Solving Equations And Total Marks Expression A Mathematical Guide

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In this comprehensive guide, we'll dive into the process of solving a system of linear equations. We'll tackle the specific system: 3y = 15x - 40 and 3y = 114. Linear equations are a fundamental concept in mathematics, and mastering the techniques to solve them is crucial for various applications. This detailed explanation aims to provide a clear understanding of the steps involved, making it easy for anyone to follow, regardless of their mathematical background. We'll explore the substitution method, which is particularly effective in solving systems like this one. By the end of this guide, you'll be equipped with the knowledge and skills to confidently solve similar problems. We will break down each step, ensuring clarity and comprehension. Let's embark on this mathematical journey together and unlock the solution to this intriguing system of equations. The goal is to provide not just the answer, but also the understanding behind it. This is the key to building a strong foundation in mathematics.

Step-by-Step Solution

Our journey begins with the given system of equations. The first equation is 3y = 15x - 40, a linear equation that relates two variables, x and y. The second equation, 3y = 114, is a simpler equation that directly gives us the value of 3y. This simplicity is our starting point, a stepping stone that allows us to unravel the values of x and y. The beauty of mathematics lies in its structured approach, where each step builds upon the previous one. In this case, we'll use the value of 3y from the second equation to substitute into the first equation. This process of substitution is a powerful technique in solving simultaneous equations. It's like fitting puzzle pieces together, where each piece of information guides us closer to the solution. So, let's take the value from the second equation and carefully place it into the first, setting the stage for the next step in our mathematical adventure. This systematic approach is what makes problem-solving in mathematics so rewarding and effective. We're not just finding numbers; we're uncovering the underlying relationships between them.

1. Isolate 3y in Both Equations

We are presented with the system of equations: 3y = 15x - 40 and 3y = 114. The first equation, 3y = 15x - 40, shows a direct relationship between 3y and 15x. The term 3y is already isolated on the left side, making it straightforward to work with. The constant -40 plays a key role in defining this relationship, shifting the line on the graph and influencing the values of x and y. The second equation, 3y = 114, is even simpler. It explicitly states the value of 3y. This is a crucial piece of information because it allows us to directly substitute this value into the first equation. The simplicity of this equation is a gift, a shortcut that simplifies the entire solving process. It's like finding a key that unlocks a door, making the path to the solution much clearer and faster. We can appreciate the elegance of mathematical structures here, where a simple equation can hold the key to solving a more complex one. So, with 3y isolated in both equations, we're perfectly positioned to move on to the next stage of our solution.

2. Substitute the Value of 3y

Now comes the critical step of substitution. We know from the second equation that 3y = 114. This knowledge is our leverage, allowing us to replace 3y in the first equation with the value 114. So, in the equation 3y = 15x - 40, we replace 3y with 114, transforming the equation into 114 = 15x - 40. This act of substitution is like plugging a known value into a formula, simplifying the equation and bringing us closer to finding the unknown. It's a powerful technique in algebra, allowing us to reduce the number of variables and solve for them one by one. By substituting, we've effectively eliminated y from the equation, leaving us with an equation in terms of x only. This is a significant advancement because we can now focus solely on solving for x. The substituted equation, 114 = 15x - 40, is a stepping stone, a bridge that connects the known value of 3y to the unknown value of x. This methodical approach of substitution highlights the elegance and efficiency of algebraic techniques.

3. Solve for x

With the substituted equation 114 = 15x - 40, our focus shifts to isolating x. Our mission is to get x by itself on one side of the equation. The first step in this isolation process is to eliminate the constant term -40 from the right side. To do this, we apply the fundamental principle of algebraic manipulation: we perform the same operation on both sides of the equation. So, we add 40 to both sides, transforming the equation. Adding 40 to both sides maintains the balance of the equation, ensuring that the equality remains valid. This is like adjusting the weights on a balance scale, keeping the scale level as we add or remove weights. The equation becomes 114 + 40 = 15x, which simplifies to 154 = 15x. Now, 15x is isolated on the right side, but x is still multiplied by 15. To isolate x completely, we need to undo this multiplication. We do this by dividing both sides of the equation by 15. This division is the final step in isolating x, freeing it from the coefficient 15. Dividing both sides by 15 gives us x = 154/15. This fraction represents the value of x. We've successfully navigated the algebraic steps, isolating x and determining its value. It's a testament to the power of algebraic manipulation, where we can systematically transform equations to reveal the values of the unknowns.

4. Substitute x to Find y

Now that we've triumphantly found the value of x as 154/15, the next phase of our mathematical journey is to determine the value of y. We'll leverage the simplicity of the second original equation, 3y = 114, for this purpose. This equation is our shortcut, a direct route to finding y without the need for further substitution of x. The beauty of this equation lies in its directness; it already isolates 3y, making it a perfect tool for finding y. To solve for y, we need to isolate y by itself. Currently, y is multiplied by 3. To undo this multiplication, we apply the inverse operation: division. We divide both sides of the equation by 3, maintaining the balance and equality of the equation. Dividing both sides by 3 transforms the equation into y = 114/3. This division is the key step in unlocking the value of y. Performing the division, we find that y = 38. This is the solution for y, the numerical value that satisfies the original system of equations. We've successfully navigated through the equations, found the value of x, and now determined the value of y. This methodical approach, step by step, has allowed us to solve for both variables, showcasing the power and precision of algebraic techniques.

Solution

Through our meticulous step-by-step process, we have successfully navigated the system of equations and arrived at the solution. The value of x is 154/15, and the value of y is 38. This solution represents the point where the two linear equations intersect, the single pair of values for x and y that satisfy both equations simultaneously. We can express this solution as an ordered pair, (154/15, 38), which pinpoints the exact location on a graph where the lines represented by the two equations meet. This ordered pair is the culmination of our efforts, the final answer that we sought. It's a testament to the power of algebraic techniques, where we can systematically manipulate equations to reveal the values of the unknowns. The solution is not just a set of numbers; it's a representation of the relationship between the two equations, a point of intersection that defines their connection. This understanding of solutions as intersections is a fundamental concept in algebra and provides a visual way to interpret the results. So, with x equal to 154/15 and y equal to 38, we have successfully solved the system of equations, demonstrating the elegance and effectiveness of mathematical problem-solving.

Now, let's shift our focus to a different kind of problem, one that involves creating an algebraic expression to represent a real-world scenario. This problem revolves around a contest where learners have scored marks, and we need to write an expression for their total marks based on given conditions. Translating word problems into algebraic expressions is a crucial skill in mathematics. It allows us to represent complex situations with concise mathematical notation, making it easier to analyze and solve them. In this case, we're given information about the scores of three learners relative to each other. We'll need to carefully translate these relationships into algebraic terms, using variables to represent the unknown scores. This process of translation is like deciphering a code, where we convert English phrases into mathematical symbols. The goal is to create an expression that accurately represents the total marks scored by the three learners, capturing the relationships between their scores. By the end of this exercise, we'll not only have the expression but also a deeper understanding of how algebra can be used to model real-world situations. This skill is invaluable in various fields, from science and engineering to economics and finance. Let's embark on this journey of translating words into algebra and create the expression that captures the total marks in the contest.

Defining the Variables

The foundation of any algebraic expression lies in clearly defining the variables. In this problem, we're dealing with the marks scored by three learners. Let's denote the marks scored by the third learner as x. This choice is strategic because the scores of the other two learners are given in relation to the third learner's score. By assigning x to the third learner's marks, we create a base from which we can easily express the other scores. This is like setting a reference point, a fixed anchor that helps us navigate the relationships between the scores. Now, let's move on to the first learner. We're told that the first learner scored 14 marks more than the third learner. Since the third learner scored x marks, the first learner's score can be represented as x + 14. This expression captures the direct relationship between the scores of the first and third learners. The addition of 14 to x reflects the