Solving Systems Of Equations: Step-by-Step Guide

Hey guys! Math can sometimes feel like deciphering a secret code, especially when we're dealing with systems of equations. But don't worry, we're going to break down these problems step by step, making them super easy to understand. We'll tackle some word problems and then dive into different methods for solving systems of equations. So grab your pencils, and let's get started!

Unraveling the Mystery of Systems of Equations

First off, what exactly is a system of equations? Well, it's simply a set of two or more equations that share the same variables. The goal is to find the values of those variables that make all the equations true simultaneously. Think of it like solving a puzzle where all the pieces need to fit perfectly.

Now, let's jump into our first problem – a real-world scenario involving money! These types of problems are super common, and once you get the hang of them, you'll feel like a math wizard. We will be using different approaches to solve systems of equations, and we will go through examples in detail. Each method has its pros and cons, and choosing the right one can make the solution process much smoother. We will be focusing on substitution, elimination, and graphical methods.

Problem 1: The Case of the Cash-Filled Bag

Imagine this: "In a bag, there's a total of R$ 640.00, made up of R$ 10.00 and R$ 50.00 bills. If there are 24 bills in total, how many of each type are there?"

This might seem tricky at first, but let's break it down. The key here is to translate the words into mathematical equations. This is a crucial skill in algebra, guys, so pay close attention! We will start by defining our variables. Let's dive deep into how we can approach such problems using systems of equations.

• Step 1: Define our variables Let's use 'x' to represent the number of R$ 10.00 bills and 'y' to represent the number of R$ 50.00 bills. This is always the first step in solving word problems – turning the unknowns into variables. Defining variables clearly helps to avoid confusion and sets the stage for building our equations. Remember, guys, clarity is key in math!
• Step 2: Formulate the equations Now, we need to translate the information from the problem into equations. We know two things:
  • The total value of the bills is R$ 640.00.
  • The total number of bills is 24. So, we can write these as:
  • 10x + 50y = 640 (The value equation: R$ 10 times the number of R$ 10 bills plus R$ 50 times the number of R$ 50 bills equals the total value)
  • x + y = 24 (The quantity equation: the number of R$ 10 bills plus the number of R$ 50 bills equals the total number of bills) See how we turned the word problem into a system of equations? This is a super powerful technique! Now we have a set of equations that we can solve using different methods. We are now ready to explore different techniques to solve this system of equations, including substitution, elimination, and even graphical methods. The choice of method often depends on the specific problem and personal preference, so understanding each approach is beneficial.
• Step 3: Solve the system of equations We have a couple of ways to tackle this. Let's use the substitution method first.
  • Solve the second equation for x: x = 24 - y
  • Substitute this expression for x into the first equation: 10(24 - y) + 50y = 640
  • Simplify and solve for y: 240 - 10y + 50y = 640 => 40y = 400 => y = 10
  • Now, substitute the value of y back into the equation x = 24 - y: x = 24 - 10 => x = 14 Substitution involves solving one equation for one variable and then substituting that expression into the other equation. This simplifies the problem to a single equation with one variable, which is easier to solve. Remember, the goal is to isolate one variable and find its value. This method is particularly effective when one equation can be easily solved for one variable in terms of the other, making the substitution process straightforward.
• Step 4: State the answer We found that x = 14 and y = 10. That means there are 14 R$ 10.00 bills and 10 R$ 50.00 bills. We've cracked the case! Always remember to state your answer in the context of the problem. It's not just about finding x and y; it's about answering the original question. A clear answer shows that you understand what the numbers represent and have successfully solved the problem. Let's look into another method we can use to solve systems of equations: Elimination.

Problem 2: Tackling Systems of Equations with Different Methods

Now, let's move on to the next part of our challenge: solving systems of equations using different methods. There are several ways to skin this cat, guys, and it's good to know a few different approaches.

Here are the systems we need to solve:

• a) x + y = 8 x - y = 6
• b) (The problem statement for 'b' is missing, so I'll demonstrate the methods with a general example)

We'll explore two primary methods: the elimination method and, briefly, the substitution method (which we already saw in action!). Understanding both methods gives you a powerful toolkit for solving systems of equations. The choice of method often depends on the specific problem, and sometimes one method is clearly more efficient than the other. For instance, elimination is particularly useful when the coefficients of one variable are opposites or easy to make opposites, while substitution is great when one variable is already isolated or can be easily isolated. Let's delve deeper into the elimination method and see how it works.

The Elimination Method

The elimination method is all about strategically adding or subtracting the equations to eliminate one of the variables. It's like magic, watching a variable disappear! This method is super effective when the coefficients of one variable are the same or opposites, or when they can easily be made that way by multiplying one or both equations by a constant.

Let's apply this to system 'a':

• x + y = 8 x - y = 6

Notice that the 'y' terms have opposite signs. That's perfect for elimination! This is a key indicator that elimination will work well. We want to look for opportunities where adding or subtracting the equations will directly cancel out one variable, making the problem simpler. Sometimes, you might need to multiply one or both equations by a constant to create matching coefficients with opposite signs before you can eliminate a variable. This is a powerful technique that greatly expands the applicability of the elimination method.

• Step 1: Add the equations If we add the two equations together, the 'y' terms will cancel out: (x + y) + (x - y) = 8 + 6 2x = 14 See how the 'y' vanished? That's the power of elimination! By carefully adding the equations, we've created a simpler equation with only one variable, which is much easier to solve. This strategic cancellation is the core idea behind the elimination method, and it's what makes it so efficient for certain types of systems of equations.
• Step 2: Solve for x Divide both sides by 2: x = 7
• Step 3: Substitute the value of x back into either original equation to solve for y Let's use the first equation: 7 + y = 8 Subtract 7 from both sides: y = 1
• Step 4: State the solution The solution to the system is x = 7 and y = 1. We've done it! Just like with the word problem, it's important to clearly state the solution as a pair of values, one for each variable. This makes it clear that you've solved for all unknowns and provides a complete answer to the problem. Remember to always double-check your solution by substituting the values back into the original equations to ensure they hold true. This verification step is crucial to avoid errors and gain confidence in your answer.

Let's say we had a system like this for 'b' (since the original problem statement is missing):

• 2x + 3y = 13
• x - y = -1

In this case, we can't directly add or subtract the equations to eliminate a variable. But don't worry, we have a trick up our sleeves! Here, we need to manipulate the equations before we can eliminate a variable. The key is to find a way to make the coefficients of either x or y opposites or identical. This often involves multiplying one or both equations by a constant. The goal is to create a situation where adding or subtracting the equations will cancel out one of the variables, leading us closer to the solution.

• Step 1: Multiply the second equation by 3 This will give us a '3y' term that we can eliminate: 3(x - y) = 3(-1) 3x - 3y = -3 By multiplying the entire equation, we maintain the equality and set up the elimination of the 'y' variable. This is a crucial step in the elimination method when the coefficients don't directly allow for cancellation. Remember, whatever operation you perform on one side of the equation, you must also perform on the other side to keep the equation balanced.
• Step 2: Add the modified second equation to the first equation Now we can add the equations: (2x + 3y) + (3x - 3y) = 13 + (-3) 5x = 10 Perfect! The 'y' terms canceled out, leaving us with a simple equation in terms of 'x'. This is the desired outcome of the elimination method – reducing the system to a single equation with one unknown. The strategic manipulation of the equations has paid off, and we're now well on our way to solving for the variables.
• Step 3: Solve for x Divide both sides by 5: x = 2
• Step 4: Substitute the value of x back into either original equation to solve for y Let's use the second equation: 2 - y = -1 Subtract 2 from both sides: -y = -3 Multiply both sides by -1: y = 3
• Step 5: State the solution The solution to this system is x = 2 and y = 3. Great job! We have successfully used the elimination method to solve another system of equations. The elimination method shines when the coefficients of one variable are either the same or opposites, or when they can easily be made so through multiplication. This method simplifies the system by eliminating one variable, making it easier to solve for the remaining variable. Remember to always check your solution by substituting the values back into the original equations to ensure they satisfy both equations.

A Quick Look at Substitution (Again!)

We used substitution in the first word problem, but let's recap the general idea. Substitution is excellent when one equation is already solved for a variable or can be easily solved. We solve one equation for one variable and then substitute that expression into the other equation.

Mastering Systems of Equations: Your Journey Continues

So there you have it, guys! We've explored how to tackle systems of equations, both from word problems and in their pure algebraic form. We've seen how to translate real-world scenarios into equations and how to use both substitution and elimination methods to find the solutions. This is a fundamental skill in math, and with practice, you'll become a pro at solving these problems.

Remember, the key to mastering systems of equations is practice. Work through different types of problems, try different methods, and don't be afraid to make mistakes – that's how we learn! Keep practicing, and you'll be solving systems of equations like a boss in no time. Good luck, and happy problem-solving!