Solving Linear-Quadratic Equations By Elimination
Hey math enthusiasts! Today, we're going to dive into a cool method for solving systems of equations where one equation is linear (a straight line) and the other is quadratic (a curve, like a parabola). We're talking about the elimination method, a powerful tool that simplifies things by strategically removing one of the variables. Let's tackle the system: y = x² + 10 and y = -7x - 2. This system contains a quadratic equation and a linear equation. The goal is to find the points where the line and the parabola intersect. This is where the y-values and x-values satisfy both equations simultaneously. So, grab your pencils and let's get started!
Understanding the Elimination Method
Before we jump into the problem, let's quickly recap what the elimination method is all about. This method aims to eliminate one of the variables (either x or y) by adding or subtracting the equations. We do this to reduce the system into a single equation with only one variable, which we can then solve. Once we find the value of that variable, we can substitute it back into either of the original equations to find the value of the other variable. In our case, the equations are already set up perfectly for elimination because both equations are solved for y. This is the beauty of this problem; it's already set up for success! We'll manipulate the equations, so one variable cancels out when we combine them. This simplification is key to solving the system efficiently. This is the whole idea of the elimination method. The method is used for solving a system of equations by eliminating one variable. This is usually done by adding or subtracting the equations. The goal is to get one equation with a single variable, which can be easily solved. Once we find the value of that variable, we can substitute it back into either of the original equations to find the value of the other variable. It is a powerful method for solving systems of equations and is particularly useful when dealing with linear and quadratic equations. It is also good for more complex systems. The process involves manipulating the equations in such a way that when the equations are combined, either the x or y variable cancels out. This leaves us with a single equation containing only one variable, which is then solved. The value obtained is substituted back into one of the original equations to find the value of the remaining variable. This gives us the solution to the system of equations, representing the points where the graphs of the equations intersect. This method is often preferred due to its ability to streamline the solution process. It can be more direct than other methods. This is why it is preferred by many.
Preparing the Equations for Elimination
Notice that both equations have y isolated. Because both equations equal y, we can set them equal to each other. We start with:
y = x² + 10 y = -7x - 2
Since both equations equal y, we can set them equal to each other: x² + 10 = -7x - 2
Now, we need to arrange this into a standard quadratic form (ax² + bx + c = 0). Let's move all the terms to one side of the equation:
x² + 7x + 12 = 0
We did this by adding 7x and 2 to both sides. Now, we have a simple quadratic equation that we can easily solve.
Solving for x
With our quadratic equation in place, x² + 7x + 12 = 0, we can now solve for x. There are a few methods to do this: factoring, completing the square, or using the quadratic formula. In this case, factoring is the simplest approach. We need to find two numbers that multiply to 12 and add up to 7. Those numbers are 3 and 4.
So, we can factor the quadratic equation as follows: (x + 3)(x + 4) = 0
This means that either (x + 3) = 0 or (x + 4) = 0. Solving these simple linear equations gives us: x = -3 or x = -4
We've found our x-values! These are the x-coordinates of the points where the line and the parabola intersect. That's a huge step forward!
Solving for y
Now that we have the x-values, we need to find the corresponding y-values. We'll do this by substituting each x-value back into either of the original equations. I recommend using the linear equation y = -7x - 2 because it's simpler. Let's start with x = -3:
y = -7(-3) - 2 y = 21 - 2 y = 19
So, when x = -3, y = 19. Now, let's do the same for x = -4:
y = -7(-4) - 2 y = 28 - 2 y = 26
So, when x = -4, y = 26. We now have our x and y values for both points of intersection. The points of intersection are (-3, 19) and (-4, 26).
Verifying the Solutions
It's always a good idea to check our answers. We can do this by substituting our solutions back into both original equations to make sure they are valid. Let's check the first solution, (-3, 19), with the first equation, y = x² + 10:
19 = (-3)² + 10 19 = 9 + 10 19 = 19
It works! Now let's try the second equation, y = -7x - 2:
19 = -7(-3) - 2 19 = 21 - 2 19 = 19
This checks out as well! Now, let's verify the second solution, (-4, 26). With the first equation, y = x² + 10:
26 = (-4)² + 10 26 = 16 + 10 26 = 26
And with the second equation, y = -7x - 2:
26 = -7(-4) - 2 26 = 28 - 2 26 = 26
Both solutions check out perfectly! This confirms that we have correctly found the points of intersection between the line and the parabola. Always verify your answers to ensure accuracy. It helps identify any calculation errors and builds confidence in your results.
Conclusion: The Power of Elimination
In this tutorial, we successfully used the elimination method to solve a system of linear-quadratic equations. We found that the line and the parabola intersect at the points (-3, 19) and (-4, 26). The elimination method simplifies the process by strategically manipulating the equations to eliminate a variable. This allows us to reduce the system to a single variable equation, making it easier to solve. The process involves setting the equations equal to each other, rearranging them into standard quadratic form, factoring the quadratic equation, and then solving for the variables. By finding the x-values, we can substitute them back into one of the original equations to find the corresponding y-values. To make sure that we have accurate results, we can verify our answers by substituting the solutions back into both original equations. Remember that the elimination method is a powerful tool in your math toolbox and can be applied to various types of equations. Practice more examples to enhance your skills and become more confident. This is the cornerstone of problem-solving. It's a key skill for more advanced math concepts. Keep practicing and exploring different types of problems, and you'll find that the elimination method becomes second nature.
If you want to delve deeper into solving systems of equations, I recommend checking out resources on Khan Academy. They offer excellent tutorials and practice problems to hone your skills. Remember, practice makes perfect!
