Solving Systems Of Equations A Step-by-Step Guide

Introduction

In the realm of mathematics, solving systems of equations stands as a fundamental skill with wide-ranging applications. From determining the intersection points of curves to modeling real-world scenarios, the ability to find solutions that satisfy multiple equations simultaneously is crucial. This article delves into a specific example of solving a system of equations, providing a step-by-step guide and exploring the underlying concepts. In this context, we will be focusing on a system containing a quadratic equation and a linear equation. These types of systems often arise in various fields, including physics, engineering, and economics. Mastering the techniques to solve them is essential for anyone pursuing these disciplines. The solution to a system of equations represents the set of points where the graphs of the equations intersect. For a system involving a quadratic and a linear equation, there can be zero, one, or two intersection points. These points, when expressed as coordinate pairs (x, y), constitute the solution set of the system. The process of solving involves finding these x and y values that satisfy both equations. Understanding the nature of the equations involved is the first step towards devising a solution strategy. In our case, we have a quadratic equation, which represents a parabola, and a linear equation, which represents a straight line. The solutions will correspond to the points where the line intersects the parabola. This article aims to provide a clear and concise explanation of how to solve such systems, equipping readers with the necessary tools to tackle similar problems. By the end of this guide, you will be able to confidently approach systems of equations involving quadratic and linear expressions and determine their solutions.

Problem Statement

Let's consider the following system of equations:

$ y = x^2 - 2x - 19 $$ y + 4x = 5 $

The objective is to find the pair of points (x, y) that satisfy both equations simultaneously. These points represent the solution set of the system. This means that if we substitute the x and y values of these points into both equations, the equations will hold true. To solve this system, we will employ a common algebraic technique known as substitution. This method involves solving one equation for one variable and then substituting that expression into the other equation. This process reduces the system to a single equation with a single variable, which can then be solved using standard algebraic methods. In our specific case, we can observe that the second equation, y + 4x = 5, can be easily solved for y. This will allow us to express y in terms of x, which we can then substitute into the first equation. This substitution will result in a quadratic equation in x, which we can solve using factoring, the quadratic formula, or completing the square. The solutions for x will then be used to find the corresponding y values. The solution set will consist of the (x, y) pairs that satisfy both equations. This problem provides a good illustration of how to combine algebraic techniques to solve systems of equations. It also highlights the importance of understanding the relationships between different types of equations, such as quadratic and linear equations. By carefully applying the substitution method and solving the resulting equations, we can determine the solution set of the system.

Step-by-Step Solution

Step 1 Solve the linear equation for y

We begin by isolating y in the second equation:

$ y + 4x = 5 $$ y = 5 - 4x $

This step is crucial as it allows us to express y in terms of x, which is essential for the substitution method. By subtracting 4x from both sides of the equation, we effectively isolate y, obtaining the expression y = 5 - 4x. This expression now represents the relationship between y and x as defined by the linear equation. We will use this expression in the next step to substitute for y in the quadratic equation. The goal here is to reduce the system of two equations into a single equation with only one variable, making it easier to solve. The substitution method is a powerful technique for solving systems of equations, and this first step lays the groundwork for its successful application. By carefully isolating y, we have created an expression that will allow us to eliminate y from the system and solve for x. This step demonstrates the importance of algebraic manipulation in solving mathematical problems. The ability to rearrange equations and isolate variables is a fundamental skill that is used extensively in various areas of mathematics and science.

Step 2 Substitute the expression for y into the quadratic equation

Now, we substitute $ 5 - 4x $ for y in the first equation:

$ 5 - 4x = x^2 - 2x - 19 $

This substitution is the heart of the method, allowing us to eliminate one variable and create an equation in terms of only one variable. By replacing y with its equivalent expression in terms of x, we have effectively transformed the system of two equations into a single equation that we can solve for x. The resulting equation is a quadratic equation, which we can solve using various techniques such as factoring, the quadratic formula, or completing the square. This step highlights the power of substitution in simplifying complex problems. By strategically replacing one variable with its equivalent expression, we can reduce the complexity of the system and make it more manageable. The equation we have obtained now represents the intersection points of the parabola and the line. Solving this equation will give us the x-coordinates of these intersection points. This substitution is a critical step in solving the system of equations and demonstrates the interconnectedness of algebraic expressions. By understanding how to manipulate and substitute expressions, we can solve a wide range of mathematical problems.

Step 3 Simplify and rearrange the equation

Move all terms to one side to set the equation to zero:

$ 0 = x^2 + 2x - 24 $

This step is essential for solving quadratic equations. By setting the equation to zero, we can utilize techniques such as factoring or the quadratic formula to find the solutions for x. Rearranging the terms involves adding 4x and subtracting 5 from both sides of the equation. This process ensures that all terms are on one side, leaving zero on the other side. The resulting equation is a standard quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = 2, and c = -24. This form is particularly useful for applying the quadratic formula or attempting to factor the equation. The goal of this step is to transform the equation into a form that is conducive to solving for the unknown variable. By carefully manipulating the terms, we have created an equation that can be readily solved using standard algebraic techniques. This step demonstrates the importance of algebraic manipulation in simplifying equations and preparing them for further analysis. The ability to rearrange equations and set them to zero is a fundamental skill in mathematics and is used extensively in various problem-solving scenarios.

Step 4 Factor the quadratic equation

Factor the quadratic:

$ 0 = (x + 6)(x - 4) $

Factoring the quadratic equation is a key step in finding the solutions for x. By expressing the quadratic as a product of two binomials, we can easily identify the values of x that make the equation equal to zero. In this case, the quadratic factors into (x + 6)(x - 4). This means that the equation will be true if either (x + 6) = 0 or (x - 4) = 0. Factoring is a powerful technique for solving quadratic equations, but it is not always possible. In some cases, the quadratic may not factor easily, and other methods such as the quadratic formula may be necessary. The ability to factor quadratic equations is a valuable skill in algebra and is used extensively in various mathematical applications. This step demonstrates the importance of recognizing patterns and applying appropriate algebraic techniques to solve problems. By successfully factoring the quadratic, we have simplified the process of finding the solutions for x. This step highlights the interconnectedness of different algebraic concepts and the importance of mastering fundamental skills.

Step 5 Solve for x

Set each factor equal to zero and solve for x:

$ x + 6 = 0 => x = -6 $$ x - 4 = 0 => x = 4 $

This step utilizes the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. By setting each factor equal to zero, we can find the values of x that make the equation true. This gives us two possible solutions for x: x = -6 and x = 4. These values represent the x-coordinates of the points where the parabola and the line intersect. Solving for x is a critical step in finding the solution set of the system of equations. These solutions will be used in the next step to find the corresponding y-values. The zero-product property is a fundamental concept in algebra and is used extensively in solving various types of equations. By applying this property, we have successfully found the x-values that satisfy the quadratic equation. This step demonstrates the importance of understanding mathematical principles and applying them to solve problems. The ability to solve for x is a crucial skill in algebra and is essential for solving systems of equations and other mathematical problems.

Step 6 Substitute the values of x back into the linear equation to find the corresponding y values

For $ x = -6 $:

$ y = 5 - 4(-6) = 5 + 24 = 29 $

For $ x = 4 $:

$ y = 5 - 4(4) = 5 - 16 = -11 $

Substituting the x-values back into the linear equation is the final step in finding the solution set of the system of equations. By plugging in each x-value, we can find the corresponding y-value that satisfies both equations. This gives us the complete coordinates of the points where the parabola and the line intersect. For x = -6, we find that y = 29, and for x = 4, we find that y = -11. These (x, y) pairs represent the solutions to the system of equations. This step demonstrates the importance of carefully substituting values and performing calculations to arrive at the correct answer. The solutions we have found represent the points where the graphs of the two equations intersect. This step highlights the interconnectedness of different algebraic concepts and the importance of verifying solutions. By substituting the x-values, we have successfully found the corresponding y-values and completed the solution set of the system of equations. This final step reinforces the importance of accuracy and attention to detail in mathematical problem-solving.

Step 7 Write the solution set

The solution set is:

$ (-6, 29) $ and $ (4, -11) $

This step is the culmination of the entire solution process. We have successfully found the two points that satisfy both equations in the system. These points, (-6, 29) and (4, -11), represent the intersection points of the parabola and the line. The solution set is typically written as a set of ordered pairs, where each pair represents a solution. This step emphasizes the importance of clearly presenting the final answer in a concise and organized manner. The solution set represents the complete solution to the problem and provides the values of x and y that satisfy both equations simultaneously. By writing the solution set, we have effectively communicated the answer to the problem. This step demonstrates the importance of mathematical communication and the ability to express solutions clearly and accurately. The solution set is the ultimate goal of solving a system of equations, and this step represents the successful completion of the problem-solving process. The ability to find and express the solution set is a crucial skill in mathematics and is essential for various applications in science, engineering, and other fields.

Final Answer

The pair of points representing the solution set of this system are (-6, 29) and (4, -11).

Conclusion

In conclusion, solving systems of equations is a fundamental skill in mathematics with broad applications. This article has demonstrated a step-by-step approach to solving a system consisting of a quadratic and a linear equation using the substitution method. By carefully isolating variables, substituting expressions, and solving the resulting equations, we were able to find the solution set of the system. The key steps involved solving the linear equation for one variable, substituting that expression into the quadratic equation, simplifying and rearranging the equation, factoring the quadratic, solving for the unknown variable, and substituting the values back into the linear equation to find the corresponding values. The solution set represents the points where the graphs of the equations intersect, providing valuable insights into the relationships between the equations. Mastering these techniques is essential for success in various mathematical and scientific disciplines. The ability to solve systems of equations is a valuable skill that empowers individuals to tackle complex problems and make informed decisions. By understanding the underlying concepts and practicing problem-solving techniques, anyone can become proficient in solving systems of equations. This skill is not only applicable in academic settings but also in real-world scenarios where mathematical modeling is used to analyze and solve problems. The process of solving systems of equations fosters critical thinking, problem-solving skills, and a deeper understanding of mathematical relationships. This article serves as a comprehensive guide to solving systems of equations involving quadratic and linear expressions, equipping readers with the necessary tools to confidently approach similar problems. By following the steps outlined in this article, readers can develop a solid foundation in solving systems of equations and enhance their mathematical abilities.