Solving Linear Equations: A Step-by-Step Discussion

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Hey guys! Let's dive into the fascinating world of linear equations! Today, we're going to tackle a system of linear equations, break it down step-by-step, and discuss the methods we can use to find a solution. Think of it like solving a puzzle – we have a set of clues (equations) and we need to figure out the values of the unknowns (variables). So, grab your thinking caps, and let’s get started!

The System of Equations

First, let's take a look at the system of linear equations we're going to work with. It's crucial to understand what we're dealing with before we jump into solving it. We have three equations, each containing three variables: a, b, and c. Our goal is to find the values of these variables that satisfy all three equations simultaneously. Think of it like finding a secret code that unlocks all the locks at once!

Here’s the system:

  • a + b + c = 12
  • 3a - b + 2c = 14
  • 2a - 2b + c = -3

Each of these equations represents a plane in a 3D space. The solution to the system, if it exists, is the point where all three planes intersect. This intersection point gives us the values of a, b, and c that work for all equations. Systems of linear equations like this pop up in all sorts of real-world scenarios, from engineering and physics to economics and computer science. For example, they can help us model the flow of electricity in a circuit, balance chemical equations, or optimize resource allocation. So, understanding how to solve them is a valuable skill!

Before we dive into specific methods, let's think about what a solution means. A solution is a set of values for a, b, and c that, when plugged into each equation, make the equation true. If we find such values, we've cracked the code! If not, it might mean the system is inconsistent – like a puzzle with no solution.

Methods for Solving Linear Equations

Now, let's talk about the methods we can use to solve this system. There are several tried-and-true techniques, each with its own strengths and weaknesses. The choice of method often depends on the specific system we're facing and our personal preference. We'll explore a couple of the most common and versatile methods: substitution and elimination (also sometimes called Gaussian elimination).

1. Substitution Method

The substitution method is like a detective's strategy – we isolate one variable in one equation and then substitute its expression into the other equations. This reduces the number of variables in those equations, making them easier to solve. It's like narrowing down the suspects in a crime case!

Here's the basic idea:

  1. Choose one equation and solve it for one of the variables. For example, if we have an equation like x + y = 5, we could solve for x to get x = 5 - y.
  2. Substitute the expression you found in step 1 into the other equations. This will eliminate one variable from those equations.
  3. You'll now have a system with one fewer equation and one fewer variable. Solve this new system.
  4. Once you've found the value(s) of the remaining variable(s), substitute them back into the expression you found in step 1 to find the value of the variable you initially isolated.

Let's think about applying this to our system. Looking at the equations, the first one (a + b + c = 12) seems like a good candidate for substitution. We could easily solve it for a, b, or c. The key is to choose the variable that will make the subsequent substitutions as simple as possible. This might take a bit of trial and error, but with practice, you'll get a feel for it!

2. Elimination Method

The elimination method, also known as Gaussian elimination, is like a strategic game of subtraction. We manipulate the equations to eliminate one variable at a time by adding or subtracting multiples of the equations. Think of it like strategically removing pieces in a game to reveal the final solution.

Here's how it works:

  1. Look for pairs of equations where the coefficients of one variable are either the same or opposites. If they're not, multiply one or both equations by constants to make them so.
  2. Add or subtract the equations to eliminate that variable. If the coefficients are the same, subtract. If they're opposites, add.
  3. You'll now have a new equation with one fewer variable. Repeat steps 1 and 2 with different pairs of equations until you have a system that's easier to solve.
  4. Solve the simplified system. You might end up with a single equation with one variable, which is easy to solve.
  5. Substitute the values you find back into the previous equations to find the values of the other variables.

The elimination method is particularly powerful for larger systems of equations, as it provides a systematic way to reduce the complexity. It's like having a well-organized strategy for tackling a big problem!

Think about our system again. Notice how the b terms in the first two equations (a + b + c = 12 and 3a - b + 2c = 14) have opposite coefficients (+1 and -1). This makes them perfect candidates for elimination! If we add these two equations together, the b terms will cancel out, leaving us with an equation involving only a and c. This is a great first step in simplifying the system.

Solving the System: A Practical Example

Okay, guys, enough theory! Let's roll up our sleeves and actually solve the system of equations we started with. We'll use a combination of the elimination and substitution methods to show you how it's done.

Here's the system again for easy reference:

  • a + b + c = 12 (Equation 1)
  • 3a - b + 2c = 14 (Equation 2)
  • 2a - 2b + c = -3 (Equation 3)

Step 1: Eliminate b from Equations 1 and 2

As we discussed earlier, the b terms in Equations 1 and 2 are perfectly set up for elimination. Let's add the two equations together:

(a + b + c) + (3a - b + 2c) = 12 + 14

This simplifies to:

4a + 3c = 26 (Equation 4)

Step 2: Eliminate b from Equations 1 and 3

Now, let's eliminate b from Equations 1 and 3. To do this, we need to make the coefficients of b opposites. We can multiply Equation 1 by 2:

2 * (a + b + c) = 2 * 12

This gives us:

2a + 2b + 2c = 24 (Equation 5)

Now, add Equation 3 and Equation 5:

(2a - 2b + c) + (2a + 2b + 2c) = -3 + 24

This simplifies to:

4a + 3c = 21 (Equation 6)

Step 3: Solve for a and c

Notice that we now have two equations (Equation 4 and Equation 6) that both involve a and c:

  • 4a + 3c = 26
  • 4a + 3c = 21

Wait a minute! This is interesting. We have the same expression on the left-hand side (4a + 3c) but different values on the right-hand side (26 and 21). This means the system is inconsistent! There is no solution that satisfies both equations simultaneously.

Discussion and Conclusion

So, what does it mean that our system is inconsistent? It means that the three planes represented by our original equations don't intersect at a single point. They might intersect in pairs, or they might not intersect at all. Think of it like trying to find a place where three roads all cross – sometimes it just doesn't exist!

This highlights an important point: not all systems of linear equations have solutions. Sometimes they are inconsistent, and sometimes they have infinitely many solutions (if the planes intersect along a line). The methods we've discussed can help us not only find solutions but also identify when solutions don't exist.

We walked through an example where we combined the elimination and substitution methods to tackle a system of three linear equations. We learned how to manipulate equations to eliminate variables and simplify the system. However, in this specific case, we stumbled upon an inconsistent system, meaning there's no single set of values for a, b, and c that satisfies all three equations. This is a valuable lesson – not all problems have a solution, and it's important to recognize when that's the case!

Understanding how to solve systems of linear equations is a crucial skill in many fields. Whether you're an engineer designing a bridge, an economist modeling market trends, or a computer scientist developing algorithms, you'll likely encounter these types of problems. The techniques we've discussed today – substitution and elimination – are powerful tools in your problem-solving arsenal. Keep practicing, guys, and you'll become masters of linear equations in no time!