Solving Absolute Value Equations A Step By Step Guide

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In the realm of mathematics, translating verbal statements into algebraic equations is a fundamental skill. This ability allows us to represent real-world scenarios and solve for unknown variables. This article will delve into the process of converting a given statement into an equation and subsequently solving it, with a particular focus on equations involving absolute values. Absolute value equations introduce an additional layer of complexity, as we must consider both positive and negative scenarios. This comprehensive guide will walk you through the steps, ensuring a clear understanding of the underlying concepts and techniques.

Understanding the Statement

The given statement is: "Negative three times the absolute value of the quantity two minus five times pp is -42." Our first task is to dissect this statement and identify the key components that will form our equation. The phrase "negative three times" indicates a multiplication by -3. The term "absolute value" signifies that we are dealing with the magnitude of the expression inside the absolute value bars, regardless of its sign. The phrase "the quantity two minus five times pp " translates to 2βˆ’5p2 - 5p. Finally, "is -42" tells us that the entire expression equals -42. By carefully breaking down the statement, we can begin to construct the corresponding equation.

Translating Words into Mathematical Symbols

To effectively translate a verbal statement into a mathematical equation, it is crucial to identify the mathematical operations and relationships being described. Key phrases and their corresponding mathematical symbols are shown bellow:

  • "times" or "product" indicates multiplication
  • "minus" or "difference" indicates subtraction
  • "is" or "equals" indicates equality (=)
  • "absolute value" is represented by vertical bars ||

Using these guidelines, we can translate "Negative three times the absolute value of the quantity two minus five times pp is -42" into the equation: βˆ’3∣2βˆ’5p∣=βˆ’42-3|2 - 5p| = -42. This equation now represents the given statement in a concise mathematical form, making it easier to manipulate and solve for the unknown variable pp. The process of translating verbal statements into equations is a crucial skill in algebra, as it allows us to model and solve real-world problems. Accurate translation is the first step towards finding a solution.

Forming the Equation

Now that we have dissected the statement, we can construct the equation. "Negative three times" translates to -3 multiplied by something. The "absolute value of the quantity two minus five times pp " is represented as ∣2βˆ’5p∣|2 - 5p|. The entire phrase "Negative three times the absolute value of the quantity two minus five times pp " becomes βˆ’3∣2βˆ’5p∣-3|2 - 5p|. Finally, "is -42" indicates that this expression equals -42. Therefore, the equation is:

βˆ’3∣2βˆ’5p∣=βˆ’42-3|2 - 5p| = -42

This equation accurately represents the given statement in mathematical form. The next step is to solve this equation for the variable pp. Solving absolute value equations requires a specific approach, which we will explore in the subsequent sections.

Detailed Breakdown of the Equation Formation

To solidify the understanding of equation formation, let's delve deeper into the step-by-step process. The statement we are working with is: "Negative three times the absolute value of the quantity two minus five times pp is -42." We begin by identifying the main components and their mathematical equivalents. "Negative three times" signifies the multiplication of -3 with the subsequent expression. The phrase "the absolute value of the quantity two minus five times pp " is the core of the equation. First, "five times pp " translates to 5p5p. Then, "two minus five times pp " becomes 2βˆ’5p2 - 5p. The absolute value of this expression is represented as ∣2βˆ’5p∣|2 - 5p|. Combining these, "negative three times the absolute value of the quantity two minus five times pp " is expressed as βˆ’3∣2βˆ’5p∣-3|2 - 5p|. The final part of the statement, "is -42," indicates that the entire expression equals -42. Thus, we arrive at the complete equation: βˆ’3∣2βˆ’5p∣=βˆ’42-3|2 - 5p| = -42. This detailed breakdown illustrates how each part of the verbal statement is meticulously translated into its mathematical counterpart, resulting in the final equation. Accurate formation of the equation is paramount for successful problem-solving. This process requires a careful and methodical approach, ensuring that each component of the statement is correctly represented in the equation.

Solving the Equation

To solve the equation βˆ’3∣2βˆ’5p∣=βˆ’42-3|2 - 5p| = -42, we first need to isolate the absolute value term. We can do this by dividing both sides of the equation by -3:

∣2βˆ’5p∣=14|2 - 5p| = 14

Now, we have an absolute value equation in the form ∣ax+b∣=c|ax + b| = c. To solve this type of equation, we need to consider two cases:

Case 1: The expression inside the absolute value is positive or zero:

2βˆ’5p=142 - 5p = 14

Case 2: The expression inside the absolute value is negative:

βˆ’(2βˆ’5p)=14-(2 - 5p) = 14

Solving Case 1: 2βˆ’5p=142 - 5p = 14

To solve 2βˆ’5p=142 - 5p = 14, we first subtract 2 from both sides:

βˆ’5p=12-5p = 12

Then, we divide both sides by -5:

p=βˆ’12/5=βˆ’2.4p = -12/5 = -2.4

So, one solution is p=βˆ’2.4p = -2.4.

Solving Case 2: βˆ’(2βˆ’5p)=14-(2 - 5p) = 14

To solve βˆ’(2βˆ’5p)=14-(2 - 5p) = 14, we first distribute the negative sign:

βˆ’2+5p=14-2 + 5p = 14

Next, we add 2 to both sides:

5p=165p = 16

Finally, we divide both sides by 5:

p=16/5=3.2p = 16/5 = 3.2

Thus, the second solution is p=3.2p = 3.2.

Comprehensive Explanation of Solving Absolute Value Equations

Solving absolute value equations requires a systematic approach that accounts for the nature of absolute values. The absolute value of a number is its distance from zero, which means it is always non-negative. Therefore, an absolute value equation ∣ax+b∣=c|ax + b| = c implies that the expression inside the absolute value bars, ax+bax + b, can be either cc or βˆ’c-c, as both have the same distance from zero. To solve the equation βˆ’3∣2βˆ’5p∣=βˆ’42-3|2 - 5p| = -42, we begin by isolating the absolute value term. Dividing both sides by -3 gives us ∣2βˆ’5p∣=14|2 - 5p| = 14. Now, we recognize that the expression 2βˆ’5p2 - 5p can be either 14 or -14. This leads us to consider two separate cases:

Case 1: The expression inside the absolute value is equal to 14:

2βˆ’5p=142 - 5p = 14

To solve this linear equation, we first subtract 2 from both sides:

βˆ’5p=14βˆ’2-5p = 14 - 2

βˆ’5p=12-5p = 12

Next, we divide both sides by -5 to isolate pp:

p = rac{12}{-5}

p=βˆ’2.4p = -2.4

Case 2: The expression inside the absolute value is equal to -14:

2βˆ’5p=βˆ’142 - 5p = -14

Again, we solve this linear equation by first subtracting 2 from both sides:

βˆ’5p=βˆ’14βˆ’2-5p = -14 - 2

βˆ’5p=βˆ’16-5p = -16

Now, we divide both sides by -5 to find pp:

p = rac{-16}{-5}

p=3.2p = 3.2

Therefore, the solutions to the equation βˆ’3∣2βˆ’5p∣=βˆ’42-3|2 - 5p| = -42 are p=βˆ’2.4p = -2.4 and p=3.2p = 3.2. This comprehensive approach ensures that we account for both possibilities arising from the absolute value, leading to a complete and accurate solution.

Solutions

Therefore, the solutions for pp are:

p=βˆ’2.4p = -2.4 and p=3.2p = 3.2

These values satisfy the original equation, meaning that when we substitute either -2.4 or 3.2 for pp in the equation βˆ’3∣2βˆ’5p∣=βˆ’42-3|2 - 5p| = -42, the equation holds true. Absolute value equations often have two solutions because the absolute value represents the distance from zero, and both a positive and a negative value can have the same distance.

Verification of Solutions

To ensure the accuracy of our solutions, it is essential to verify them by substituting each value back into the original equation. This process confirms that the solutions satisfy the equation and that no errors were made during the solving process. The original equation is βˆ’3∣2βˆ’5p∣=βˆ’42-3|2 - 5p| = -42. Let's verify the solutions p=βˆ’2.4p = -2.4 and p=3.2p = 3.2 one by one.

Verification for p=βˆ’2.4p = -2.4:

Substitute p=βˆ’2.4p = -2.4 into the equation:

βˆ’3∣2βˆ’5(βˆ’2.4)∣=βˆ’42-3|2 - 5(-2.4)| = -42

First, calculate the expression inside the absolute value:

2βˆ’5(βˆ’2.4)=2+12=142 - 5(-2.4) = 2 + 12 = 14

Now, take the absolute value:

∣14∣=14|14| = 14

Multiply by -3:

βˆ’3(14)=βˆ’42-3(14) = -42

Since -42 equals -42, the solution p=βˆ’2.4p = -2.4 is verified.

Verification for p=3.2p = 3.2:

Substitute p=3.2p = 3.2 into the equation:

βˆ’3∣2βˆ’5(3.2)∣=βˆ’42-3|2 - 5(3.2)| = -42

Calculate the expression inside the absolute value:

2βˆ’5(3.2)=2βˆ’16=βˆ’142 - 5(3.2) = 2 - 16 = -14

Take the absolute value:

βˆ£βˆ’14∣=14|-14| = 14

Multiply by -3:

βˆ’3(14)=βˆ’42-3(14) = -42

Again, -42 equals -42, confirming that the solution p=3.2p = 3.2 is also valid.

By verifying both solutions, we have confirmed that p=βˆ’2.4p = -2.4 and p=3.2p = 3.2 are indeed the correct solutions to the equation βˆ’3∣2βˆ’5p∣=βˆ’42-3|2 - 5p| = -42. This verification step is a crucial practice in mathematics to ensure the accuracy and reliability of the results.

Conclusion

In this article, we have successfully translated the statement "Negative three times the absolute value of the quantity two minus five times pp is -42" into the equation βˆ’3∣2βˆ’5p∣=βˆ’42-3|2 - 5p| = -42. We then solved this equation by considering two cases, which arise due to the absolute value. The solutions we found are p=βˆ’2.4p = -2.4 and p=3.2p = 3.2. This process demonstrates the importance of understanding absolute value equations and the steps required to solve them. By carefully breaking down the statement, forming the equation, and considering both positive and negative scenarios, we can confidently find the solutions. This skill is crucial in algebra and various other mathematical contexts. Understanding how to handle absolute value equations is a valuable tool for solving a wide range of problems.

Key Takeaways and Further Practice

Mastering the solution of equations involving absolute values is a fundamental skill in algebra with broad applications in various mathematical and real-world contexts. Throughout this comprehensive guide, we have explored the step-by-step process of translating verbal statements into algebraic equations, solving absolute value equations, and verifying the solutions. To reinforce these concepts, let's summarize the key takeaways and suggest additional practice exercises.

Key Takeaways:

  1. Translating Verbal Statements: Accurate translation of verbal statements into mathematical equations is the cornerstone of problem-solving. Pay close attention to keywords and phrases such as "times," "minus," "is," and "absolute value," and understand their corresponding mathematical symbols and operations.

  2. Forming Absolute Value Equations: When dealing with absolute values, recognize that the expression inside the absolute value bars can be either positive or negative. This leads to two separate cases that need to be considered.

  3. Solving Absolute Value Equations: To solve an absolute value equation ∣ax+b∣=c|ax + b| = c, isolate the absolute value term and then set up two equations: ax+b=cax + b = c and ax+b=βˆ’cax + b = -c. Solve each equation independently to find the possible values of the variable.

  4. Verifying Solutions: Always verify your solutions by substituting them back into the original equation. This step ensures that the solutions are correct and that no errors were made during the solving process.

  5. Understanding Absolute Value: The absolute value of a number is its distance from zero, which is always non-negative. This property is crucial for understanding why absolute value equations often have two solutions.

Further Practice:

To solidify your understanding and skills in solving absolute value equations, consider working through the following practice exercises:

  1. Solve: ∣3xβˆ’4∣=8|3x - 4| = 8
  2. Solve: 2∣x+1βˆ£βˆ’5=92|x + 1| - 5 = 9
  3. Solve: βˆ’4∣2pβˆ’3∣=βˆ’20-4|2p - 3| = -20
  4. Write the equation and solve: "Five times the absolute value of the quantity one plus two times xx is 15."
  5. Write the equation and solve: "Negative two times the absolute value of the quantity three minus four times yy is -24."

By consistently practicing these types of problems, you will develop confidence and proficiency in solving absolute value equations. Remember to break down each problem into manageable steps, carefully consider both positive and negative cases, and always verify your solutions. This methodical approach will serve you well in more advanced mathematical studies and real-world applications.