Solving Jordan's System Of Equations A Quadratic And Linear Analysis

Introduction

In this article, we will delve into the system of equations that Jordan is attempting to solve: $y = 2x^2 + 3$ and $y - x = 6$. We will analyze the characteristics of these equations, explore the application of substitution as a solution method, and determine the true statements about Jordan's system. Our comprehensive analysis will provide a clear understanding of the system's properties and solution strategies. To effectively tackle this problem, we need to grasp the nuances of quadratic equations and systems of equations. Let’s begin by scrutinizing each equation individually and then explore how they interact within the system. Understanding quadratic equations and linear equations is fundamental to solving systems like Jordan's. Our goal is to provide a detailed explanation that not only answers the immediate questions but also enhances your grasp of algebraic concepts. So, let's dive deep into the world of equations and unravel the intricacies of this mathematical puzzle.

Analyzing the Quadratic Equation

The first equation in Jordan's system is $y = 2x^2 + 3$. This is a quadratic equation, which is characterized by the presence of a term with $x$ raised to the power of 2. The general form of a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants. In Jordan's equation, we can rewrite it in a form that highlights its quadratic nature. The equation $y = 2x^2 + 3$ represents a parabola when graphed. The coefficient of the $x^2$ term, which is 2 in this case, determines the parabola's direction and width. Since the coefficient is positive, the parabola opens upwards. The constant term, 3, represents the y-intercept of the parabola. In the context of the general quadratic form, we can rearrange Jordan's equation to $2x^2 + 0x + 3 - y = 0$. Here, $a = 2$, $b = 0$, and $c = 3 - y$. Understanding the coefficients is essential for analyzing the parabola's characteristics. The absence of an $x$ term (i.e., $b = 0$) indicates that the parabola's axis of symmetry is the y-axis. This simplifies the analysis of the vertex and overall shape of the curve. To determine if the quadratic equation is in standard form, we compare it to the standard form $ax^2 + bx + c = 0$. While $y = 2x^2 + 3$ is a valid representation of a parabola, it's not technically in the standard form because it's expressed as $y$ in terms of $x$ rather than being set to zero. To get it into standard form for solving, one might rearrange or substitute it into another equation. The key takeaway here is the recognition of the quadratic nature of the equation and its graphical representation as a parabola.

Examining the Linear Equation

The second equation in Jordan's system is $y - x = 6$. This equation is linear, meaning it represents a straight line when graphed. A linear equation can generally be written in the form $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. To better understand Jordan's linear equation, we can rearrange it into slope-intercept form: $y = x + 6$. From this form, we can easily identify the slope and y-intercept. The slope, $m$, is 1, indicating that the line rises one unit for every one unit it moves to the right. The y-intercept, $b$, is 6, meaning the line crosses the y-axis at the point (0, 6). Linear equations are straightforward to analyze and visualize. The slope and y-intercept provide all the necessary information to graph the line. In the context of a system of equations, the linear equation can intersect the quadratic equation (parabola) at zero, one, or two points. These points of intersection represent the solutions to the system. Understanding the properties of linear equations is crucial for solving systems, especially when combined with other types of equations like quadratics. The simplicity of the linear equation often makes it a convenient choice for substitution or elimination methods when solving systems. Therefore, recognizing the linear nature of $y - x = 6$ is the first step towards effectively finding the solution to Jordan's system.

Applying Substitution to Solve the System

The substitution method is a powerful technique for solving systems of equations, particularly when one equation can be easily expressed in terms of one variable. In Jordan's system, we have $y = 2x^2 + 3$ and $y - x = 6$. We've already rearranged the linear equation to $y = x + 6$, which makes it ideal for substitution. The core idea behind substitution is to replace one variable in one equation with its equivalent expression from the other equation. In this case, we can substitute the expression for $y$ from the linear equation ($y = x + 6$) into the quadratic equation ($y = 2x^2 + 3$). This yields the new equation: $x + 6 = 2x^2 + 3$. Now, we have a single equation with only one variable, $x$. This equation can be further rearranged into the standard quadratic form to facilitate solving: $2x^2 - x - 3 = 0$. Substitution simplifies the system by reducing it to a single-variable equation. This equation can then be solved using various methods, such as factoring, completing the square, or the quadratic formula. The solutions for $x$ can then be substituted back into either of the original equations to find the corresponding values of $y$. The process of applying substitution is a fundamental skill in algebra, and it's particularly effective when dealing with systems involving linear and quadratic equations. It allows us to transform a complex system into a manageable single equation, paving the way for finding the solutions.

Solving the Quadratic Equation After Substitution

After applying substitution, we arrived at the quadratic equation $2x^2 - x - 3 = 0$. To solve this equation, we can employ several methods, including factoring, completing the square, or using the quadratic formula. Let's explore factoring first, as it's often the quickest method when applicable. We look for two numbers that multiply to the product of the leading coefficient (2) and the constant term (-3), which is -6, and add up to the middle coefficient (-1). The numbers -3 and 2 satisfy these conditions. We can rewrite the middle term using these numbers: $2x^2 - 3x + 2x - 3 = 0$. Now, we factor by grouping: $x(2x - 3) + 1(2x - 3) = 0$. This simplifies to $(x + 1)(2x - 3) = 0$. Setting each factor equal to zero gives us the solutions for $x$: $x + 1 = 0$ or $2x - 3 = 0$. Solving these equations, we find $x = -1$ and x = rac{3}{2}. Factoring is an efficient method when the quadratic equation can be easily factored. However, if factoring is not straightforward, we can resort to the quadratic formula. The quadratic formula is given by: x = rac{-b extit{±} ext{\sqrt{b^2 - 4ac}}}{2a}, where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$. In our case, $a = 2$, $b = -1$, and $c = -3$. Plugging these values into the formula, we get the same solutions for $x$ as we did by factoring. The solutions for x are crucial, as they represent the x-coordinates of the points where the parabola and the line intersect. To find the corresponding y-coordinates, we substitute these x-values back into either the original linear or quadratic equation.

Finding the Solutions for y

Now that we have the solutions for $x$, which are $x = -1$ and x = rac{3}{2}, we need to find the corresponding values for $y$ to fully solve the system of equations. We can substitute these $x$-values into either the linear equation $y = x + 6$ or the quadratic equation $y = 2x^2 + 3$. The linear equation is generally simpler to work with, so let's use that. For $x = -1$, substituting into $y = x + 6$ gives us $y = -1 + 6 = 5$. So, one solution to the system is the point (-1, 5). For x = rac{3}{2}, substituting into $y = x + 6$ gives us y = rac{3}{2} + 6 = rac{3}{2} + rac{12}{2} = rac{15}{2}. Therefore, the other solution to the system is the point $\left(\frac{3}{2}, \frac{15}{2}\right)$. Substituting the x-values back into the equations allows us to find the corresponding y-values, thus completing the solution process. The solutions represent the points of intersection between the parabola and the line when graphed. These points satisfy both equations simultaneously. Finding the solutions for y is the final step in solving the system of equations. By substituting the x-values into the linear equation, we efficiently determined the corresponding y-values, providing us with the complete solutions to Jordan's system.

Determining the True Statements about Jordan's System

Now that we have thoroughly analyzed Jordan's system of equations, let's revisit the initial statements and determine which ones are true. The system consists of the equations $y = 2x^2 + 3$ and $y - x = 6$. We've identified the first equation as a quadratic equation representing a parabola and the second as a linear equation representing a straight line. We successfully applied the substitution method to solve the system, finding the solutions to be $(-1, 5)$ and $\left(\frac{3}{2}, \frac{15}{2}\right)$. Let's evaluate the statements:

1. The quadratic equation is in standard form: As discussed earlier, the standard form of a quadratic equation is $ax^2 + bx + c = 0$. While $y = 2x^2 + 3$ is a valid equation, it's not technically in standard form until it's rearranged to equal zero. Therefore, this statement is not entirely accurate.

2. Using substitution, the system of equations can be solved: We demonstrated the successful application of the substitution method to solve the system, so this statement is true. The true statements accurately reflect the properties and solutions of the system. Understanding the characteristics of each equation and the effectiveness of different solution methods is crucial for making these determinations. Our comprehensive analysis has equipped us with the knowledge to confidently identify the accurate statements about Jordan's system.

Conclusion

In conclusion, we have conducted a thorough analysis of Jordan's system of equations, which includes a quadratic equation $y = 2x^2 + 3$ and a linear equation $y - x = 6$. We explored the properties of each equation, applied the substitution method to solve the system, and found the solutions to be $(-1, 5)$ and $\left(\frac{3}{2}, \frac{15}{2}\right)$. Through this detailed examination, we were able to determine the true statements about the system. This exercise highlights the importance of understanding the characteristics of different types of equations and the various techniques available for solving systems. The key takeaways from this analysis include the ability to recognize quadratic and linear equations, apply the substitution method effectively, and interpret the solutions in the context of the system. By mastering these concepts, one can confidently tackle a wide range of algebraic problems and gain a deeper appreciation for the beauty and power of mathematics.