Solving The System Of Equations: X^2 + 4y = 4 And -15x + 5 = 5y
Hey guys! Today, we're diving into the fascinating world of system of equations. Specifically, we're going to tackle the system:
- x² + 4y = 4
- -15x + 5 = 5y
Don't worry, it might look a bit intimidating at first glance, but we'll break it down step by step. So, grab your pencils, and let's get started!
Understanding the Equations
Before we jump into solving, let's make sure we understand what we're dealing with. The first equation, x² + 4y = 4, is a quadratic equation. Notice the x² term? That's your clue! This equation represents a parabola. The second equation, -15x + 5 = 5y, is a linear equation. It's a straight line when graphed. Our goal is to find the points (x, y) where these two graphs intersect. These intersection points are the solutions to our system of equations. These solutions are the values of x and y that satisfy both equations simultaneously. Imagine the parabola and the line drawn on the same graph; the points where they cross are the solutions we're after. Understanding the nature of each equation helps us strategize our approach to solving the system. For instance, knowing that one equation is linear allows us to easily isolate one variable and substitute it into the other equation. This is a common and effective technique for solving systems where one equation is linear and the other is not. In the case of a quadratic system, we anticipate that there could be up to two solutions, as a line can intersect a parabola at zero, one, or two points. The geometry of the equations gives us valuable insight into the possible solutions we're trying to find.
Step 1: Simplify the Linear Equation
Okay, so let's simplify the linear equation first. It'll make things easier down the road. We have -15x + 5 = 5y. To isolate y, we'll divide both sides of the equation by 5:
(-15x + 5) / 5 = 5y / 5
This simplifies to:
-3x + 1 = y
Great! Now we have y expressed in terms of x. This is super useful because we can now substitute this expression for y into the quadratic equation. This step is a crucial part of the solution process because it allows us to reduce the system of two equations into a single equation with one variable. By expressing y in terms of x, we eliminate y from the system, making it solvable. Simplifying the linear equation first is a smart move because it avoids dealing with more complex expressions later on. It keeps the algebra cleaner and reduces the chances of making errors. Also, having y isolated makes the substitution process straightforward. We now have a clear expression for y that we can plug into the other equation, which is a key step in finding the solutions to the system. So, by dividing both sides of the equation by 5, we've set ourselves up for the next phase of the solution. Remember, this is a classic technique in solving systems of equations: simplify, isolate, and substitute.
Step 2: Substitute into the Quadratic Equation
Now comes the fun part – substitution! We're going to take our simplified expression for y (y = -3x + 1) and plug it into the quadratic equation, x² + 4y = 4. So, wherever we see y in the quadratic equation, we'll replace it with (-3x + 1). This gives us:
x² + 4(-3x + 1) = 4
See what we did there? We've successfully eliminated y from the equation, and now we have an equation with only x. This is a major breakthrough because we can now solve for x. This substitution method is a powerful technique because it transforms a system of equations into a single equation that we can solve. It's like magic, but it's just math! By substituting the expression for y into the quadratic equation, we've essentially combined the information from both equations into one. This is a critical step in solving simultaneous equations. We're now working with a single equation in one variable, which is a much simpler problem to tackle. The key to successful substitution is to make sure you replace y correctly with the entire expression (-3x + 1). It's essential to use parentheses to ensure that the 4 is properly distributed in the next step. So, we've taken a significant step forward by using substitution, and we're now on the home stretch to finding the values of x that satisfy both equations.
Step 3: Solve for x
Alright, let's solve for x. First, we need to distribute the 4 in our equation:
x² + 4(-3x + 1) = 4 becomes x² - 12x + 4 = 4
Next, we subtract 4 from both sides to set the equation to zero:
x² - 12x = 0
Now we can factor out an x:
x(x - 12) = 0
This gives us two possible solutions for x:
x = 0 or x - 12 = 0
So,
x = 0 or x = 12
Woohoo! We've found our x values. This is a huge accomplishment! By setting the equation to zero and factoring, we've used a classic technique for solving quadratic equations. Each factor gives us a potential solution for x, and in this case, we found two. This means there are potentially two points of intersection between the parabola and the line, which is consistent with what we expected. The zero-product property is the underlying principle here: if the product of two factors is zero, then at least one of the factors must be zero. Factoring out the x made it clear what those factors were and allowed us to easily find the solutions. Remember, these x values are only half the battle. We still need to find the corresponding y values for each of these x values. But for now, let's celebrate this victory – we've successfully solved for x!
Step 4: Solve for y
Now that we have our x values (x = 0 and x = 12), we need to find the corresponding y values. We'll use the simplified linear equation we found earlier: y = -3x + 1.
Let's start with x = 0:
y = -3(0) + 1 y = 1
So, when x = 0, y = 1. This gives us our first solution point (0, 1).
Now let's find y when x = 12:
y = -3(12) + 1 y = -36 + 1 y = -35
So, when x = 12, y = -35. This gives us our second solution point (12, -35).
Fantastic! We've found the y values for both x values. This step is crucial because it completes the solution process. We're not just looking for x values; we're looking for the points (x, y) that satisfy both equations. Using the simplified linear equation to find y is the most efficient approach, as it avoids having to substitute back into the quadratic equation. This method leverages the work we did earlier in isolating y in the linear equation. Each x value has a corresponding y value that makes the pair a solution to the system. We now have two distinct points, (0, 1) and (12, -35), which represent the intersections of the parabola and the line. These points are the solutions to our original system of equations. Remember, a solution to a system of equations must satisfy all equations in the system simultaneously. We've checked that these points work in both equations, giving us confidence in our solution.
Step 5: Check Your Answers
It's always a good idea to check our answers to make sure they're correct. We'll plug each solution point back into both original equations.
Let's check (0, 1):
- x² + 4y = 4 becomes 0² + 4(1) = 4, which is 4 = 4 (True)
- -15x + 5 = 5y becomes -15(0) + 5 = 5(1), which is 5 = 5 (True)
So, (0, 1) is indeed a solution.
Now let's check (12, -35):
- x² + 4y = 4 becomes 12² + 4(-35) = 4, which is 144 - 140 = 4, which is 4 = 4 (True)
- -15x + 5 = 5y becomes -15(12) + 5 = 5(-35), which is -180 + 5 = -175, which is -175 = -175 (True)
So, (12, -35) is also a solution.
Excellent! Both points check out. This confirms that our solutions are correct and that we've successfully solved the system of equations. Checking our answers is a critical step in the problem-solving process. It ensures that we haven't made any errors in our calculations and that our solutions truly satisfy the original equations. Plugging the solutions back into the original equations is the most reliable way to verify their correctness. This step can save us from submitting incorrect answers and helps build confidence in our problem-solving abilities. Remember, accuracy is key in mathematics, and checking our work is an essential part of achieving accuracy. We've now not only solved the system but also verified our solutions, completing the problem with confidence. So, always take the time to check your answers – it's worth the effort!
Final Answer
The solutions to the system of equations are (0, 1) and (12, -35).
Conclusion
There you have it! We've successfully solved the system of equations. We simplified, substituted, solved, and checked our answers. Remember, solving systems of equations is a fundamental skill in algebra, and it's used in many different areas of mathematics and science. So, keep practicing, and you'll become a pro in no time! Solving systems of equations might seem daunting at first, but with a systematic approach, it becomes manageable. We started by understanding the nature of the equations, then simplified and used substitution to reduce the problem to a single equation. We then solved for x, found the corresponding y values, and most importantly, we checked our answers. This step-by-step process is a valuable strategy for tackling any math problem. Remember, math isn't just about getting the right answer; it's about the journey of problem-solving. Each step we took, from simplifying the equation to checking our answers, contributed to our understanding and confidence in solving these types of problems. Practice is key, and the more systems of equations you solve, the more comfortable and proficient you'll become. So, keep up the great work, and remember that every problem solved is a step forward in your mathematical journey. You've got this!
