{2}$ and $b = 1$. Therefore, the solution is a closed interval centered at $\frac{1}{2}$ with a radius of 1.

Related Topics

• Solving linear inequalities
• Solving absolute value inequalities
• Interval notation

References

• [1] "Algebra and Trigonometry" by Michael Sullivan
• [2] "College Algebra" by James Stewart
• [3] "Precalculus" by Michael Sullivan