Solving Gillian's Book Purchase A System Of Equations Word Problem

Hey there, bookworms and math enthusiasts! Ever been to a library book sale? It's like stepping into a treasure trove, isn't it? Gillian certainly thought so when she snagged a whopping 25 books! But here's the twist – a mix of hardcover and paperback, each with a different price tag. And that's where our little mathematical puzzle begins. Let's dive into this intriguing scenario, break down the numbers, and figure out exactly how many of each type of book Gillian added to her collection. We'll explore how a simple system of equations can help us solve real-world problems, making math not just a subject, but a super useful tool in our everyday lives. So, grab your thinking caps, and let's get started!

Decoding the Bookworm's Budget: Setting Up the Equations

So, Gillian's book-buying adventure presents us with a classic problem that can be beautifully solved using a system of equations. The core of the issue? Gillian bought a total of 25 books, a mix of hardcovers and paperbacks, and spent a total of $26.50. Hardcovers were priced at $1.50 each, while paperbacks were a steal at $0.50 each. Our mission is to figure out how many of each type she purchased. To get started, we need to translate this word problem into the language of mathematics. The beauty of algebra lies in its ability to represent unknowns with variables, and relationships with equations. Let's define our variables: let 'h' represent the number of hardcover books Gillian bought, and 'p' represent the number of paperback books. This simple step is crucial; it lays the foundation for our entire solution. Now, with our variables defined, we can start constructing the equations that will lead us to the answer. The first piece of information we have is the total number of books. Gillian bought 25 books in total, which includes both hardcovers and paperbacks. This translates directly into our first equation: h + p = 25. This equation is a straightforward representation of the total quantity, and it's a cornerstone of our solution. The second piece of information is the total amount Gillian spent, which was $26.50. This involves the cost of each type of book. Each hardcover costs $1.50, so the total cost of hardcovers is 1.50h. Similarly, each paperback costs $0.50, so the total cost of paperbacks is 0.50p. The sum of these costs must equal the total amount Gillian spent. This gives us our second equation: 1. 50h + 0.50p = 26.50. This equation captures the financial aspect of Gillian's book-buying spree. Together, these two equations form a system of linear equations. This system is a powerful tool because it allows us to solve for two unknowns (h and p) using two related equations. The equations, h + p = 25 and 1.50h + 0.50p = 26.50, perfectly encapsulate the scenario. They represent the two key constraints of the problem: the total number of books and the total amount spent. With this system in place, we're well on our way to uncovering the exact breakdown of Gillian's literary haul. The next step is to choose a method to solve this system, and that's where things get even more interesting!

Cracking the Code: Solving the System of Equations

Alright guys, we've got our system of equations all set up, shining like a mathematical beacon, ready to guide us to the solution. Now comes the fun part – actually solving it! We have a couple of trusty methods in our algebraic toolkit: substitution and elimination. Each has its own charm, and for this particular problem, both could work swimmingly. But let's walk through the substitution method first, just to keep things crystal clear. The substitution method is like a clever algebraic dance. The main idea? We isolate one variable in one equation, and then we substitute that expression into the other equation. This transforms our two-variable problem into a single-variable problem, which is much easier to handle. Looking at our system, h + p = 25 and 1.50h + 0.50p = 26.50, the first equation seems simpler to manipulate. Let's isolate 'p' in the first equation. To do this, we simply subtract 'h' from both sides, giving us p = 25 - h. See how neatly we've expressed 'p' in terms of 'h'? This is our golden ticket for the substitution step. Now, we take this expression for 'p' (which is 25 - h) and substitute it into the second equation, replacing 'p' wherever it appears. So, 1.50h + 0.50p = 26.50 becomes 1.50h + 0.50(25 - h) = 26.50. Notice what we've done? We've eliminated 'p' from the second equation, and we're left with an equation that only involves 'h'. This is a huge step forward! With our equation in a single variable, we can now solve for 'h'. First, we need to distribute the 0.50 across the parentheses: 1.50h + 12.5 - 0.50h = 26.50. Next, we combine like terms. We have 1.50h and -0.50h, which combine to give us 1.00h (or just 'h'). So our equation simplifies to h + 12.5 = 26.50. Now, to isolate 'h', we subtract 12.5 from both sides: h = 26.50 - 12.5. This gives us h = 14. Hooray! We've found the value of 'h', the number of hardcover books Gillian bought. But we're not done yet. We still need to find 'p', the number of paperback books. But finding 'p' is now a breeze because we already have an expression for 'p' in terms of 'h': p = 25 - h. We simply plug in the value we found for 'h', which is 14. So, p = 25 - 14, which gives us p = 11. And there we have it! We've successfully solved the system of equations. Gillian bought 14 hardcover books and 11 paperback books. We've cracked the code of Gillian's book-buying adventure using the magic of algebra! But just to be super sure, let's take a moment to verify our solution. This is a crucial step in any problem-solving process. It's like a final checkmark to ensure we're on the right track. So, how do we verify our solution? We simply plug our values for 'h' and 'p' back into our original equations and see if they hold true. Let's start with the first equation: h + p = 25. If we substitute h = 14 and p = 11, we get 14 + 11 = 25, which is indeed true. So far, so good! Now, let's check the second equation: 1.50h + 0.50p = 26.50. Substituting our values, we get 1. 50(14) + 0.50(11) = 26.50. Let's break this down: 1.50 multiplied by 14 is 21, and 0.50 multiplied by 11 is 5.5. So the equation becomes 21 + 5.5 = 26.50, which is also true! Both equations hold true with our values for 'h' and 'p', giving us confidence that our solution is correct. We've not only solved the problem, but we've also verified our answer, ensuring accuracy. This methodical approach is key to successful problem-solving in mathematics and beyond.

The Literary Tally: Gillian's Hardcover vs. Paperback Collection

Okay, math detectives, let's recap the case of Gillian's book sale haul! We started with a seemingly simple scenario: Gillian bought 25 books, a mix of hardcovers and paperbacks, spending a total of $26.50. The prices were clear – $1.50 for each hardcover and a sweet $0.50 for each paperback. But hidden within these details was a mathematical puzzle, a system of equations waiting to be solved. We translated the words into equations, defining 'h' for the number of hardcovers and 'p' for the number of paperbacks. This crucial step of turning real-world information into mathematical language is the cornerstone of problem-solving. We crafted two equations that beautifully captured the essence of the situation: h + p = 25 (the total number of books) and 1.50h + 0.50p = 26.50 (the total amount spent). These equations were our map, guiding us towards the solution. Then came the exciting part – solving the system! We chose the substitution method, a clever algebraic dance where we isolated one variable and substituted its expression into the other equation. It was like untangling a knot, one step at a time. We isolated 'p' in the first equation, expressing it as p = 25 - h. Then, we substituted this into the second equation, eliminating 'p' and giving us an equation solely in terms of 'h'. This was a breakthrough moment! With a single variable equation in our hands, we meticulously solved for 'h'. We combined like terms, simplified, and isolated 'h', finally arriving at h = 14. Gillian bought 14 hardcover books! But we didn't stop there. Our mission wasn't complete until we knew the number of paperbacks as well. Fortunately, we had an expression for 'p' ready and waiting: p = 25 - h. Plugging in our value for 'h', we found p = 11. Gillian also bought 11 paperback books! We had our solution: 14 hardcovers and 11 paperbacks. But a good detective always verifies their findings. We plugged our values for 'h' and 'p' back into our original equations, ensuring that they held true. And they did! This step not only confirmed our solution but also gave us confidence in our process. So, what does this solution tell us about Gillian's literary preferences and budget? Well, it reveals a balanced mix of hardcovers and paperbacks. Gillian likely appreciated the durability and prestige of hardcovers, but also the affordability and portability of paperbacks. The numbers also reflect a thoughtful approach to her book-buying. Gillian managed to acquire a substantial collection without exceeding her budget. This problem demonstrates the power of mathematics in everyday scenarios. It's not just about abstract numbers and formulas; it's about using tools to make sense of the world around us. Whether you're planning a budget, managing resources, or simply satisfying your book-buying cravings, mathematical thinking can be a powerful ally.

Real-World Math: Beyond the Book Sale

So, we've conquered Gillian's book-buying dilemma, and it's a victory worth celebrating! But let's zoom out for a moment and appreciate the bigger picture. This isn't just about books and prices; it's about the power of math to solve real-world problems. The system of equations we used is a versatile tool that can be applied in countless situations. Think about it: any time you have two related unknowns and two pieces of information connecting them, you can potentially use a system of equations to find the answer. It's like having a secret decoder ring for the complexities of life! Let's brainstorm some other scenarios where this approach could come in handy. Imagine you're planning a party and need to buy snacks. You want to get a mix of chips and cookies, and you have a budget to stick to. Each item has a different price, and you have a certain amount you want to spend in total. Sound familiar? This is essentially the same structure as Gillian's book problem! You can define variables for the number of bags of chips and boxes of cookies, set up equations based on the total quantity and the total cost, and solve for the exact number of each item to buy. Or how about a business context? Suppose a company sells two different products. They know the total number of products sold and the total revenue generated. If they also know the price of each product, they can use a system of equations to determine how many units of each product were sold. This kind of analysis is crucial for understanding sales trends and making informed business decisions. Even in the realm of science and engineering, systems of equations are indispensable. They're used to model physical systems, analyze circuits, optimize designs, and much more. From calculating the trajectory of a rocket to designing a bridge, systems of equations are the backbone of many scientific and engineering applications. What makes this mathematical tool so powerful? It's the ability to break down complex problems into manageable parts. By representing unknowns with variables and relationships with equations, we can create a structured framework for solving even the most challenging puzzles. And the beauty is, the techniques we've used today – substitution, elimination – are applicable across a wide range of problems. Mastering these skills opens doors to a world of problem-solving possibilities. But here's the most important takeaway: math isn't just about numbers and formulas; it's about thinking logically and creatively. It's about translating real-world situations into mathematical models and using those models to gain insights and make decisions. Gillian's book sale adventure was a fun example, but it's just the tip of the iceberg. The power of mathematics extends far beyond the classroom, shaping our understanding of the world and empowering us to solve problems in all areas of life. So, the next time you encounter a problem that seems daunting, remember the power of systems of equations. Break it down, define your variables, set up your equations, and solve with confidence. You might be surprised at what you can achieve! And who knows, maybe you'll even stumble upon your own literary treasure trove along the way.

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