Solving For Initial Cheese Supply A Math Problem With Fractions

This is a classic word problem that combines fractions and basic algebra to determine an unknown quantity. In this case, we need to figure out the total amount of cheese Sarah had initially, given that she used a fraction of it for making pizza and another fraction for making lasagna. The key here is to translate the word problem into a mathematical equation that we can then solve. We will break down the problem step-by-step, identify the knowns and unknowns, set up an equation, and then solve for the unknown. By understanding the core concepts of fractions and algebraic equations, you'll be able to tackle similar problems with confidence. This type of problem is commonly encountered in middle school mathematics and serves as a foundational step for more complex algebraic concepts. The practical application of this problem is evident in everyday scenarios involving cooking, baking, and resource management. Solving this problem effectively requires careful reading, a clear understanding of fractions, and the ability to manipulate algebraic equations. It's not just about getting the right answer, but also about understanding the process and reasoning behind the solution. By mastering this type of problem, you'll not only improve your math skills but also enhance your problem-solving abilities in various real-world situations. Remember, practice is crucial in mathematics, and working through similar problems will solidify your understanding and build your confidence. Before we delve into the solution, let's consider the core concepts involved. We need to understand how to add fractions, how to convert mixed numbers into improper fractions, and how to solve a basic linear equation. These are the building blocks that will allow us to unravel the problem and arrive at the correct answer. So, let's embark on this mathematical journey with a clear mind and a focus on understanding every step of the way.

Step 1: Identify the Fractions of Cheese Used

In this crucial step, we will meticulously examine the problem statement to pinpoint the exact fractions representing the cheese Sarah utilized for her culinary creations. The problem clearly states that Sarah used $\frac{2}{3}$ of her total cheese supply for crafting a delicious pizza. This fraction, $\frac{2}{3}$, is a key piece of information that we will incorporate into our calculations later on. It represents a significant portion of the total cheese supply, and understanding its role is vital for solving the problem. Additionally, the problem mentions that Sarah used another fraction of her cheese supply, specifically $\frac{1}{9}$, to prepare a mouthwatering lasagna. This fraction, $\frac{1}{9}$, is equally important as it represents another portion of the total cheese supply that was consumed. By identifying these two fractions, $\frac{2}{3}$ and $\frac{1}{9}$, we are laying the foundation for the subsequent steps in our problem-solving process. These fractions will be used to determine the total fraction of cheese that Sarah used, which will then help us calculate the initial amount of cheese she had. The ability to accurately identify the relevant fractions from a word problem is a fundamental skill in mathematics. It requires careful reading and comprehension of the given information. Often, word problems are designed to test your understanding of mathematical concepts in a real-world context, and the first step in solving them is to extract the key numerical values and their relationships. This is precisely what we have done in this step by identifying the fractions of cheese used for pizza and lasagna. Moving forward, we will use these fractions to calculate the total fraction of cheese used. This will involve adding the two fractions together, which requires understanding the concept of finding a common denominator. Once we have the total fraction of cheese used, we can then relate it to the given amount of cheese in pounds, which will ultimately lead us to the answer. So, let's proceed to the next step with a clear understanding of the fractions we have identified and their significance in the problem.

Step 2: Calculate the Total Fraction of Cheese Used

Now that we've identified the individual fractions of cheese used for pizza ($\frac{2}{3}$) and lasagna ($\frac{1}{9}$), our next step is to determine the total fraction of cheese Sarah used in her cooking endeavors. To achieve this, we need to perform a simple addition operation, combining the two fractions. However, a crucial aspect of adding fractions is that they must share a common denominator. In our case, the denominators are 3 and 9. The least common multiple of 3 and 9 is 9, making it the ideal common denominator for our calculations. To convert $\frac{2}{3}$ to an equivalent fraction with a denominator of 9, we multiply both the numerator and the denominator by 3. This gives us $\frac{2 * 3}{3 * 3} = \frac{6}{9}$. Now we have two fractions with the same denominator: $\frac{6}{9}$ (equivalent to $\frac{2}{3}$) and $\frac{1}{9}$. We can now add these fractions together by adding their numerators while keeping the denominator the same: $\frac{6}{9} + \frac{1}{9} = \frac{6 + 1}{9} = \frac{7}{9}$. This result, $\frac{7}{9}$, represents the total fraction of Sarah's cheese supply that she used for making both pizza and lasagna. This is a significant milestone in our problem-solving journey because it allows us to relate the fraction of cheese used to the actual amount of cheese in pounds. The ability to add fractions with different denominators is a fundamental skill in mathematics and is frequently encountered in various contexts, both academic and real-world. Mastering this skill requires understanding the concept of equivalent fractions and finding the least common multiple. In this specific problem, we successfully converted the fractions to a common denominator and then added them together to find the total fraction of cheese used. This fraction, $\frac{7}{9}$, is a crucial link between the amount of cheese used and the total cheese supply. In the subsequent steps, we will use this fraction along with the given information about the amount of cheese used in pounds to calculate the total amount of cheese Sarah initially had. So, let's move forward with a clear understanding of the total fraction of cheese used and its significance in solving the problem.

Step 3: Convert the Mixed Number to an Improper Fraction

Moving forward, we encounter a mixed number, $2\frac{1}{3}$ pounds, which represents the total amount of cheese Sarah used. To effectively incorporate this value into our calculations, we need to convert it from a mixed number to an improper fraction. A mixed number combines a whole number and a fraction, while an improper fraction has a numerator that is greater than or equal to its denominator. The conversion process involves multiplying the whole number part of the mixed number by the denominator of the fractional part, and then adding the result to the numerator. The denominator remains the same. In our case, the mixed number is $2\frac{1}{3}$. The whole number part is 2, the numerator of the fractional part is 1, and the denominator is 3. Applying the conversion process, we get: (2 * 3) + 1 = 6 + 1 = 7. So, the new numerator is 7, and the denominator remains 3. Therefore, the improper fraction equivalent of $2\frac{1}{3}$ is $\frac{7}{3}$. This conversion is crucial because it allows us to perform algebraic operations more easily. Working with improper fractions in equations is often simpler than working with mixed numbers. In the context of our problem, converting the mixed number to an improper fraction makes it easier to relate the total fraction of cheese used (which we calculated in the previous step) to the total amount of cheese in pounds. The ability to convert between mixed numbers and improper fractions is a fundamental skill in mathematics. It's essential for performing various arithmetic operations, such as addition, subtraction, multiplication, and division, involving fractions. Understanding this conversion process ensures that we can manipulate numbers in different forms and apply them effectively in problem-solving scenarios. Now that we have converted the mixed number $2\frac{1}{3}$ to the improper fraction $\frac{7}{3}$, we have another key piece of information in the correct format for our calculations. In the next step, we will use this improper fraction along with the total fraction of cheese used ($\frac{7}{9}$) to set up an equation that will allow us to solve for the total amount of cheese Sarah initially had. So, let's proceed with this crucial conversion in mind and prepare to formulate our equation.

Step 4: Set Up the Equation

With the total fraction of cheese used calculated as $\frac{7}{9}$ and the amount of cheese used expressed as the improper fraction $\frac{7}{3}$ pounds, we are now in a position to formulate a mathematical equation that will help us determine the total amount of cheese Sarah initially had. Let's represent the unknown total amount of cheese by the variable 'x'. According to the problem statement, Sarah used $\frac{7}{9}$ of her total cheese supply, which is represented by 'x'. This amount is equal to $\frac{7}{3}$ pounds. Therefore, we can express this relationship as an equation: $\frac{7}{9} * x = \frac{7}{3}$. This equation is the cornerstone of our solution. It mathematically represents the information provided in the word problem and sets the stage for solving for the unknown variable 'x'. The equation states that $\frac{7}{9}$ of the total cheese (x) is equal to $\frac{7}{3}$ pounds. By solving this equation, we will find the value of 'x', which represents the total amount of cheese Sarah initially had. Setting up the correct equation is a critical step in solving any word problem. It requires a clear understanding of the relationships between the given quantities and the unknown quantity. In this case, we carefully translated the information about the fraction of cheese used and the amount of cheese used into a mathematical equation. The ability to translate word problems into equations is a fundamental skill in algebra and is essential for solving a wide range of mathematical problems. This skill involves identifying the key variables, constants, and relationships described in the problem and expressing them in mathematical notation. Now that we have successfully set up the equation, the next step is to solve it for 'x'. This will involve using algebraic techniques to isolate 'x' on one side of the equation and determine its value. So, let's proceed to the next step with our equation in hand and prepare to solve for the total amount of cheese Sarah initially had.

Step 5: Solve for the Unknown (x)

Having established the equation $\frac{7}{9} * x = \frac{7}{3}$, our focus now shifts to solving for the unknown variable 'x', which represents the total amount of cheese Sarah initially had. To isolate 'x' on one side of the equation, we need to undo the multiplication by $\frac{7}{9}$. The inverse operation of multiplication is division, but when dealing with fractions, dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of $\frac{7}{9}$ is $\frac{9}{7}$. Therefore, we will multiply both sides of the equation by $\frac{9}{7}$: ($\frac{7}{9}$ * x) * $\frac{9}{7}$ = $\frac{7}{3}$ * $\frac{9}{7}$. On the left side of the equation, the $\frac{7}{9}$ and $\frac{9}{7}$ cancel each other out, leaving us with just 'x': x = $\frac{7}{3}$ * $\frac{9}{7}$. On the right side of the equation, we can simplify by canceling out the common factor of 7 in the numerators and denominators: x = $\frac{1}{3}$ * 9. Now we multiply the remaining fractions: x = $\frac{9}{3}$. Finally, we simplify the fraction by dividing both the numerator and the denominator by 3: x = 3. Therefore, the value of 'x' is 3, which means Sarah initially had 3 pounds of cheese. Solving for an unknown variable in an equation is a fundamental skill in algebra. It involves using inverse operations to isolate the variable on one side of the equation and determine its value. In this case, we used the concept of reciprocals to undo the multiplication by a fraction and successfully solved for 'x'. The ability to manipulate equations and solve for unknowns is essential for a wide range of mathematical problems and applications. Now that we have found the value of 'x', we have the answer to our problem. Sarah initially had 3 pounds of cheese. In the next step, we will verify our solution to ensure its accuracy and completeness. So, let's proceed with confidence, knowing that we have successfully navigated the algebraic process and arrived at a potential solution.

Step 6: Verify the Solution

Having arrived at a solution of x = 3 pounds for the initial amount of cheese Sarah had, it's crucial to verify our answer to ensure its accuracy and consistency with the information provided in the problem statement. To verify, we will substitute the value of x (3 pounds) back into the original equation we established: $\frac{7}{9} * x = \frac{7}{3}$. Substituting x = 3, we get: $\frac{7}{9} * 3 = \frac{7}{3}$. Now, let's simplify the left side of the equation: $\frac{7}{9} * 3 = \frac{7 * 3}{9} = \frac{21}{9}$. We can further simplify $\frac{21}{9}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 3: $\frac{21}{9} = \frac{21 ÷ 3}{9 ÷ 3} = \frac{7}{3}$. So, the left side of the equation simplifies to $\frac{7}{3}$, which is exactly the same as the right side of the equation. This confirms that our solution of x = 3 pounds is correct. Verifying the solution is an essential step in the problem-solving process. It helps to catch any potential errors in our calculations or reasoning and ensures that our answer is consistent with the problem statement. In this case, by substituting our solution back into the original equation, we were able to confirm its accuracy. The process of verification reinforces our understanding of the problem and the steps we took to solve it. It also builds confidence in our problem-solving abilities. Now that we have verified our solution, we can confidently state that Sarah initially had 3 pounds of cheese. This completes our problem-solving journey, from identifying the key information to setting up an equation, solving for the unknown, and verifying the solution. By following this structured approach, we were able to successfully navigate the problem and arrive at the correct answer. In conclusion, the solution to the problem is that Sarah initially had 3 pounds of cheese. This corresponds to option A. This problem demonstrates the application of fractions and basic algebra in a practical scenario. By understanding the concepts and following a systematic approach, we can solve similar problems with confidence.