Hey guys! Today, we're diving into a classic math problem: solving a system of equations. Specifically, we're going to tackle the system:
y = -3x² + 2y = 3
This involves finding the point(s) where these two equations intersect. One equation is a parabola (y = -3x² + 2), and the other is a horizontal line (y = 3). Let's break down the process step by step.
Setting Up the Equations
The core concept here is that at the point(s) of intersection, both equations will have the same y value. Since we already know that y = 3 from the second equation, we can substitute this value into the first equation. This is the substitution method in action! It’s a super handy technique for solving systems of equations. So, we replace the y in the first equation with 3, giving us:
3 = -3x² + 2
Now we have a single equation with only one variable, x. This is a quadratic equation, which we can solve using various methods. Let's get into the nitty-gritty of solving this equation.
Solving the Quadratic Equation
To solve the quadratic equation 3 = -3x² + 2, we need to rearrange it into a standard form that we can work with. The standard form of a quadratic equation is ax² + bx + c = 0. Our first step is to get all the terms on one side of the equation. Subtracting 3 from both sides gives us:
0 = -3x² - 1
Now, to make things a little easier, let's multiply both sides of the equation by -1. This will get rid of the negative sign in front of the x² term, making it easier to work with. Remember, multiplying both sides of an equation by the same number doesn't change the solution. So, we get:
0 = 3x² + 1
This equation is in a simplified quadratic form. Notice that the bx term is missing (or rather, it has a coefficient of 0). This makes it even easier to solve. We can isolate the x² term by subtracting 1 from both sides:
-1 = 3x²
Next, we divide both sides by 3 to isolate x²:
x² = -1/3
Now, this is where things get interesting! We have x² equal to a negative number. Think about what this means. When you square any real number (positive, negative, or zero), the result is always non-negative. You can't square a real number and get a negative number. This is a crucial point! Because x² cannot be negative for any real number x, this equation has no real solutions.
Key Takeaway: When solving equations, always pay attention to situations like this where you might encounter a contradiction. In this case, finding x² = -1/3 tells us that there are no real values of x that satisfy the equation.
Graphical Interpretation
Let's take a step back and visualize what's happening. Remember our original equations:
y = -3x² + 2y = 3
The first equation represents a parabola that opens downwards (because of the negative coefficient in front of the x² term). The vertex of the parabola is at the point (0, 2). The second equation, y = 3, represents a horizontal line that intersects the y-axis at y = 3. Graphically, we're looking for the points where the parabola and the line intersect.
Since we found that there are no real solutions to the equation x² = -1/3, it means that the parabola and the line do not intersect in the real number plane. The line y = 3 is above the vertex of the parabola, and because the parabola opens downwards, it will never intersect the line. This graphical interpretation reinforces our algebraic solution.
Complex Solutions (Optional)
While there are no real solutions, there are complex solutions. This is a more advanced topic, but let's briefly touch on it. Remember that the imaginary unit i is defined as the square root of -1 (i.e., i² = -1). We can use this to find complex solutions for x² = -1/3.
Taking the square root of both sides, we get:
x = ±√(-1/3)
We can rewrite the square root of a negative number using the imaginary unit:
x = ±√(1/3 * -1)
x = ±√(1/3) * √(-1)
x = ±√(1/3) * i
So the complex solutions are:
x = √(1/3)i and x = -√(1/3)i
These solutions are complex numbers, which involve the imaginary unit i. They represent points in the complex plane, but they don't correspond to any real intersections on the graph of the parabola and the line in the Cartesian plane.
Conclusion
In summary, the system of equations:
y = -3x² + 2y = 3
has no real solutions. This means the parabola and the line do not intersect in the real number plane. We reached this conclusion by substituting y = 3 into the first equation, solving the resulting quadratic equation, and finding that x² would have to be a negative number, which is impossible for real numbers. We also saw how a graphical interpretation can help visualize this result. While there are complex solutions, they don't represent real-world intersections on a graph.
Understanding how to solve systems of equations, especially when quadratic equations are involved, is a fundamental skill in algebra. Keep practicing, and you'll become a pro at spotting these types of solutions (or lack thereof!). Remember, no real solutions means the graphs don't intersect in the real plane!
Let's clarify the original question to make it super easy to understand. The question was essentially asking: "What values of x and y satisfy both equations y = -3x² + 2 and y = 3 simultaneously?" Or, in simpler terms, "Where do these two graphs intersect?"
So, a better way to phrase the question, making it crystal clear, would be:
"Find the solutions (if any) to the system of equations:
y = -3x² + 2
y = 3
Explain the steps involved in finding these solutions and interpret the results graphically."
This revised question is more explicit about what we're looking for – the solutions – and it also prompts a discussion about the graphical interpretation, which is crucial for understanding the problem fully. Breaking down math problems into clear, understandable questions is the first step in conquering them! Remember, clear questions lead to clear solutions.
Systems of equations can seem daunting at first, but they're actually a powerful tool for solving real-world problems. This SEO title will guide you through solving a specific system involving a parabola and a horizontal line. We'll break down the process step-by-step, so you can understand the underlying concepts and apply them to other problems. Let’s dive deep into the fascinating world of solving systems of equations, specifically focusing on the interaction between a parabola and a horizontal line. We're going to explore the system defined by the equations y = -3x² + 2 and y = 3. Mastering these techniques will empower you to tackle a wide range of mathematical challenges. Understanding how different types of equations interact within a system of equations is a fundamental skill in algebra. This particular system, involving a parabola and a horizontal line, provides an excellent opportunity to explore concepts like substitution, quadratic equations, and graphical interpretation. We will meticulously examine each step, ensuring clarity and providing a solid foundation for more advanced mathematical concepts.
Understanding the Equations
Before we jump into solving, let's make sure we understand what these equations represent graphically. The first equation, y = -3x² + 2, is a quadratic equation, and its graph is a parabola. The negative coefficient in front of the x² term tells us that the parabola opens downwards. The + 2 at the end shifts the parabola upwards by two units. So, we have a downward-opening parabola with its vertex at the point (0, 2). Now, let’s talk about the parabola itself. It's a U-shaped curve, symmetrical around its vertex. The equation y = ax² + bx + c defines a parabola, where the sign of a determines whether the parabola opens upwards (if a > 0) or downwards (if a < 0). In our case, a = -3, confirming that it's a downward-opening parabola. The vertex of a parabola is its highest or lowest point, depending on whether it opens downwards or upwards, respectively. Finding the vertex is crucial for understanding the parabola's position and shape. In the given equation, the vertex is easily identifiable as (0, 2) due to the absence of a bx term and the constant term + 2. Visualizing the parabola's shape and position helps in predicting the number of solutions when solving systems of equations.
The second equation, y = 3, is much simpler. It's a linear equation, and its graph is a horizontal line that crosses the y-axis at y = 3. Understanding this horizontal line is key to visualizing the system. A horizontal line is defined by the equation y = c, where c is a constant. This means that the y-coordinate is the same for all points on the line, making it perfectly horizontal. In our case, the horizontal line is at y = 3, which means it intersects the y-axis at the point (0, 3). The simplicity of the equation for a horizontal line makes it a useful tool for solving systems of equations, especially when combined with other types of equations like parabolas. Visualizing the horizontal line as a straight, flat line across the graph helps in understanding where it might intersect with other curves, like the parabola in this example. The position of the horizontal line relative to the parabola's vertex is crucial in determining the number of solutions.
Solving the System: Substitution Method
To find the solutions to this system of equations, we'll use the substitution method. This method is particularly useful when one of the equations is already solved for one variable (in this case, y). Since we know that y = 3, we can substitute this value into the first equation:
3 = -3x² + 2
Now we have a single equation with only one variable, x. Let's solve for x. The substitution method is a powerful technique in algebra for solving systems of equations. It involves replacing one variable in an equation with its equivalent expression from another equation. This simplifies the system, allowing us to solve for the remaining variable. The effectiveness of the substitution method lies in its ability to reduce a multi-variable problem into a single-variable equation, making it easier to handle. In our case, substituting y = 3 into the equation y = -3x² + 2 allows us to eliminate y and focus solely on solving for x. The key to successful substitution is identifying which equation is easiest to manipulate and which variable to substitute, streamlining the solution process. Once we find the value of x, we can use it to find the corresponding value of y, thereby completing the solution set for the system.
Isolating x²
First, subtract 2 from both sides of the equation:
1 = -3x²
Next, divide both sides by -3:
x² = -1/3
Now we have x² isolated on one side of the equation. Isolating the variable is a fundamental technique in solving equations. It involves performing algebraic operations on both sides of the equation to get the variable term by itself. The goal of isolating x² in this case is to simplify the equation to a point where we can easily take the square root of both sides and solve for x. However, it's crucial to maintain balance in the equation by performing the same operations on both sides, ensuring that the equality remains valid. In this particular equation, the steps to isolate x² involve first subtracting a constant and then dividing by a coefficient. These operations gradually simplify the equation, making it more manageable to analyze and solve. Once x² is isolated, we can better understand the nature of the solutions, or in this case, the lack thereof in the real number system.
Analyzing the Result
Here's where things get interesting. We have x² = -1/3. Think about what this means. Can a real number, when squared, ever be negative? The answer is no. Squaring any real number (positive, negative, or zero) will always result in a non-negative number. This is a critical concept in algebra: the square of a real number cannot be negative. This principle stems from the fundamental properties of real numbers and their interactions under multiplication. A positive number times itself is positive, a negative number times itself is also positive, and zero squared is zero. There's no real number that, when multiplied by itself, yields a negative result. This mathematical certainty is what allows us to conclude that the equation x² = -1/3 has no real solutions. Recognizing that the square of a real number cannot be negative is a crucial skill in solving various types of equations, including quadratic equations and systems of equations. It helps us identify situations where solutions do not exist within the realm of real numbers.
Therefore, there are no real solutions for x that satisfy this equation. This means that the original system of equations has no real solutions. The absence of real solutions signifies that the two graphs do not intersect in the real coordinate plane. This means that there are no real number pairs (x, y) that simultaneously satisfy both equations y = -3x² + 2 and y = 3. The concept of no real solutions in a system of equations has significant implications in both algebra and graphical interpretations. Algebraically, it means that the equation derived from solving the system leads to a contradiction, such as x² equaling a negative number. Graphically, it translates to the curves or lines represented by the equations not having any points of intersection on the real coordinate plane. Understanding when a system has no real solutions is crucial in applications where real-world scenarios are modeled using mathematical equations. It helps us determine if a proposed solution is feasible or if the conditions of the problem are incompatible.
Graphical Interpretation: No Intersection
Let's visualize this. The parabola y = -3x² + 2 opens downwards and has its highest point (vertex) at (0, 2). The horizontal line y = 3 is a straight line that runs across the graph at a height of 3. Since the parabola's highest point is below the line, and it opens downwards, the two graphs never intersect. Graphical interpretation is a powerful tool for understanding and verifying the solutions (or lack thereof) in a system of equations. By visually representing the equations on a graph, we can quickly assess whether the curves or lines intersect, and if so, at what points. In this case, visualizing the downward-opening parabola and the horizontal line clearly shows that they do not intersect. The vertex of the parabola is at (0, 2), which lies below the horizontal line at y = 3. Since the parabola opens downwards, it moves further away from the horizontal line, confirming that there are no intersection points. The graphical interpretation provides an intuitive understanding of why there are no real solutions, reinforcing the algebraic conclusion and enhancing overall comprehension of the problem.
This graphical representation aligns perfectly with our algebraic solution. When we solve the system algebraically and find no real solutions, the graph confirms that the curves do not intersect. The consistency between the graphical interpretation and the algebraic solution is a crucial aspect of mathematical problem-solving. It provides a comprehensive understanding of the system, ensuring that the results are both accurate and meaningful. The graphical interpretation not only validates the algebraic solution but also helps in identifying potential errors or inconsistencies. It offers a visual check on the work, making the problem-solving process more robust and reliable. For example, if the algebraic solution suggested intersections that were not present on the graph, it would indicate a mistake in the calculations. Therefore, the graphical interpretation is an essential component of a thorough mathematical analysis.
Conclusion: No Real Solutions
In conclusion, the system of equations y = -3x² + 2 and y = 3 has no real solutions. This means that there are no points on the coordinate plane where these two graphs intersect. We arrived at this conclusion both algebraically and graphically. The algebraic solution showed that we would need to take the square root of a negative number to solve for x, which is not possible within the real number system. The graphical interpretation confirmed this, as the downward-opening parabola never intersects the horizontal line. Understanding that a system can have no real solutions is just as important as knowing how to find solutions when they exist. It signifies that the mathematical model represented by the equations does not have any feasible outcomes within the real-world context. The concept of no real solutions is common in various applications, such as physics, engineering, and economics, where constraints or conditions may lead to incompatible scenarios. Recognizing when a system has no real solutions allows us to refine the model, identify alternative approaches, or conclude that the problem may not have a solution within the defined parameters. Therefore, it's a crucial skill in mathematical problem-solving.