Solving The System Of Equations Y=-x^2+6x+16 And Y=-4x+37 Algebraically
Hey guys! Today, we're diving into the exciting world of systems of equations, specifically those involving a quadratic and a linear equation. Our mission? To figure out the solution set algebraically for the system:
So, buckle up, grab your thinking caps, and let's get started!
Setting the Stage: Understanding the Equations
Before we jump into the nitty-gritty of solving, let's take a moment to understand what these equations represent. The first equation, y = -x² + 6x + 16, is a quadratic equation. If we were to graph it, we'd see a parabola – a U-shaped curve. The negative sign in front of the x² term tells us the parabola opens downwards, like a frown. This parabola embodies a myriad of points, each a potential solution if we were just dealing with this equation alone.
Now, let's turn our attention to the second equation, y = -4x + 37. This, my friends, is a linear equation. Graphically, it's a straight line. The '-4' is the slope, indicating the line slopes downwards, and the '+37' is the y-intercept, where the line crosses the vertical axis. Like its quadratic counterpart, this line is also a collection of points, each a solution in its own right.
But here's where the magic happens: we're not interested in the solutions to these equations individually. We want the solution(s) to the system of equations. This means we're looking for the point(s) where the parabola and the line intersect. These intersection points are the only points that satisfy both equations simultaneously. They're the common ground, the shared solutions, the sweet spot where our quadratic and linear worlds collide.
Think of it like this: imagine you have two friends, each with their own set of preferences (represented by our equations). The solution to the system is like finding the activities they both enjoy – the common ground where they can hang out and have a good time. Similarly, the solution to our system of equations is the (x, y) coordinate(s) that make both equations happy.
The Algebraic Showdown: Solving for x
Okay, enough with the metaphors! Let's get down to business and solve this thing algebraically. The key to solving a system like this is to realize that at the point(s) of intersection, the y-values of both equations are equal. This gives us the green light to set the right-hand sides of our equations equal to each other. So, we have:
Now we have a single equation with a single variable (x), which is something we can definitely handle. Our goal is to rearrange this equation into a standard quadratic form, which looks like ax² + bx + c = 0. This form is our ticket to using some powerful tools like factoring or the quadratic formula.
To get there, let's add x² to both sides of the equation. This gets rid of the negative x² term on the left, which is always a good move. We now have:
Next, let's subtract 6x and 16 from both sides. This will bring all the terms to the right side, leaving us with zero on the left. Doing so, we get:
Now, let's simplify by combining like terms. We have -4x and -6x, which combine to -10x, and 37 and -16, which combine to 21. This gives us our quadratic equation in standard form:
Awesome! We've successfully transformed our equation into the familiar quadratic form. Now, the million-dollar question: how do we solve it? Well, we have a couple of options: factoring or the quadratic formula. Factoring is generally faster if we can spot the factors easily. The quadratic formula, on the other hand, is a more general method that works for any quadratic equation, but it can be a bit more computationally intensive.
Let's try factoring first. We're looking for two numbers that multiply to 21 and add up to -10. Think about the factors of 21: 1 and 21, 3 and 7. Aha! 3 and 7 look promising. Since we need them to add up to -10, we'll use -3 and -7. These numbers satisfy both conditions: (-3) * (-7) = 21 and (-3) + (-7) = -10. So, we can factor our quadratic equation as:
Now, we use the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This gives us two possible equations:
Solving these simple equations, we find our x-values:
Woohoo! We've found our x-coordinates for the solution(s) to the system. But hold on, we're not quite done yet. Remember, a solution to a system of equations is a point (x, y), not just an x-value. So, we still need to find the corresponding y-values.
Finding the Y: Completing the Coordinates
To find the y-values, we simply plug our x-values back into either of the original equations. It doesn't matter which one we choose, as both equations should give us the same y-value for a given x-value at the points of intersection. However, to make our lives easier, let's choose the simpler equation, the linear one: y = -4x + 37.
Let's start with x = 3. Plugging this into our linear equation, we get:
So, when x = 3, y = 25. This gives us our first solution point: (3, 25).
Now, let's do the same for x = 7. Plugging this into our linear equation, we get:
So, when x = 7, y = 9. This gives us our second solution point: (7, 9).
And there you have it, guys! We've successfully found the solution set for the system of equations algebraically. Our solutions are the points (3, 25) and (7, 9).
The Grand Finale: Solution Set and Verification
To formally state our solution, we can write it as a set of ordered pairs:
Solution Set: {(3, 25), (7, 9)}
This means that the parabola and the line intersect at these two points, and these are the only points that satisfy both equations simultaneously.
But before we declare victory and pat ourselves on the back, let's take a moment to verify our solutions. This is a crucial step to ensure we haven't made any silly mistakes along the way. To verify, we simply plug our solution points back into both original equations and see if they hold true.
Let's start with the point (3, 25). For the quadratic equation, y = -x² + 6x + 16, we have:
This checks out! Now, let's check the linear equation, y = -4x + 37:
This also checks out! So, the point (3, 25) is definitely a solution.
Now, let's do the same for the point (7, 9). For the quadratic equation, we have:
This checks out as well! And for the linear equation:
This also checks out! So, the point (7, 9) is also a solution.
Since both points satisfy both equations, we can confidently say that our solution set {(3, 25), (7, 9)} is correct.
Wrapping Up: Mastering the Art of Systems
And there you have it! We've successfully navigated the world of systems of equations, solved a quadratic-linear system algebraically, and verified our solutions. You guys are now one step closer to becoming math whizzes!
Solving systems of equations is a fundamental skill in algebra and has applications in various fields, from physics and engineering to economics and computer science. The key takeaway here is the power of combining different mathematical concepts – in this case, quadratic and linear equations – to solve a problem. By setting the equations equal to each other, we transformed the system into a single equation that we could then solve using our algebraic toolbox.
Remember, practice makes perfect. So, keep tackling those systems of equations, and don't be afraid to try different approaches. The more you practice, the more comfortable and confident you'll become in your problem-solving abilities. And who knows, maybe you'll even start seeing the world as a giant system of equations just waiting to be solved!
