The Hardcover Version Of A Book Weighs Twice As Much As Its Paperback Version. The Hardcover Book And The Paperback Together Weigh 4.2 Pounds. Which System Of Equations Can Be Used To Find H H H , The Weight Of The Hardcover Book, And

Introduction

In the world of mathematics, problems often arise in the form of equations, which can be solved using various techniques. In this article, we will delve into a classic problem involving the weights of hardcover and paperback books. We will explore the system of equations that can be used to find the weight of the hardcover book, denoted as $h$, and the weight of the paperback book, denoted as $p$.

The Problem

The problem states that the hardcover version of a book weighs twice as much as its paperback version. Additionally, the combined weight of the hardcover and paperback books is 4.2 pounds. Mathematically, this can be represented as:

• $h = 2p$ (the hardcover book weighs twice as much as the paperback book)
• $h + p = 4.2$ (the combined weight of the hardcover and paperback books is 4.2 pounds)

System of Equations

To find the weight of the hardcover book, denoted as $h$, and the weight of the paperback book, denoted as $p$, we can use a system of equations. A system of equations is a set of two or more equations that are solved simultaneously to find the values of the variables.

In this case, we have two equations:

1. $h = 2p$
2. $h + p = 4.2$

We can solve this system of equations using substitution or elimination. Let's use substitution to find the values of $h$ and $p$.

Substitution Method

To solve the system of equations using substitution, we can substitute the expression for $h$ from the first equation into the second equation. This gives us:

$h + p = 4.2$

Substituting $h = 2p$ into the equation, we get:

$2p + p = 4.2$

Combine like terms:

$3p = 4.2$

Divide both sides by 3:

$p = 1.4$

Now that we have found the value of $p$, we can substitute it back into the first equation to find the value of $h$:

$h = 2p$

Substituting $p = 1.4$ into the equation, we get:

$h = 2(1.4)$

$h = 2.8$

Elimination Method

To solve the system of equations using elimination, we can multiply the first equation by 1 and the second equation by -2. This gives us:

1. $h = 2p$
2. $-2h - 2p = -8.4$

Add the two equations to eliminate the variable $h$:

$-h = -8.4$

Divide both sides by -1:

$h = 8.4$

Now that we have found the value of $h$, we can substitute it back into the first equation to find the value of $p$:

$h = 2p$

Substituting $h = 8.4$ into the equation, we get:

$8.4 = 2p$

Divide both sides by 2:

$p = 4.2$

Conclusion

In this article, we have explored the system of equations that can be used to find the weight of the hardcover book, denoted as $h$, and the weight of the paperback book, denoted as $p$. We have used both the substitution and elimination methods to solve the system of equations and find the values of $h$ and $p$. The final answer is:

• $h = 2.8$ pounds
• $p = 1.4$ pounds

Final Answer

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Q&A: Frequently Asked Questions

Q: What is the main difference between the hardcover and paperback versions of a book? A: The main difference between the hardcover and paperback versions of a book is the weight of the book. The hardcover book weighs twice as much as the paperback book.

Q: How can we represent the weight of the hardcover book and the paperback book mathematically? A: We can represent the weight of the hardcover book and the paperback book mathematically using the following equations:

• $h = 2p$ (the hardcover book weighs twice as much as the paperback book)
• $h + p = 4.2$ (the combined weight of the hardcover and paperback books is 4.2 pounds)

Q: What is a system of equations? A: A system of equations is a set of two or more equations that are solved simultaneously to find the values of the variables.

Q: How can we solve the system of equations using substitution? A: To solve the system of equations using substitution, we can substitute the expression for $h$ from the first equation into the second equation. This gives us:

$h + p = 4.2$

Substituting $h = 2p$ into the equation, we get:

$2p + p = 4.2$

Combine like terms:

$3p = 4.2$

Divide both sides by 3:

$p = 1.4$

Now that we have found the value of $p$, we can substitute it back into the first equation to find the value of $h$:

$h = 2p$

Substituting $p = 1.4$ into the equation, we get:

$h = 2(1.4)$

$h = 2.8$

Q: How can we solve the system of equations using elimination? A: To solve the system of equations using elimination, we can multiply the first equation by 1 and the second equation by -2. This gives us:

1. $h = 2p$
2. $-2h - 2p = -8.4$

Add the two equations to eliminate the variable $h$:

$-h = -8.4$

Divide both sides by -1:

$h = 8.4$

Now that we have found the value of $h$, we can substitute it back into the first equation to find the value of $p$:

$h = 2p$

Substituting $h = 8.4$ into the equation, we get:

$8.4 = 2p$

Divide both sides by 2:

$p = 4.2$

Q: What are the final answers for the weight of the hardcover book and the paperback book? A: The final answers for the weight of the hardcover book and the paperback book are:

• $h = 2.8$ pounds
• $p = 1.4$ pounds

Q: Can you provide a summary of the article? A: In this article, we explored the system of equations that can be used to find the weight of the hardcover book, denoted as $h$, and the weight of the paperback book, denoted as $p$. We used both the substitution and elimination methods to solve the system of equations and find the values of $h$ and $p$. The final answers are $h = 2.8$ pounds and $p = 1.4$ pounds.

Q: What is the main takeaway from this article? A: The main takeaway from this article is that systems of equations can be used to solve problems involving multiple variables. By using substitution or elimination, we can find the values of the variables and solve the problem.