Solving Systems Of Equations Algebraically The First Two Steps
In mathematics, solving a system of equations is a fundamental skill with applications across various fields. A system of equations involves two or more equations with the same set of variables. The solution set of a system of equations represents the values of the variables that satisfy all equations simultaneously. One common method for finding the solution set is the algebraic approach, which involves manipulating the equations to isolate the variables and determine their values. This article will delve into the initial steps of solving a system of equations algebraically, focusing on a specific example to illustrate the process clearly. Let's explore the first two crucial steps in determining the solution set of the system of equations, y = x² - 2x - 3 and y = -x + 3, algebraically.
Understanding the Problem
Before diving into the steps, it's essential to understand the nature of the problem. We are given two equations:
- y = x² - 2x - 3 (a quadratic equation representing a parabola)
- y = -x + 3 (a linear equation representing a straight line)
The solution set we seek consists of the points (x, y) that satisfy both equations. Geometrically, these points represent the intersections of the parabola and the line. Algebraically, we aim to find the values of x and y that make both equations true.
Step 1: Setting the Equations Equal
The first crucial step in solving this system algebraically is to recognize that since both equations are solved for y, we can set them equal to each other. This is based on the principle that if two expressions are equal to the same quantity, they are equal to each other. In this case, both x² - 2x - 3 and -x + 3 are equal to y, so we can write:
x² - 2x - 3 = -x + 3
This step is significant because it eliminates one variable (y) and leaves us with a single equation in one variable (x). This equation now represents the x-coordinates of the intersection points of the parabola and the line. By solving this equation for x, we will find the x-values where the two graphs intersect. This transformation is a key technique in solving systems of equations, as it simplifies the problem into a more manageable form. The equation x² - 2x - 3 = -x + 3 is a quadratic equation, which we can solve using various methods, such as factoring, completing the square, or the quadratic formula. The next step will focus on rearranging this equation into a standard quadratic form to facilitate solving for x. This initial step of equating the expressions is a powerful technique that lays the foundation for finding the solution set of the system.
Step 2: Rearranging into Standard Quadratic Form
Having set the equations equal in Step 1, the next step is to rearrange the resulting equation into the standard form of a quadratic equation, which is ax² + bx + c = 0, where a, b, and c are constants. This form is essential because it allows us to apply various methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula.
To rearrange the equation x² - 2x - 3 = -x + 3, we need to move all terms to one side, leaving zero on the other side. We can achieve this by adding x and subtracting 3 from both sides of the equation:
x² - 2x - 3 + x - 3 = -x + 3 + x - 3
Simplifying this, we get:
x² - x - 6 = 0
This equation is now in the standard quadratic form, with a = 1, b = -1, and c = -6. This form is crucial for solving the equation because it allows us to identify the coefficients needed for the quadratic formula or to factor the quadratic expression. Factoring involves finding two binomials that multiply to give the quadratic expression, while the quadratic formula provides a direct solution for x in terms of a, b, and c. Completing the square is another method that can be used, which involves manipulating the equation to create a perfect square trinomial. By rearranging the equation into the standard quadratic form, we have set the stage for applying these techniques to find the values of x that satisfy the equation, which will then lead us to the complete solution set of the system of equations. This step is a critical bridge between the initial equation and the final solutions, making it a vital part of the algebraic solution process.
Solving the Quadratic Equation
With the quadratic equation in standard form (x² - x - 6 = 0), we can now proceed to solve for x. There are several methods to accomplish this, including factoring, completing the square, and using the quadratic formula. In this case, factoring is a straightforward approach.
Factoring
Factoring involves finding two binomials that multiply to give the quadratic expression. We look for two numbers that multiply to c (-6) and add up to b (-1). These numbers are -3 and 2. Therefore, we can factor the quadratic equation as follows:
(x - 3)(x + 2) = 0
For the product of two factors to be zero, at least one of them must be zero. This gives us two possible solutions for x:
- x - 3 = 0 => x = 3
- x + 2 = 0 => x = -2
The Quadratic Formula
Alternatively, we could have used the quadratic formula, which is a general solution for any quadratic equation in the form ax² + bx + c = 0:
x = (-b ± √(b² - 4ac)) / (2a)
In our case, a = 1, b = -1, and c = -6. Plugging these values into the formula, we get:
x = (1 ± √((-1)² - 4 * 1 * -6)) / (2 * 1) x = (1 ± √(1 + 24)) / 2 x = (1 ± √25) / 2 x = (1 ± 5) / 2
This gives us the same two solutions:
- x = (1 + 5) / 2 = 3
- x = (1 - 5) / 2 = -2
Finding the Corresponding y-values
Now that we have the x-values, we need to find the corresponding y-values to complete the solution set. We can do this by substituting each x-value back into either of the original equations. The linear equation y = -x + 3 is simpler, so we'll use that.
For x = 3
y = -3 + 3 = 0
So, one solution is (3, 0).
For x = -2
y = -(-2) + 3 = 2 + 3 = 5
So, the other solution is (-2, 5).
The Solution Set
Therefore, the solution set for the system of equations is:
{(3, 0), (-2, 5)}
These are the points where the parabola and the line intersect on the graph.
Conclusion
In this article, we have demonstrated the first two steps in solving a system of equations algebraically, using the example of y = x² - 2x - 3 and y = -x + 3. These steps involve setting the equations equal to each other and rearranging the resulting equation into the standard quadratic form. These steps are crucial for simplifying the problem and setting the stage for solving for the variables. By understanding and mastering these steps, you can effectively solve a wide range of systems of equations. The process of solving the quadratic equation and finding the corresponding y-values further illustrates the complete algebraic approach, providing a comprehensive understanding of how to determine the solution set for a system of equations. The solution set, consisting of the points (3, 0) and (-2, 5), represents the intersection points of the parabola and the line, thus satisfying both equations simultaneously. This algebraic method provides a precise and reliable way to find these solutions, making it an essential tool in mathematics and its applications.
