In Exercises 35-44, Solve The Equation. Check Your Solutions.35. $|4n - 15| = |n|$36. $|2c + 8| = |10c|$37. $|2b - 9| = |b - 6|$38. $|3k - 2| = 2|k + 2|$39. $4|p - 3| = |2p + 8|$40. $2|4w - 1| = 3|4w +

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Introduction

Absolute value equations are a fundamental concept in algebra, and solving them requires a deep understanding of the properties of absolute value. In this article, we will explore the solutions to a series of absolute value equations, from 35 to 44, and provide a comprehensive guide on how to approach these types of problems.

Understanding Absolute Value

Before we dive into the solutions, let's take a moment to understand what absolute value is. The absolute value of a number is its distance from zero on the number line. In other words, it is the magnitude of the number without considering its direction. For example, the absolute value of 5 is 5, and the absolute value of -5 is also 5.

Solving Absolute Value Equations

Now that we have a basic understanding of absolute value, let's move on to solving absolute value equations. There are two main cases to consider when solving absolute value equations:

  • Case 1: The expressions inside the absolute value symbols are equal.
  • Case 2: The expressions inside the absolute value symbols are opposite.

Equation 35: ∣4nβˆ’15∣=∣n∣|4n - 15| = |n|

To solve this equation, we need to consider both cases.

Case 1: 4nβˆ’15=n4n - 15 = n

We can start by subtracting nn from both sides of the equation:

4nβˆ’15βˆ’n=nβˆ’n4n - 15 - n = n - n

This simplifies to:

3nβˆ’15=03n - 15 = 0

Next, we can add 15 to both sides of the equation:

3nβˆ’15+15=0+153n - 15 + 15 = 0 + 15

This simplifies to:

3n=153n = 15

Finally, we can divide both sides of the equation by 3:

3n3=153\frac{3n}{3} = \frac{15}{3}

This simplifies to:

n=5n = 5

Case 2: 4nβˆ’15=βˆ’n4n - 15 = -n

We can start by adding nn to both sides of the equation:

4nβˆ’15+n=βˆ’n+n4n - 15 + n = -n + n

This simplifies to:

5nβˆ’15=05n - 15 = 0

Next, we can add 15 to both sides of the equation:

5nβˆ’15+15=0+155n - 15 + 15 = 0 + 15

This simplifies to:

5n=155n = 15

Finally, we can divide both sides of the equation by 5:

5n5=155\frac{5n}{5} = \frac{15}{5}

This simplifies to:

n=3n = 3

Equation 36: ∣2c+8∣=∣10c∣|2c + 8| = |10c|

To solve this equation, we need to consider both cases.

Case 1: 2c+8=10c2c + 8 = 10c

We can start by subtracting 2c2c from both sides of the equation:

2c+8βˆ’2c=10cβˆ’2c2c + 8 - 2c = 10c - 2c

This simplifies to:

8=8c8 = 8c

Next, we can divide both sides of the equation by 8:

88=8c8\frac{8}{8} = \frac{8c}{8}

This simplifies to:

1=c1 = c

Case 2: 2c+8=βˆ’10c2c + 8 = -10c

We can start by adding 10c10c to both sides of the equation:

2c+8+10c=βˆ’10c+10c2c + 8 + 10c = -10c + 10c

This simplifies to:

12c+8=012c + 8 = 0

Next, we can subtract 8 from both sides of the equation:

12c+8βˆ’8=0βˆ’812c + 8 - 8 = 0 - 8

This simplifies to:

12c=βˆ’812c = -8

Finally, we can divide both sides of the equation by 12:

12c12=βˆ’812\frac{12c}{12} = \frac{-8}{12}

This simplifies to:

c=βˆ’23c = -\frac{2}{3}

Equation 37: ∣2bβˆ’9∣=∣bβˆ’6∣|2b - 9| = |b - 6|

To solve this equation, we need to consider both cases.

Case 1: 2bβˆ’9=bβˆ’62b - 9 = b - 6

We can start by subtracting bb from both sides of the equation:

2bβˆ’9βˆ’b=bβˆ’bβˆ’62b - 9 - b = b - b - 6

This simplifies to:

bβˆ’9=βˆ’6b - 9 = -6

Next, we can add 9 to both sides of the equation:

bβˆ’9+9=βˆ’6+9b - 9 + 9 = -6 + 9

This simplifies to:

b=3b = 3

Case 2: 2bβˆ’9=βˆ’(bβˆ’6)2b - 9 = -(b - 6)

We can start by distributing the negative sign to the expression inside the parentheses:

2bβˆ’9=βˆ’b+62b - 9 = -b + 6

Next, we can add bb to both sides of the equation:

2bβˆ’9+b=βˆ’b+b+62b - 9 + b = -b + b + 6

This simplifies to:

3bβˆ’9=63b - 9 = 6

Next, we can add 9 to both sides of the equation:

3bβˆ’9+9=6+93b - 9 + 9 = 6 + 9

This simplifies to:

3b=153b = 15

Finally, we can divide both sides of the equation by 3:

3b3=153\frac{3b}{3} = \frac{15}{3}

This simplifies to:

b=5b = 5

Equation 38: ∣3kβˆ’2∣=2∣k+2∣|3k - 2| = 2|k + 2|

To solve this equation, we need to consider both cases.

Case 1: 3kβˆ’2=2(k+2)3k - 2 = 2(k + 2)

We can start by distributing the 2 to the expression inside the parentheses:

3kβˆ’2=2k+43k - 2 = 2k + 4

Next, we can subtract 2k2k from both sides of the equation:

3kβˆ’2βˆ’2k=2kβˆ’2k+43k - 2 - 2k = 2k - 2k + 4

This simplifies to:

kβˆ’2=4k - 2 = 4

Next, we can add 2 to both sides of the equation:

kβˆ’2+2=4+2k - 2 + 2 = 4 + 2

This simplifies to:

k=6k = 6

Case 2: 3kβˆ’2=βˆ’2(k+2)3k - 2 = -2(k + 2)

We can start by distributing the negative sign to the expression inside the parentheses:

3kβˆ’2=βˆ’2kβˆ’43k - 2 = -2k - 4

Next, we can add 2k2k to both sides of the equation:

3kβˆ’2+2k=βˆ’2k+2kβˆ’43k - 2 + 2k = -2k + 2k - 4

This simplifies to:

5kβˆ’2=βˆ’45k - 2 = -4

Next, we can add 2 to both sides of the equation:

5kβˆ’2+2=βˆ’4+25k - 2 + 2 = -4 + 2

This simplifies to:

5k=βˆ’25k = -2

Finally, we can divide both sides of the equation by 5:

5k5=βˆ’25\frac{5k}{5} = \frac{-2}{5}

This simplifies to:

k=βˆ’25k = -\frac{2}{5}

Equation 39: 4∣pβˆ’3∣=∣2p+8∣4|p - 3| = |2p + 8|

To solve this equation, we need to consider both cases.

Case 1: 4(pβˆ’3)=2p+84(p - 3) = 2p + 8

We can start by distributing the 4 to the expression inside the parentheses:

4pβˆ’12=2p+84p - 12 = 2p + 8

Next, we can subtract 2p2p from both sides of the equation:

4pβˆ’12βˆ’2p=2pβˆ’2p+84p - 12 - 2p = 2p - 2p + 8

This simplifies to:

2pβˆ’12=82p - 12 = 8

Next, we can add 12 to both sides of the equation:

2pβˆ’12+12=8+122p - 12 + 12 = 8 + 12

This simplifies to:

2p=202p = 20

Finally, we can divide both sides of the equation by 2:

2p2=202\frac{2p}{2} = \frac{20}{2}

This simplifies to:

p=10p = 10

Case 2: 4(pβˆ’3)=βˆ’(2p+8)4(p - 3) = -(2p + 8)

We can start by distributing the negative sign to the expression inside the parentheses:

4(pβˆ’3)=βˆ’2pβˆ’84(p - 3) = -2p - 8

Next, we can distribute the 4 to the expression inside the parentheses:

4pβˆ’12=βˆ’2pβˆ’84p - 12 = -2p - 8

Next, we can add 2p2p to both sides of the equation:

4pβˆ’12+2p=βˆ’2p+2pβˆ’84p - 12 + 2p = -2p + 2p - 8

This simplifies to:

6pβˆ’12=βˆ’86p - 12 = -8

Next, we can add 12 to both sides of the equation:

6pβˆ’12+12=βˆ’8+126p - 12 + 12 = -8 + 12

This simplifies to:

6p=46p = 4

Finally, we can divide both sides of the equation by 6:

6p6=46\frac{6p}{6} = \frac{4}{6}

This simplifies to:

p =<br/> **Solving Absolute Value Equations: A Comprehensive Guide** =========================================================== **Q&A: Frequently Asked Questions** ----------------------------------- **Q: What is an absolute value equation?** ----------------------------------------- A: An absolute value equation is an equation that contains an absolute value expression. The absolute value of a number is its distance from zero on the number line. **Q: How do I solve an absolute value equation?** ---------------------------------------------- A: To solve an absolute value equation, you need to consider two cases: * **Case 1:** The expressions inside the absolute value symbols are equal. * **Case 2:** The expressions inside the absolute value symbols are opposite. **Q: What is the difference between Case 1 and Case 2?** ------------------------------------------------ A: In Case 1, the expressions inside the absolute value symbols are equal. In Case 2, the expressions inside the absolute value symbols are opposite. **Q: How do I determine which case to use?** -------------------------------------------- A: To determine which case to use, you need to look at the equation and see if the expressions inside the absolute value symbols are equal or opposite. **Q: What if I get stuck on a problem?** -------------------------------------- A: If you get stuck on a problem, try breaking it down into smaller steps. Start by simplifying the equation and then work your way up to the solution. **Q: Can I use a calculator to solve absolute value equations?** --------------------------------------------------------- A: Yes, you can use a calculator to solve absolute value equations. However, it's always a good idea to check your work by plugging the solution back into the original equation. **Q: What are some common mistakes to avoid when solving absolute value equations?** -------------------------------------------------------------------------------- A: Some common mistakes to avoid when solving absolute value equations include: * **Not considering both cases:** Make sure to consider both Case 1 and Case 2 when solving an absolute value equation. * **Not simplifying the equation:** Make sure to simplify the equation before solving it. * **Not checking the solution:** Make sure to check the solution by plugging it back into the original equation. **Q: How do I know if my solution is correct?** ------------------------------------------------ A: To know if your solution is correct, you need to plug it back into the original equation and check if it's true. **Q: Can I use absolute value equations in real-life situations?** --------------------------------------------------------- A: Yes, you can use absolute value equations in real-life situations. For example, you can use them to model distance, temperature, and other quantities that can be positive or negative. **Conclusion** ---------- Solving absolute value equations can be challenging, but with practice and patience, you can master it. Remember to consider both cases, simplify the equation, and check the solution. With these tips and tricks, you'll be able to solve absolute value equations like a pro! **Additional Resources** ------------------------- * **Absolute Value Equations Worksheet:** Try this worksheet to practice solving absolute value equations. * **Absolute Value Equations Video:** Watch this video to learn more about solving absolute value equations. * **Absolute Value Equations Online Course:** Take this online course to learn more about solving absolute value equations. **Final Thoughts** ---------------- Solving absolute value equations is an important skill to have in mathematics. With practice and patience, you can master it and use it to solve real-life problems. Remember to consider both cases, simplify the equation, and check the solution. Happy solving!