The First Two Steps In Determining The Solution Set Of The System Of Equations, Y = X 2 − 2 X − 3 Y = X^2 - 2x - 3 Y = X 2 − 2 X − 3 And Y = − X + 3 Y = -x + 3 Y = − X + 3 , Algebraically Are Shown In The Table.[\begin{tabular}{|c|c|}\hlineStep & Equation \\hlineStep 1 & $x^2 - 2x - 3

Introduction

Solving a system of equations is a fundamental concept in algebra, and it requires a step-by-step approach to determine the solution set. In this article, we will focus on the first two steps in determining the solution set of the system of equations, $y = x^2 - 2x - 3$ and $y = -x + 3$, algebraically.

Understanding the System of Equations

A system of equations is a set of two or more equations that contain the same variables. In this case, we have two equations:

1. $y = x^2 - 2x - 3$
2. $y = -x + 3$

To solve this system of equations, we need to find the values of $x$ and $y$ that satisfy both equations simultaneously.

Step 1: Setting the Equations Equal to Each Other

The first step in solving a system of equations is to set the equations equal to each other. This is done by equating the two equations, which gives us:

$x^2 - 2x - 3 = -x + 3$

This equation is obtained by substituting the expression for $y$ from the first equation into the second equation.

Step 2: Simplifying the Equation

The next step is to simplify the equation obtained in Step 1. We can do this by combining like terms and rearranging the equation to get:

$x^2 - 3x - 6 = 0$

This equation is a quadratic equation, and it can be solved using various methods such as factoring, completing the square, or using the quadratic formula.

Solving the Quadratic Equation

To solve the quadratic equation $x^2 - 3x - 6 = 0$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = -3$, and $c = -6$. Plugging these values into the quadratic formula, we get:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-6)}}{2(1)}$

Simplifying this expression, we get:

$x = \frac{3 \pm \sqrt{9 + 24}}{2}$

$x = \frac{3 \pm \sqrt{33}}{2}$

This gives us two possible values for $x$: $x = \frac{3 + \sqrt{33}}{2}$ and $x = \frac{3 - \sqrt{33}}{2}$.

Conclusion

In this article, we have discussed the first two steps in determining the solution set of a system of equations. We have set the equations equal to each other and simplified the resulting equation to obtain a quadratic equation. We have then solved the quadratic equation using the quadratic formula to obtain two possible values for $x$. In the next article, we will discuss the final steps in determining the solution set of the system of equations.

References

• [1] "Algebra and Trigonometry" by Michael Sullivan
• [2] "College Algebra" by James Stewart

Table of Contents

1. Introduction
2. Understanding the System of Equations
3. Step 1: Setting the Equations Equal to Each Other
4. Step 2: Simplifying the Equation
5. Solving the Quadratic Equation
6. Conclusion
7. References
8. Table of Contents
   Frequently Asked Questions (FAQs) about Solving a System of Equations ====================================================================

Introduction

Solving a system of equations can be a challenging task, especially for those who are new to algebra. In this article, we will address some of the most frequently asked questions about solving a system of equations.

Q: What is a system of equations?

A system of equations is a set of two or more equations that contain the same variables. In this case, we have two equations:

1. $y = x^2 - 2x - 3$
2. $y = -x + 3$

Q: How do I solve a system of equations?

To solve a system of equations, you need to find the values of $x$ and $y$ that satisfy both equations simultaneously. You can do this by setting the equations equal to each other and simplifying the resulting equation.

Q: What is the first step in solving a system of equations?

The first step in solving a system of equations is to set the equations equal to each other. This is done by equating the two equations, which gives us:

$x^2 - 2x - 3 = -x + 3$

Q: How do I simplify the equation obtained in Step 1?

To simplify the equation obtained in Step 1, you need to combine like terms and rearrange the equation to get:

$x^2 - 3x - 6 = 0$

Q: What is the quadratic formula?

The quadratic formula is a formula that is used to solve quadratic equations. It is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Q: How do I use the quadratic formula to solve a quadratic equation?

To use the quadratic formula to solve a quadratic equation, you need to plug in the values of $a$, $b$, and $c$ into the formula. In this case, $a = 1$, $b = -3$, and $c = -6$. Plugging these values into the quadratic formula, we get:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-6)}}{2(1)}$

Simplifying this expression, we get:

$x = \frac{3 \pm \sqrt{33}}{2}$

Q: What are the possible values of $x$?

The possible values of $x$ are:

$x = \frac{3 + \sqrt{33}}{2}$ and $x = \frac{3 - \sqrt{33}}{2}$

Q: How do I find the value of $y$?

To find the value of $y$, you need to plug in the value of $x$ into one of the original equations. In this case, we can plug in the value of $x$ into the first equation:

$y = x^2 - 2x - 3$

Plugging in the value of $x$, we get:

$y = \left(\frac{3 + \sqrt{33}}{2}\right)^2 - 2\left(\frac{3 + \sqrt{33}}{2}\right) - 3$

Simplifying this expression, we get:

$y = \frac{9 + 6\sqrt{33} + 33 - 6 - 2\sqrt{33} - 6}{4}$

$y = \frac{30 + 4\sqrt{33}}{4}$

$y = \frac{15 + 2\sqrt{33}}{2}$

Q: What is the solution set of the system of equations?

The solution set of the system of equations is the set of all possible values of $x$ and $y$ that satisfy both equations simultaneously. In this case, the solution set is:

$\left\{\left(\frac{3 + \sqrt{33}}{2}, \frac{15 + 2\sqrt{33}}{2}\right), \left(\frac{3 - \sqrt{33}}{2}, \frac{15 - 2\sqrt{33}}{2}\right)\right\}$

Conclusion

In this article, we have addressed some of the most frequently asked questions about solving a system of equations. We have discussed the first step in solving a system of equations, how to simplify the equation obtained in Step 1, and how to use the quadratic formula to solve a quadratic equation. We have also discussed how to find the value of $y$ and the solution set of the system of equations.