Solving Linear-Quadratic Systems: The Elimination Method

Hey everyone! Today, we're diving into a cool math trick: solving linear-quadratic systems using the elimination method. We'll walk through how to find where a line and a parabola meet. Specifically, we will solve the system of equations $y=x^2+x-42$ and $y=-x+6$. Don't worry, it's not as scary as it sounds. We will break it down step-by-step to make it super easy to understand. So, grab your pencils and let's get started!

Understanding Linear-Quadratic Systems

Alright, before we jump into the nitty-gritty, let's chat about what a linear-quadratic system actually is. Imagine two equations: one that makes a straight line (that's the linear equation) and another that makes a curve, like a U-shape (that's the quadratic equation). When we talk about solving the system, we're basically asking: where do these two graphs cross each other? Those crossing points are the solutions to our system. There can be zero, one, or even two points where the line and parabola intersect, which means zero, one, or two solutions to the system. Understanding this concept is critical before moving on to solving the actual system of equations.

In our case, we have a parabola represented by the equation $y = x^2 + x - 42$ and a line represented by the equation $y = -x + 6$. The goal here is to determine the $x$ and $y$ values that satisfy both equations simultaneously. Graphically, it means finding the point or points where the parabola and the line intersect. The elimination method is a neat way to crack this problem. It allows us to manipulate the equations in a way that helps us isolate the variables and find the solutions. Basically, we manipulate the equations until one of the variables cancels out when we add or subtract the equations. This simplifies the system, letting us solve for the remaining variable. Once we know one variable's value, we can easily find the other. Pretty neat, right?

The elimination method is a fantastic tool because it's systematic and reduces the complexity of solving these systems. We're turning two equations into one, which is much easier to manage. Now, we'll see the power of this method when applied to the equations mentioned earlier. You will appreciate how this methodical approach simplifies what might initially seem complex. This is why learning the elimination method is super valuable for solving various types of algebraic problems. The method is versatile and not limited to just these types of problems. You'll find it can be applied to various situations where you need to solve systems of equations.

Step-by-Step: The Elimination Process

Alright, let's get down to business and solve the system using the elimination method. Here's the breakdown, step by step, so you can follow along easily. Remember our equations: $y = x^2 + x - 42$ and $y = -x + 6$. We can rewrite these to facilitate the elimination process. Our goal is to eliminate one variable by adding or subtracting the equations. Since both equations are already solved for $y$, we can set the expressions equal to each other. This is the first, but crucial step. By doing this, we create a new equation with only one variable, which we can solve.

Step 1: Set the Equations Equal

Since both equations are solved for $y$, we can set them equal to each other: $x^2 + x - 42 = -x + 6$. This simple step sets the stage for solving for $x$. By equating the two expressions, we've essentially transformed our two equations into one, making it much easier to solve. The equations are now in a format that we can easily manipulate to find the $x$ values where the line and parabola intersect. Understanding this step is super crucial as it's the core of simplifying the system.

Step 2: Simplify and Rearrange

Next, we'll simplify and rearrange the equation to get everything on one side and set it equal to zero. This is a standard move when solving quadratic equations. Add $x$ and subtract $6$ from both sides to get: $x^2 + 2x - 48 = 0$. This step is all about getting the equation into a standard quadratic form. The standard form makes it easier to identify the coefficients of the quadratic equation. Once we have the standard form, we can then factor the equation or use the quadratic formula to find the roots (solutions). This rearrangement is key to making the equation solvable.

Step 3: Solve for $x$ (Factoring)

Now, let's solve for $x$. We can do this by factoring the quadratic equation. We need to find two numbers that multiply to $-48$ and add up to $2$. Those numbers are $8$ and $-6$. So, we can factor the equation as $(x + 8)(x - 6) = 0$. Remember, we are trying to find where the line and the parabola meet, and these $x$ values are the $x$-coordinates of those points. Knowing this, we can set each factor equal to zero: $x + 8 = 0$ or $x - 6 = 0$. Solving these gives us $x = -8$ or $x = 6$. So, we've found our $x$ values! We have identified the $x$ coordinates of the intersection points. This factoring step is crucial because it directly gives us the solutions for $x$. It's a quick and efficient way to find the roots of the quadratic equation. Also, if you can't factor, don't sweat it. You could use the quadratic formula too, and you'll get the same results.

Step 4: Solve for $y$

Now that we have the $x$ values, we need to find the corresponding $y$ values. We can plug these $x$ values back into either of the original equations. Since the linear equation $y = -x + 6$ is easier, let's use that. When $x = -8$, we have $y = -(-8) + 6 = 8 + 6 = 14$. When $x = 6$, we have $y = -(6) + 6 = 0$. Therefore, we have two points of intersection: $(-8, 14)$ and $(6, 0)$. This step is pretty straightforward. We're using the $x$ values we found in the previous step to find the corresponding $y$ values. By plugging the $x$ values back into the equation, we determine where the graphs intersect on the $y$-axis. The result is the coordinates of the points where the line and the parabola cross each other. This is how we find the complete solution of the system.

Step 5: Check the Solutions

It's always a good idea to check your solutions. Plug the $x$ and $y$ values of both points into both original equations to make sure they work. For the point $(-8, 14)$: In the first equation, $14 = (-8)^2 + (-8) - 42$ simplifies to $14 = 64 - 8 - 42$, which is $14 = 14$. In the second equation, $14 = -(-8) + 6$, which simplifies to $14 = 8 + 6$, so $14 = 14$. For the point $(6, 0)$: In the first equation, $0 = (6)^2 + (6) - 42$, which is $0 = 36 + 6 - 42$, so $0 = 0$. In the second equation, $0 = -(6) + 6$, which simplifies to $0 = 0$. This confirms that our solutions are correct. The checking step is super important to ensure we have not made any errors. This step is about verifying that the points you have found actually satisfy both original equations. This double-checking step can save you a lot of trouble, especially in exams or when solving more complex systems. Always verify to ensure your answer is correct. This is a very good practice to follow.

Conclusion: You Did It!

And there you have it, guys! We've successfully used the elimination method to solve a linear-quadratic system. We found the points where the line and parabola intersect: $(-8, 14)$ and $(6, 0)$. Isn't that neat? The elimination method is a powerful tool. It simplifies the process of finding solutions to systems of equations. Keep practicing, and you'll get the hang of it in no time. If you got stuck at any step, just go back and review it. Understanding and practicing these concepts will make solving more complex systems much easier down the road. Keep up the great work. You're doing awesome!

This method is a fundamental skill in algebra. Mastering it can help you tackle more complicated problems. It also sets a solid foundation for more advanced topics in mathematics. Remember, practice is the key to mastering any math concept. So, keep solving problems, and don't be afraid to make mistakes. Mistakes are great learning opportunities. Keep up the good work, and you will become proficient at solving these systems. Keep practicing, and you'll master this technique in no time. You will find that these skills will also make it easier for you to learn other types of math topics, and in many areas of life.