Solving Systems Of Equations By Substitution: A Step-by-Step Guide
Hey Plastik Magazine readers! Today, we're diving into the fascinating world of systems of equations and learning how to solve them using a method called substitution. If you've ever felt a little lost when faced with multiple equations and variables, don't worry; we're here to break it down in a way that's super easy to understand. We'll walk through the process step-by-step, and by the end, you'll be a substitution pro! So, let's jump right in and make math a little less intimidating and a lot more fun, shall we?
Understanding Systems of Equations
Before we dive into the substitution method, let's make sure we're all on the same page about what a system of equations actually is. Simply put, a system of equations is a set of two or more equations that share the same variables. The goal when solving a system of equations is to find the values for those variables that make all the equations true simultaneously. Think of it like finding the perfect combination that unlocks all the equations at once! For instance, in our example, we have two equations: a - 1.2b = -3 and 0.2b + 0.6a = 12. We need to find values for a and b that satisfy both of these equations. There are several methods to solve these systems, and today we are focusing on the substitution method. The beauty of systems of equations lies in their versatility. They pop up in all sorts of real-world scenarios, from calculating the break-even point for a business to determining the optimal mix of ingredients in a recipe. So, mastering the art of solving them is a seriously valuable skill.
What is the Substitution Method?
The substitution method is a clever technique for solving systems of equations by, you guessed it, substituting! The main idea behind this method involves solving one equation for one variable and then substituting that expression into the other equation. This effectively eliminates one variable, leaving us with a single equation that we can easily solve. Once we've found the value of one variable, we can then substitute it back into either of the original equations to find the value of the other variable. It's like a mathematical relay race, where we pass the baton of information from one equation to the next. The substitution method is particularly useful when one of the equations is already solved for a variable or can be easily manipulated to do so. This makes the process smoother and more efficient. It's all about finding the path of least resistance and using the information at hand to our advantage. In essence, the substitution method transforms a tricky problem with two variables into a simpler one with just one, making it much more manageable. This is a core concept in algebra and will be helpful in more advanced mathematical applications as well.
Step-by-Step Solution Using Substitution
Alright, let's get down to business and solve the system of equations using the substitution method. We'll break it down into manageable steps so you can follow along easily. Remember, practice makes perfect, so don't hesitate to try this out with other problems too! Now, let’s take a look at our system of equations once more:
a - 1.2b = -3
0.2b + 0.6a = 12
Step 1: Solve One Equation for One Variable
The first move in our substitution strategy is to pick one of the equations and solve it for one of the variables. The goal here is to isolate a variable on one side of the equation. Looking at our system, the first equation, a - 1.2b = -3, seems like a good candidate because it's relatively easy to solve for a. To isolate a, we simply add 1.2b to both sides of the equation. This gives us:
a = 1.2b - 3
Now we have a expressed in terms of b. This is a crucial step because we can now substitute this expression into the other equation. If we had chosen the second equation, 0.2b + 0.6a = 12, solving for either a or b would involve a few more steps. That's why it's often a good idea to look for the easiest path forward. By solving for a in the first equation, we've set ourselves up for a smoother substitution process. This is a key skill in problem-solving: spotting the most efficient approach.
Step 2: Substitute the Expression into the Other Equation
Now that we've solved the first equation for a, giving us a = 1.2b - 3, it's time to put that knowledge to use in the second equation. This is where the substitution magic happens! We're going to take the expression we found for a and plug it into the second equation, 0.2b + 0.6a = 12. Wherever we see a in the second equation, we'll replace it with (1.2b - 3). This gives us:
0. 2b + 0.6(1.2b - 3) = 12
Notice how we've now transformed the second equation into one that involves only the variable b. This is a major victory because we can now solve for b directly. This step is the heart of the substitution method – it's where we eliminate one variable and simplify the problem. By substituting, we've turned a two-variable equation into a single-variable equation, which is much easier to handle. Remember, the goal is to reduce complexity and make the problem solvable, and substitution is a powerful tool for doing just that.
Step 3: Solve for the Remaining Variable
Okay, we've made excellent progress! We've substituted and now we have an equation with just one variable: 0.2b + 0.6(1.2b - 3) = 12. Our next task is to solve for b. This involves a little bit of algebraic maneuvering, but don't worry, we'll take it step by step. First, we need to distribute the 0.6 across the terms inside the parentheses:
0. 2b + 0.72b - 1.8 = 12
Now, let's combine the like terms involving b:
0. 92b - 1.8 = 12
Next, we want to isolate the term with b, so we add 1.8 to both sides of the equation:
0. 92b = 13.8
Finally, to solve for b, we divide both sides by 0.92:
b = 13.8 / 0.92
b = 15
Woo-hoo! We've found the value of b: it's 15. This is a significant milestone in our substitution journey. We're halfway to solving the system of equations. Now that we know b, we can use that information to find the value of a. This is where we loop back and use our earlier work to complete the puzzle.
Step 4: Substitute Back to Find the Other Variable
Now that we know b = 15, it's time to find the value of a. Remember that earlier, we solved the first equation for a and got a = 1.2b - 3. This is exactly what we need to find a! We simply substitute the value of b into this equation:
a = 1.2(15) - 3
Now, let's do the math:
a = 18 - 3
a = 15
So, we've found that a = 15 as well. Fantastic! We now have values for both a and b. This is the culmination of our substitution efforts. We've untangled the system of equations and found the solution. But before we celebrate too much, there's one crucial step left: checking our solution. This is our way of making sure we haven't made any mistakes along the way and that our solution truly works for the original system of equations.
Step 5: Check Your Solution
We've got our solution: a = 15 and b = 15. But before we declare victory, it's super important to check our work. This ensures we didn't make any sneaky errors along the way. To check our solution, we'll substitute these values back into both of the original equations and see if they hold true. Let's start with the first equation, a - 1.2b = -3:
15 - 1.2(15) = -3
15 - 18 = -3
-3 = -3
Great! The first equation checks out. Now, let's plug our values into the second equation, 0.2b + 0.6a = 12:
0. 2(15) + 0.6(15) = 12
3 + 9 = 12
12 = 12
Awesome! The second equation also holds true. Since our values for a and b satisfy both equations, we can confidently say that our solution is correct. Checking our solution is like putting a lock on our answer – it gives us the assurance that we've cracked the code correctly. This step is often overlooked, but it's a crucial part of the problem-solving process. It's always better to be safe than sorry, especially in math!
The Solution
After all our hard work, we've successfully solved the system of equations using the substitution method. Our solution is a = 15 and b = 15. This means the correct answer is A. (15, 15). We've not only found the solution but also verified it, giving us complete confidence in our answer. Remember, solving systems of equations is a fundamental skill in algebra, and mastering the substitution method opens doors to tackling more complex problems. Keep practicing, and you'll become a pro at solving these types of questions.
Why is this the correct answer?
It's worth emphasizing why (15, 15) is indeed the correct solution. We've shown that when we substitute a = 15 and b = 15 into both original equations, they both hold true. This is the very definition of a solution to a system of equations – it's the set of values that makes all the equations in the system true simultaneously. Other options, like B. (10, 12), C. (13, 12), and D. (7, 9), might satisfy one of the equations, but they won't satisfy both. This is why it's crucial to check your solution in both equations. The process of substitution led us to this unique solution, and our verification step confirmed its accuracy. This highlights the importance of a systematic approach and the value of checking your work in mathematics. It's not just about getting an answer; it's about understanding why that answer is correct.
Practice Makes Perfect
So, there you have it, guys! We've walked through the process of solving a system of equations using the substitution method. Remember, the key is to break it down into manageable steps: solve for one variable, substitute, solve for the remaining variable, and then substitute back to find the other variable. And most importantly, always check your solution! Solving systems of equations is a fundamental skill in math, and the more you practice, the better you'll get. So, grab some practice problems and give it a try. You'll be a substitution superstar in no time! Keep shining, mathletes! See you in the next breakdown.