Solving Systems Of Equations A Step-by-Step Guide

Hey guys! Ever stumbled upon a set of equations that seem impossible to crack? Don't worry, you're not alone. Systems of equations can be tricky, but with the right approach, you can solve them like a pro. In this guide, we'll dive into how to solve a specific system of equations. This system combines a linear equation and a circle equation, offering a fantastic opportunity to explore different solution techniques. So, buckle up and let's get started!

Understanding the Equations

First, let's break down the equations we're dealing with:

1. 2y - 3x = 6
2. x² + y² = 9

The first equation is a linear equation. When you graph it, you'll get a straight line. Linear equations are the bread and butter of algebra, and they're relatively straightforward to work with. We can rearrange this equation to express y in terms of x or vice versa, which will be super handy later on.

The second equation is where things get a little more interesting. x² + y² = 9 represents a circle centered at the origin (0, 0) with a radius of 3. Think of it as all the points that are exactly 3 units away from the center of a coordinate plane. This equation introduces a curve into our system, which means we'll need a slightly different approach to find the solutions.

In essence, solving this system means finding the points where the line and the circle intersect. These intersection points are the (x, y) pairs that satisfy both equations simultaneously. Imagine plotting the line and the circle on a graph; the points where they cross are our solutions. This gives us a visual way to think about the problem, which can be really helpful.

To tackle this, we're going to use a method called substitution. It involves solving one equation for one variable and plugging that expression into the other equation. This will give us a single equation with one variable, which we can then solve. Once we have the value(s) for that variable, we can substitute back into either of the original equations to find the corresponding value(s) for the other variable. It sounds like a lot, but trust me, it’s a methodical process that breaks down the problem into manageable steps.

Solving for y in the Linear Equation

Alright, let's start by tackling the linear equation: 2y - 3x = 6. Our goal here is to isolate y on one side of the equation. This will allow us to express y in terms of x, which we can then substitute into the circle equation. It’s like preparing one piece of the puzzle so it fits nicely into the other.

Here’s how we do it step-by-step:

1. Add 3x to both sides: This gets rid of the -3x term on the left side, moving it to the right side of the equation. We now have 2y = 3x + 6. Remember, we need to keep the equation balanced, so whatever we do to one side, we do to the other.
2. Divide both sides by 2: This isolates y on the left side. Dividing both sides by 2 gives us y = (3/2)x + 3. Ta-da! We've successfully solved for y.

So, we now have y expressed in terms of x: y = (3/2)x + 3. This is a crucial step because it allows us to replace y in the circle equation with this expression. Think of it as translating from one language to another; we're rewriting y in a way that the circle equation can understand.

Now that we have y in terms of x, we’re ready to substitute this expression into the circle equation. This is where the magic happens – we'll be combining the two equations into one, making it solvable. By doing this, we’re essentially finding the points where the line and the circle meet, which are the solutions to our system of equations. This substitution step is a powerful technique in solving systems, and it's all about making the equations work together.

Substituting into the Circle Equation

Now comes the exciting part where we combine our equations! We have y = (3/2)x + 3 from the linear equation, and we’re going to substitute this into the circle equation, x² + y² = 9. This step is like fitting the pieces of a puzzle together, and it’s where we start to see the solution take shape.

Here’s how the substitution works:

1. Replace y in the circle equation: We substitute (3/2)x + 3 for y in the equation x² + y² = 9. This gives us x² + ((3/2)x + 3)² = 9. Notice how we’ve replaced y with an entire expression involving x. This is the key to reducing the system to a single equation with one variable.
2. Expand the squared term: We need to expand ((3/2)x + 3)². Remember, (a + b)² = a² + 2ab + b². Applying this to our expression, we get ((3/2)x + 3)² = (9/4)x² + 9x + 9. This expansion is crucial because it allows us to combine like terms and simplify the equation.

After expanding, our equation looks like this: x² + (9/4)x² + 9x + 9 = 9. We've now successfully transformed the system into a single equation that only involves x. This is a major step forward because we can use algebraic techniques to solve for x. It might look a bit intimidating at first, but don't worry, we'll break it down and simplify it in the next step.

The next step is to simplify this equation and solve for x. This will involve combining like terms and rearranging the equation into a standard form that we can work with. Once we’ve found the values of x, we can plug them back into our expression for y to find the corresponding y values. This process of substitution is a powerful way to solve systems of equations, and it’s a fundamental technique in algebra and beyond.

Simplifying and Solving for x

Okay, let's tackle the equation we got after the substitution: x² + (9/4)x² + 9x + 9 = 9. It might look a bit messy, but don't worry, we'll clean it up step by step. Our goal here is to simplify the equation and get it into a form where we can easily solve for x. Think of it as decluttering a room – once everything is in its place, it's much easier to see what you're working with.

Here’s how we simplify and solve for x:

1. Combine like terms: We have x² and (9/4)x² terms. To combine them, we need a common denominator. We can rewrite x² as (4/4)x². So, (4/4)x² + (9/4)x² = (13/4)x². Our equation now looks like (13/4)x² + 9x + 9 = 9.
2. Subtract 9 from both sides: This simplifies the equation further by getting rid of the constant term on the right side. Subtracting 9 from both sides gives us (13/4)x² + 9x = 0. Now, we have a quadratic equation that’s set equal to zero, which is a great starting point for solving.
3. Factor out x: Notice that both terms on the left side have x in them. We can factor out an x to simplify the equation. Factoring out x gives us x((13/4)x + 9) = 0. This is a crucial step because it allows us to 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.
4. Apply the zero-product property: We have two factors: x and ((13/4)x + 9). Setting each factor equal to zero gives us two separate equations: x = 0 and (13/4)x + 9 = 0. These are much simpler equations to solve.
5. Solve for x in each equation:
   • For x = 0, we already have one solution: x = 0.
   • For (13/4)x + 9 = 0, we subtract 9 from both sides to get (13/4)x = -9. Then, we multiply both sides by 4/13 to isolate x. This gives us x = -36/13.

So, we’ve found two possible values for x: x = 0 and x = -36/13. These are the x-coordinates of the points where the line and the circle intersect. But we’re not done yet! We still need to find the corresponding y values for each of these x values. To do that, we’ll substitute each x value back into our expression for y that we found earlier.

Finding these x values is a significant achievement. We've taken a complex system of equations and broken it down into manageable steps. Now that we have the x values, we’re on the home stretch. The next step is to find the corresponding y values, which will give us the complete solutions to our system of equations. This process of simplifying and factoring is a testament to the power of algebraic techniques, and it's a skill that will serve you well in many mathematical contexts.

Finding the Corresponding y Values

Great job, guys! We've found the x values where the line and circle intersect: x = 0 and x = -36/13. Now, to complete our solution, we need to find the y values that go with each of these x values. Think of it like finding the matching piece to a puzzle; we have half the coordinate, and now we need to find the other half.

To find the corresponding y values, we'll use the equation we derived earlier: y = (3/2)x + 3. This equation expresses y in terms of x, making it perfect for this task. We'll substitute each x value into this equation and solve for y. It’s a straightforward process, but it’s a crucial step to ensure we have the complete solution.

Here’s how we find the y values:

1. Substitute x = 0:
   • Plug x = 0 into y = (3/2)x + 3.
   • y = (3/2)(0) + 3
   • y = 0 + 3
   • y = 3
   • So, when x = 0, y = 3. This gives us one solution point: (0, 3).
2. Substitute x = -36/13:
   • Plug x = -36/13 into y = (3/2)x + 3.
   • y = (3/2)(-36/13) + 3
   • y = -54/13 + 3
   • To add these, we need a common denominator. We can rewrite 3 as 39/13.
   • y = -54/13 + 39/13
   • y = -15/13
   • So, when x = -36/13, y = -15/13. This gives us another solution point: (-36/13, -15/13).

We've now found two sets of (x, y) values that satisfy both the linear equation and the circle equation. These are the points where the line and the circle intersect, and they represent the solutions to our system of equations. It’s like we’ve successfully navigated a maze and found the hidden treasure!

Finding these y values completes our solution. We’ve taken the x values we found earlier and used them to determine the corresponding y values. This process of substitution is a powerful tool in algebra, and it allows us to connect the variables in our equations and find the points that satisfy all the conditions. Now that we have the complete solutions, let’s take a moment to summarize our findings and present the final answer.

Summarizing the Solutions

Alright, let’s recap what we’ve done and present our final answer. We started with a system of equations: a linear equation (2y - 3x = 6) and a circle equation (x² + y² = 9). Our goal was to find the points (x, y) that satisfy both equations simultaneously. We've gone through a step-by-step process of substitution, simplification, and solving, and now we have our solutions. It’s like we’ve completed a journey, and now we can look back and see the path we took to get here.

Here’s a summary of the steps we took:

1. Solved the linear equation for y: We rewrote 2y - 3x = 6 as y = (3/2)x + 3.
2. Substituted into the circle equation: We replaced y in x² + y² = 9 with (3/2)x + 3, giving us x² + ((3/2)x + 3)² = 9.
3. Simplified and solved for x: We simplified the equation and found two solutions for x: x = 0 and x = -36/13.
4. Found the corresponding y values: We substituted each x value back into y = (3/2)x + 3 to find the corresponding y values: y = 3 when x = 0, and y = -15/13 when x = -36/13.

So, the solutions to the system of equations are the points (0, 3) and (-36/13, -15/13). These are the coordinates where the line and the circle intersect. If you were to graph these equations, you would see that the line and the circle cross at these two points. This gives us a visual confirmation that our solutions are correct.

To put it all together, the solutions to the system of equations are:

• (0, 3)
• (-36/13, -15/13)

We’ve successfully solved a system of equations that combines a linear equation and a circle equation. This process involved several algebraic techniques, including substitution, simplification, factoring, and solving for variables. By breaking down the problem into smaller, manageable steps, we were able to find the solutions. This is a testament to the power of systematic problem-solving in mathematics.

Visualizing the Solution

To really drive home what we’ve accomplished, let’s talk about visualizing the solution. We’ve found the points (0, 3) and (-36/13, -15/13) as the solutions to our system of equations. But what does this actually mean in terms of the graphs of the equations? Well, guys, it’s like this: these points are where the line and the circle literally intersect on the coordinate plane.

Imagine plotting the line 2y - 3x = 6 and the circle x² + y² = 9 on the same graph. The line is, well, a straight line, and the circle is a circle centered at the origin with a radius of 3. The points where these two shapes cross each other are the solutions to our system. It's a beautiful visual representation of what we’ve calculated algebraically.

• The point (0, 3) is easy to spot. It's where the line crosses the y-axis. If you plug x = 0 and y = 3 into both equations, you'll see that they both hold true.
• The point (-36/13, -15/13) is a bit trickier to visualize exactly, as it involves fractions. But it's located in the third quadrant, where both x and y are negative. If you were to zoom in on the graph, you’d see that this point also lies on both the line and the circle.

Visualizing the solution helps to solidify our understanding. It’s not just about manipulating equations and finding numbers; it’s about understanding the geometric interpretation of those numbers. When we solve a system of equations, we’re finding the points that satisfy all the equations simultaneously. Graphically, this means we’re finding the points where the graphs of the equations intersect. This connection between algebra and geometry is a fundamental concept in mathematics, and it’s one that’s worth grasping.

Thinking about the graphs of the equations can also give us insights into the nature of the solutions. For example, if the line and the circle didn't intersect at all, we would have no real solutions. If the line touched the circle at only one point, we would have one solution. In our case, the line intersects the circle at two points, which means we have two solutions. This visual perspective can help us anticipate and interpret our algebraic results.

Tips and Tricks for Solving Systems of Equations

Before we wrap up, let’s go over some tips and tricks that can help you tackle systems of equations in general. These are like secret weapons in your mathematical arsenal, and they can make the solving process much smoother and more efficient. These tips will help you approach any system of equations with confidence, whether it involves lines, circles, or more complex curves.

1. Choose the Right Method: Substitution is just one method for solving systems of equations. Others include elimination and graphing. Sometimes, one method is clearly better suited to a particular problem than others. For example, if one of the equations is already solved for a variable, substitution is often the way to go. If the equations are in a form where the coefficients of one variable are opposites, elimination might be easier. And graphing can give you a visual sense of the solutions, although it might not always give you exact answers.
2. Look for Simplifications: Before you start substituting or eliminating, take a moment to see if you can simplify the equations. Can you divide both sides of an equation by a common factor? Can you combine like terms? Simplifying the equations first can make the subsequent steps much easier.
3. Check Your Solutions: This is crucial! After you’ve found your solutions, plug them back into the original equations to make sure they work. This will catch any algebraic errors you might have made along the way. It’s like proofreading a document before you submit it; it’s a final check to ensure everything is correct.
4. Be Organized: Solving systems of equations can involve a lot of steps, so it’s important to stay organized. Write down each step clearly, and label your equations. This will help you keep track of what you’ve done and make it easier to spot any mistakes. Think of it as creating a roadmap for your solution; a clear roadmap makes the journey much smoother.
5. Practice, Practice, Practice: Like any skill, solving systems of equations gets easier with practice. The more problems you solve, the more comfortable you’ll become with the different techniques and the more quickly you’ll be able to spot the best approach. It’s like learning a musical instrument; the more you play, the better you get.

By keeping these tips in mind, you’ll be well-equipped to solve a wide range of systems of equations. Remember, the key is to break down the problem into smaller, manageable steps, and to stay organized and methodical in your approach.

Conclusion

Well, guys, we’ve reached the end of our journey through solving systems of equations! We’ve tackled a system involving a linear equation and a circle equation, and we’ve successfully found the solutions. We’ve covered a lot of ground, from understanding the equations to visualizing the solutions and learning tips and tricks for solving systems in general. It’s like we’ve climbed a mountain, and now we can stand at the top and admire the view.

We started by breaking down the problem into smaller, manageable steps. We solved the linear equation for y, substituted that expression into the circle equation, simplified the resulting equation, solved for x, and then found the corresponding y values. This step-by-step approach is a powerful strategy for tackling complex problems in mathematics and beyond. It's a reminder that even the most daunting tasks can be accomplished by breaking them down into smaller, more manageable pieces.

We also emphasized the importance of visualizing the solutions. By thinking about the graphs of the equations, we gained a deeper understanding of what it means to solve a system of equations. We saw that the solutions are the points where the graphs intersect, which provides a visual confirmation of our algebraic results. This connection between algebra and geometry is a recurring theme in mathematics, and it's a valuable perspective to cultivate.

And we shared some tips and tricks for solving systems of equations in general. These tips included choosing the right method, looking for simplifications, checking your solutions, staying organized, and practicing regularly. These are all valuable strategies that can help you become a more confident and effective problem-solver.

Solving systems of equations is a fundamental skill in mathematics, and it has applications in many different fields, from physics and engineering to economics and computer science. By mastering this skill, you’re opening doors to a wide range of possibilities. It's not just about getting the right answer; it's about developing the problem-solving skills and the mathematical thinking that will serve you well in all aspects of life.

So, keep practicing, keep exploring, and keep pushing your mathematical boundaries. You’ve got this! And remember, the journey of learning mathematics is a marathon, not a sprint. There will be challenges along the way, but with persistence and the right strategies, you can overcome them and achieve your goals. Keep up the great work, and I’ll see you in the next mathematical adventure!