Solving Equations With Given Solutions A Comprehensive Guide

This comprehensive guide delves into the fascinating realm of solving equations when specific solutions are provided. We will explore two distinct scenarios, each requiring a unique approach to determine the unknown variables. Our journey will begin with a linear equation involving two variables, x and y, where the solution is expressed in terms of another variable, a. The challenge lies in finding the value of a that satisfies the equation. Subsequently, we will tackle another equation, this time with a solution provided as numerical values for x and y. Our mission here is to uncover the possible values of the unknown variable a within the equation.

Finding the Value of 'a' in a Linear Equation

In this section, we embark on a quest to determine the value of the elusive variable 'a' within the context of a linear equation. We are presented with a scenario where the variables x and y are elegantly expressed in terms of a: x = 2a + 1 and y = a - 1. These expressions act as a bridge, connecting the variables x and y to the unknown a. Furthermore, we are given a linear equation, 2x - 3y + 5 = 0, which serves as the stage upon which our variables will interact. The key to unlocking the value of a lies in the realization that the given expressions for x and y represent a solution to this equation. This means that when we substitute these expressions into the equation, the equation must hold true. Our strategy, therefore, is to perform this substitution, replacing x and y with their respective expressions in terms of a. This substitution will transform the equation into a form where a is the sole unknown, allowing us to manipulate the equation and isolate a, thereby revealing its value. Let's embark on this algebraic journey, carefully substituting and simplifying, until we arrive at the solution for a. This process of substitution is a fundamental technique in algebra, allowing us to connect different parts of a problem and ultimately solve for the unknowns. The elegance of this method lies in its ability to transform a seemingly complex problem into a more manageable one, where the solution can be obtained through systematic steps. By replacing the variables with their expressions, we create a new equation that encapsulates the relationships between the variables and the unknown, paving the way for us to find the value of 'a'.

Step-by-Step Solution

Let's delve into the step-by-step solution to find the value of a. Our starting point is the equation 2x - 3y + 5 = 0, and the expressions x = 2a + 1 and y = a - 1. The first crucial step is to substitute these expressions for x and y into the equation. This yields the following: 2(2a + 1) - 3(a - 1) + 5 = 0. Now, we must carefully simplify this equation. We begin by distributing the constants: 4a + 2 - 3a + 3 + 5 = 0. Next, we combine like terms, grouping the terms with a together and the constant terms together: (4a - 3a) + (2 + 3 + 5) = 0. This simplifies to a + 10 = 0. Finally, to isolate a, we subtract 10 from both sides of the equation, resulting in a = -10. Therefore, the value of a that satisfies the given conditions is -10. This step-by-step approach highlights the importance of careful algebraic manipulation. Each step, from substitution to simplification, is crucial in ensuring the accuracy of the final result. The ability to perform these operations with precision is a cornerstone of mathematical problem-solving, and it is through such systematic approaches that we can unravel the solutions to even the most intricate equations. The solution, a = -10, is not just a numerical answer; it represents a specific value that, when plugged back into the original expressions for x and y, will satisfy the given linear equation. This underscores the interconnectedness of the variables and the importance of finding the correct value to maintain the balance of the equation.

Determining Possible Values of 'a' with Numerical Solutions

Now, we shift our focus to a different type of problem, one where we are given specific numerical values for x and y that satisfy an equation, and our task is to determine the possible values of the unknown variable a. In this scenario, we are told that x = 1 and y = 6 is a solution to the equation 8x - ay + a² = 0. Unlike the previous problem, where x and y were expressed in terms of a, here they are given as concrete numbers. This simplifies the initial substitution step, as we can directly replace x and y with their numerical values in the equation. However, the challenge now lies in the nature of the equation itself. Notice that the equation contains a term with a², which means it is a quadratic equation in a. Quadratic equations can have up to two distinct solutions, reflecting the fact that there might be two different values of a that would make the equation true when x = 1 and y = 6. Our strategy, therefore, is to substitute the given values of x and y into the equation, which will transform it into a quadratic equation in a. We will then employ the techniques of solving quadratic equations, such as factoring, completing the square, or using the quadratic formula, to find the possible values of a. Understanding the nature of the equation – in this case, recognizing it as a quadratic – is crucial in choosing the appropriate solution method. The presence of the a² term signals that we need to consider the possibility of multiple solutions, and the techniques for solving quadratic equations provide us with the tools to uncover these solutions.

Solving the Quadratic Equation

Let's proceed with solving the quadratic equation to find the possible values of a. We begin by substituting x = 1 and y = 6 into the equation 8x - ay + a² = 0. This yields: 8(1) - a(6) + a² = 0, which simplifies to 8 - 6a + a² = 0. To make it easier to work with, let's rearrange the terms to write the equation in the standard quadratic form: a² - 6a + 8 = 0. Now, we need to solve this quadratic equation. One common method is factoring. We look for two numbers that multiply to 8 and add up to -6. These numbers are -2 and -4. Thus, we can factor the quadratic as follows: (a - 2)(a - 4) = 0. For this product to be zero, at least one of the factors must be zero. Therefore, we have two possibilities: a - 2 = 0 or a - 4 = 0. Solving these linear equations, we get a = 2 or a = 4. These are the two possible values of a that satisfy the given conditions. The factoring method, as demonstrated here, is a powerful technique for solving quadratic equations, especially when the roots are integers. It allows us to break down the quadratic into simpler linear factors, making it easier to find the solutions. However, it's important to remember that not all quadratic equations can be easily factored. In such cases, other methods like completing the square or the quadratic formula may be necessary. The two solutions, a = 2 and a = 4, highlight the fact that quadratic equations can have multiple solutions. Each of these values, when substituted back into the original equation along with x = 1 and y = 6, will make the equation true. This emphasizes the importance of considering all possible solutions when dealing with quadratic equations.

Conclusion

In this comprehensive guide, we have explored the methods for solving equations when given specific solutions. We tackled two distinct scenarios, each requiring a tailored approach. In the first scenario, we successfully determined the value of a in a linear equation by employing the technique of substitution and algebraic manipulation. By carefully replacing the variables x and y with their expressions in terms of a, we transformed the equation into a form where a was the sole unknown, allowing us to isolate and solve for its value. This process highlighted the importance of substitution as a fundamental tool in algebra, enabling us to connect different parts of a problem and ultimately find the unknowns. The second scenario presented us with a quadratic equation, where we were tasked with finding the possible values of a given numerical solutions for x and y. We substituted the given values into the equation, transforming it into a quadratic equation in a. We then employed the factoring method to find the two possible values of a that satisfied the conditions. This exploration emphasized the significance of recognizing the nature of the equation (linear vs. quadratic) and choosing the appropriate solution method. Furthermore, it underscored the fact that quadratic equations can have multiple solutions, and it's crucial to consider all possibilities. The techniques and strategies discussed in this guide are fundamental to solving a wide range of mathematical problems. The ability to manipulate equations, substitute values, and solve for unknowns is a cornerstone of mathematical proficiency. By mastering these skills, you can confidently tackle more complex problems and delve deeper into the fascinating world of mathematics. Remember, practice is key to solidifying these concepts. The more you engage with these types of problems, the more adept you will become at identifying the appropriate solution strategies and executing them with precision. Mathematics is a journey of discovery, and each problem solved is a step forward in your understanding and appreciation of this beautiful and powerful subject.