Solving Systems Of Equations Algebraically A Step-by-Step Guide

Understanding how to solve systems of equations algebraically is a fundamental skill in mathematics. This article delves into a step-by-step approach to solving systems of equations, using a specific example to illustrate the process. We will explore the initial steps, the underlying principles, and the significance of each action in finding the solution set. By the end of this guide, you will have a clear understanding of how to tackle similar problems effectively.

Understanding the System of Equations

Systems of equations are sets of two or more equations that share the same variables. The solution to a system of equations is the set of values for the variables that make all equations in the system true simultaneously. In other words, it's the point (or points) where the graphs of the equations intersect. Solving systems of equations is a core concept in algebra and has wide applications in various fields, including physics, engineering, economics, and computer science. Mastering the techniques to solve these systems is crucial for anyone pursuing further studies or careers in these areas.

The system of equations presented here is:

y = x^2 - x - 3
y = -3x + 5

This system consists of a quadratic equation ($y = x^2 - x - 3$) and a linear equation ($y = -3x + 5$). The solutions to this system will be the points where the parabola represented by the quadratic equation intersects the line represented by the linear equation. These points of intersection are the values of x and y that satisfy both equations simultaneously. Solving this system algebraically involves finding these specific values without relying on graphical methods, ensuring precision and accuracy in the solutions. This algebraic approach is particularly useful when the solutions are not easily discernible from a graph or when dealing with more complex equations where graphical methods might be cumbersome.

Step 1: Setting the Equations Equal

The first crucial step in solving this system of equations algebraically is to recognize that both equations are already solved for y. This allows us to use the substitution method effectively. Since both expressions are equal to y, they must also be equal to each other. This principle forms the basis of our initial step, where we set the two equations equal to one another. By equating the expressions, we create a new equation in terms of a single variable, x, which we can then solve. This step is essential as it transforms the system of two equations with two variables into a single equation with one variable, making the problem more manageable. The resulting equation captures the relationship between x values where the two original equations intersect, paving the way for finding the specific points of intersection. This algebraic manipulation is a powerful technique in solving systems of equations and is widely applicable across various mathematical problems.

Therefore, we set the right-hand sides of the equations equal:

x^2 - x - 3 = -3x + 5

This equation now represents the x-values where the y-values of both original equations are the same. Solving for x in this new equation will give us the x-coordinates of the intersection points of the parabola and the line. This step is a fundamental application of the transitive property of equality, which states that if a = b and b = c, then a = c. In this context, since both expressions are equal to y, they are also equal to each other, allowing us to form a new equation that simplifies the problem.

Step 2: Rearranging the Equation into Standard Quadratic Form

Now that we have an equation in terms of x alone, our next step is to rearrange it into the standard quadratic form. The standard quadratic form is given by:

ax^2 + bx + c = 0

where a, b, and c are constants. This form is essential because it allows us to easily identify the coefficients and use various methods to solve for x, such as factoring, completing the square, or using the quadratic formula. Rearranging the equation into this standard form is a critical step in solving quadratic equations algebraically, as it provides a structured approach to finding the roots. Without this form, it would be significantly more challenging to apply standard solution techniques. The process involves moving all terms to one side of the equation, leaving zero on the other side, and then combining like terms to simplify the equation into the standard quadratic format.

To rearrange our equation, we add 3x to both sides and subtract 5 from both sides:

x^2 - x - 3 + 3x - 5 = -3x + 5 + 3x - 5

This manipulation ensures that all terms are brought to one side of the equation, specifically the side with the $x^2$ term, which is a standard practice when dealing with quadratic equations. The goal is to consolidate all terms and simplify the equation, making it easier to identify the coefficients a, b, and c. Adding 3x to both sides cancels out the -3x term on the right side, and subtracting 5 from both sides cancels out the +5 term on the right side, effectively moving these terms to the left side of the equation. This step is a direct application of the addition and subtraction properties of equality, which state that adding or subtracting the same value from both sides of an equation maintains the equality.

Step 3: Simplifying the Quadratic Equation

After rearranging the equation, the next logical step is to simplify it. This involves combining like terms to reduce the equation to its simplest form. Simplifying the equation makes it easier to identify the coefficients a, b, and c in the standard quadratic form, which is crucial for applying methods like factoring or the quadratic formula. A simplified equation is less prone to errors in subsequent steps and provides a clearer picture of the quadratic relationship. The process of simplification typically involves adding or subtracting like terms, such as combining the x terms and the constant terms, to arrive at a more concise equation. This step is a fundamental aspect of algebraic manipulation and is essential for solving equations efficiently.

Combining like terms, we get:

x^2 + 2x - 8 = 0

Here, we combined the -x and +3x terms to get +2x, and we combined the -3 and -5 terms to get -8. This simplified quadratic equation is now in the standard form ($ax^2 + bx + c = 0$), where a = 1, b = 2, and c = -8. This form is ideal for solving the equation using various methods. The simplification process is a critical step because it reduces the complexity of the equation, making it easier to handle in the following steps. A simpler equation is less likely to lead to mistakes and makes the solution process more straightforward. This step ensures that we are working with the most manageable form of the quadratic equation before proceeding to solve for x.

Solving the Simplified Quadratic Equation

With the quadratic equation in the simplified standard form, $x^2 + 2x - 8 = 0$, we can now solve for x. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, factoring is a straightforward approach. Factoring involves expressing the quadratic equation as a product of two binomials. This method is efficient when the quadratic equation can be easily factored, as it directly leads to the solutions.

Factoring the Quadratic Equation

To factor the quadratic equation $x^2 + 2x - 8 = 0$, we look for two numbers that multiply to -8 (the constant term) and add to 2 (the coefficient of the x term). These numbers are 4 and -2. Therefore, we can factor the equation as follows:

(x + 4)(x - 2) = 0

This factorization breaks down the quadratic expression into two linear factors. The solutions for x can be found by setting each factor equal to zero. Factoring is a powerful method for solving quadratic equations because it provides a direct path to the solutions without the need for complex formulas or procedures. This method relies on 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. This property is the foundation of solving factored quadratic equations, allowing us to find the values of x that make the equation true.

Finding the Values of x

Now that we have factored the equation, we set each factor equal to zero and solve for x:

x + 4 = 0  or  x - 2 = 0

Solving these linear equations gives us the x-values:

x = -4  or  x = 2

These are the x-coordinates of the points where the parabola and the line intersect. These values represent the solutions for x that satisfy the original system of equations. Finding these x-values is a critical step in solving the system, as they provide the foundation for determining the corresponding y-values. These x-values are the roots of the quadratic equation and represent the x-intercepts of the parabola if it were graphed. Each x-value corresponds to a point where the parabola and the line intersect, making them crucial components of the solution set for the system of equations.

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 ($y = -3x + 5$) is often simpler to use for this purpose, but we can use either equation and should obtain the same y-values.

Substituting x = -4

Substituting x = -4 into the linear equation:

y = -3(-4) + 5
y = 12 + 5
y = 17

So, when x = -4, y = 17. This gives us one solution point: (-4, 17). This substitution directly applies the principle that the solutions to the system must satisfy both equations. By plugging the x-value into one of the equations, we can determine the corresponding y-value that makes the equation true. This process effectively finds the y-coordinate of the point where the parabola and the line intersect when x is -4. The result, y = 17, forms one part of the solution set, representing a specific point of intersection between the two equations. This step is essential for completing the solution, as it provides the y-coordinate to match the x-coordinate, giving a full picture of one of the points where the equations are simultaneously satisfied.

Substituting x = 2

Substituting x = 2 into the linear equation:

y = -3(2) + 5
y = -6 + 5
y = -1

So, when x = 2, y = -1. This gives us another solution point: (2, -1). This second substitution mirrors the first, ensuring that we find all possible intersection points between the two equations. By plugging x = 2 into the linear equation, we determine the corresponding y-value that satisfies both equations at this x-value. The result, y = -1, completes the second solution point, indicating another specific location where the parabola and the line intersect. This step is crucial for ensuring a comprehensive solution to the system of equations, as it identifies the second point where both equations hold true simultaneously. Together with the first solution point, this provides a complete picture of the solution set for the system.

The Solutions of the System of Equations

The solutions of the system of equations are the points (-4, 17) and (2, -1). These points represent the intersections of the parabola and the line. They are the values of x and y that satisfy both equations simultaneously. The solution set consists of these ordered pairs, which fully describe the points where the two equations intersect on a graph. Understanding the solution set is crucial, as it provides a complete and accurate representation of the points that make both equations true. These solutions are not just numbers; they are coordinates on a plane, representing specific locations where the two equations have the same x and y values.

Therefore, the correct answer is:

• (-4, 17) and (2, -1)

Conclusion

Solving systems of equations algebraically is a fundamental skill in mathematics. By following a step-by-step approach, we can systematically find the solutions that satisfy all equations in the system. In this example, we demonstrated how to solve a system of equations involving a quadratic and a linear equation by setting the equations equal, rearranging into standard quadratic form, simplifying, factoring, and finding the corresponding y-values. This methodical approach is applicable to various systems of equations and is a crucial tool for problem-solving in mathematics and related fields. Mastering these steps allows for accurate and efficient solutions, fostering a deeper understanding of algebraic principles and their applications. The ability to solve systems of equations is not just an academic skill but a valuable asset in many real-world scenarios, making it an essential component of mathematical literacy.