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

Hey guys! Today, we're diving deep into the fascinating world of systems of equations. If you've ever felt lost trying to solve these, don't worry, I'm here to break it down for you. We're going to tackle a specific example, walking through each step with clear explanations and a friendly tone. Our focus? Understanding how to determine the solution set of a system of equations algebraically. Let's get started!

The Foundation: Understanding Systems of Equations

Before we jump into the nitty-gritty, let's establish a solid foundation. What exactly is a system of equations? Simply put, it's a set of two or more equations that share the same variables. The goal is to find the values of those variables that satisfy all equations in the system simultaneously. Think of it like a puzzle where each equation is a piece, and the solution is the picture that emerges when all the pieces fit together perfectly.

In our case, we're dealing with a system of two equations:

• y = x² - 6x + 12 (a quadratic equation, represented by a parabola)
• y = 2x - 4 (a linear equation, represented by a straight line)

The solutions to this system represent the points where these two graphs intersect. Algebraically, we're looking for the x and y values that make both equations true. This might sound intimidating, but fear not! We have a systematic approach to conquer this challenge.

Solving systems of equations algebraically involves a few key methods, the most common being substitution and elimination. Today, we'll be focusing on the substitution method, as it's directly applicable to the problem at hand. The core idea behind substitution is to solve one equation for one variable and then substitute that expression into the other equation. This eliminates one variable, leaving us with a single equation that we can solve. It's like magic, but with math! We are using an algebraic system of equations to solve for the points where the quadratic and linear equations intersect.

Step 1: Setting the Stage - Equating the Expressions

The table you provided shows the first crucial step in solving our system of equations. It states:

But why are we doing this? What's the logic behind equating x² - 6x + 12 and 2x - 4? Let's break it down. Remember, both equations are set equal to y. This means that at the points where the graphs intersect (our solutions!), the y-values must be the same. If y is equal to both x² - 6x + 12 and 2x - 4, then x² - 6x + 12 must also be equal to 2x - 4. Think of it like a transitive property in action – if a = b and b = c, then a = c.

This step effectively eliminates y from the equation, leaving us with a single equation in terms of x. This is a huge step forward because we now have a quadratic equation that we can solve using various methods, such as factoring, completing the square, or the quadratic formula. By setting the two expressions for y equal to each other, we are essentially finding the x-coordinates of the points of intersection. We are transforming a system of two equations into a single, solvable equation. This algebraic manipulation is a fundamental technique in solving systems. The step highlights the power of algebraic methods in simplifying complex problems.

Step 2: Transforming to Standard Form - Preparing for the Solution

Now that we have a single equation, x² - 6x + 12 = 2x - 4, we need to manipulate it into a form that's easier to solve. Specifically, we want to transform it into the standard form of a quadratic equation: ax² + bx + c = 0. This form makes it much easier to apply techniques like factoring or the quadratic formula. So, how do we get there?

The goal is to get all the terms on one side of the equation, leaving zero on the other side. To do this, we need to subtract 2x and add 4 to both sides of the equation. This maintains the balance of the equation (whatever you do to one side, you must do to the other!) and moves all the terms to the left side.

Let's walk through the process:

• Starting equation: x² - 6x + 12 = 2x - 4
• Subtract 2x from both sides: x² - 6x - 2x + 12 = -4
• Add 4 to both sides: x² - 6x - 2x + 12 + 4 = 0
• Combine like terms: x² - 8x + 16 = 0

And there we have it! Our equation is now in the standard quadratic form: x² - 8x + 16 = 0. This seemingly simple transformation is crucial because it sets us up for the next step: actually solving for x. By rearranging the equation into standard form, we've made it much more amenable to solution techniques. It's like organizing your tools before starting a project – having everything in its place makes the job much smoother. This algebraic rearrangement is a vital skill in solving equations. Now that we've got our quadratic equation in standard form, we're one step closer to finding the elusive x-values that will unlock our solution set.

Next Steps and the Big Picture

So, where do we go from here? Now that we have our quadratic equation in standard form (x² - 8x + 16 = 0), we can solve for x. There are a few methods we can use:

1. Factoring: If the quadratic expression can be factored, this is often the quickest method. We look for two numbers that multiply to c (16 in our case) and add up to b (-8 in our case).
2. Quadratic Formula: This is a foolproof method that works for any quadratic equation, regardless of whether it can be factored. The formula is: x = (-b ± √(b² - 4ac)) / 2a.
3. Completing the Square: This method is a bit more involved but can be useful in certain situations.

In our case, the equation x² - 8x + 16 = 0 can be factored nicely. It factors into (x - 4)(x - 4) = 0, or (x - 4)² = 0. This means that x = 4 is our solution. We have a repeated root, which means the parabola and the line touch at only one point. This factoring algebraic allows us to find the root of the equation easily.

But we're not done yet! Remember, we're solving a system of equations. We've found the x-value, but we still need to find the corresponding y-value. To do this, we simply substitute our x-value (4) into either of the original equations. Let's use the simpler linear equation, y = 2x - 4:

• y = 2(4) - 4
• y = 8 - 4
• y = 4

So, the solution to our system of equations is x = 4 and y = 4. This means the parabola and the line intersect at the point (4, 4). We have successfully used algebraic techniques to find the solution set.

Wrapping Up: The Power of Algebraic Problem-Solving

Solving systems of equations algebraically might seem daunting at first, but as you can see, it's a systematic process with clear steps. By understanding the underlying principles and mastering techniques like substitution and factoring, you can confidently tackle these problems. We've walked through the first two crucial steps in detail: equating the expressions and transforming the equation into standard form. These steps lay the groundwork for finding the solution set.

Remember, the key is to break down the problem into smaller, manageable steps. Don't be afraid to practice and experiment with different methods. And most importantly, embrace the power of algebra to unlock the solutions that lie hidden within these equations. Keep practicing, and you'll become a system-solving pro in no time! Understanding the algebraic solutions to these equations can provide insights into various real-world problems. This comprehensive algebraic system solving approach is essential for mathematical proficiency.