Elimination Method: Solving Linear Equations

Hey guys! Today, we're diving deep into solving systems of linear equations using the elimination method. This is a super handy technique, and we'll explore a specific scenario to understand why it works so well. Let's break it down step by step.

Understanding the Elimination Method

The elimination method, also known as the addition method, is a technique used to solve systems of linear equations by eliminating one of the variables. The primary goal is to manipulate the equations so that when you add or subtract them, one of the variables cancels out, leaving you with a single equation in one variable. This resulting equation can then be easily solved. Once you find the value of one variable, you can substitute it back into one of the original equations to find the value of the other variable.

The beauty of the elimination method lies in its efficiency and straightforward approach. It's particularly useful when the coefficients of one of the variables in the equations are either the same or simple multiples of each other. This makes it easy to create opposite coefficients through multiplication, which is the key to eliminating a variable. For example, if you have the equations $2x + 3y = 7$ and $4x - 3y = 5$, you can simply add the two equations to eliminate the $y$ variable because the coefficients of $y$ are already opposites (+3 and -3). This results in the equation $6x = 12$, which can be easily solved for $x$.

The elimination method is not just a mathematical trick; it's a systematic way to simplify complex systems of equations. It provides a clear and organized approach, reducing the chances of errors and making the process more manageable. Whether you're dealing with simple two-variable systems or more complex systems with multiple variables, the elimination method can be adapted to suit your needs. It's a valuable tool in any mathematician's or problem-solver's toolkit.

The Given System of Equations

Let's consider the system of equations we're working with:

Equation I: $4x - 5y = 4$ Equation II: $2x + 3y = 2$

When we look at these equations, we want to determine the best approach for solving them. The elimination method is often a great choice, but why is it particularly suitable here? What makes it stand out compared to other methods like substitution?

The key to understanding why elimination is a good fit lies in the coefficients of the variables. Notice that the coefficient of $x$ in Equation I (which is 4) is an integer multiple of the coefficient of $x$ in Equation II (which is 2). Specifically, $4 = 2 eq 2$. This relationship makes it easy to manipulate Equation II so that the $x$ terms in both equations will cancel each other out when we perform addition or subtraction.

To elaborate, we can multiply Equation II by -2. This gives us a new equation:

$-2 eq (2x + 3y) = -2 eq 2$ which simplifies to $-4x - 6y = -4$.

Now, we have:

Equation I: $4x - 5y = 4$ Modified Equation II: $-4x - 6y = -4$

When we add these two equations, the $x$ terms cancel out:

$(4x - 5y) + (-4x - 6y) = 4 + (-4)$

This simplifies to:

$-11y = 0$

From here, it's a breeze to solve for $y$: $y = 0$.

Once we have the value of $y$, we can substitute it back into either Equation I or Equation II to find the value of $x$. For example, using Equation II:

$2x + 3(0) = 2$ $2x = 2$ $x = 1$

So, the solution to the system of equations is $x = 1$ and $y = 0$.

Why Elimination Works Here: A Detailed Explanation

The primary reason why the elimination method is effective for this system of equations is the integer multiple relationship between the coefficients of $x$ in the two equations. Let's dive deeper into why this is significant.

When one coefficient is an integer multiple of another, it allows us to easily create opposite coefficients by multiplying one of the equations by a suitable constant. In our case, since 4 (the coefficient of $x$ in Equation I) is twice 2 (the coefficient of $x$ in Equation II), we can multiply Equation II by -2 to get -4 as the coefficient of $x$. This creates the additive inverse we need to eliminate $x$ when we add the two equations together.

This is much simpler than, say, trying to eliminate $y$. To eliminate $y$, we'd have to find a common multiple of 5 and 3, which is 15. This would require multiplying Equation I by 3 and Equation II by 5, resulting in larger coefficients and more complex calculations. While it's certainly doable, it's not as efficient as simply multiplying Equation II by -2.

The efficiency of the elimination method in this scenario translates to less work and a lower chance of making errors. By targeting the variable with coefficients that are integer multiples, we streamline the process and make it more manageable. This is a key consideration when choosing the best method for solving a system of equations.

Furthermore, this approach highlights the strategic thinking involved in solving systems of equations. It's not just about blindly applying a method; it's about analyzing the equations and choosing the method that will be most efficient and effective. Recognizing the integer multiple relationship between coefficients is a valuable skill that can save time and effort.

Comparing Elimination with Substitution

Now, let's briefly compare the elimination method with another common technique: substitution. While substitution can be used to solve any system of equations, it's not always the most efficient choice. In our specific scenario, elimination has a clear advantage.

In the substitution method, you solve one equation for one variable and then substitute that expression into the other equation. For example, we could solve Equation II for $x$:

$2x + 3y = 2$ $2x = 2 - 3y$ $x = 1 - "(3/2)y"$

Then, we would substitute this expression for $x$ into Equation I:

$4(1 - "(3/2)y") - 5y = 4$

This simplifies to:

$4 - 6y - 5y = 4$ $-11y = 0$ $y = 0$

While this works, notice that we introduced a fraction when solving for $x$. Dealing with fractions can make the calculations more complex and increase the likelihood of errors. In contrast, the elimination method allowed us to avoid fractions altogether, making the process cleaner and simpler.

In general, substitution is often a better choice when one of the equations is already solved for one variable or when it's easy to isolate a variable without introducing fractions. However, when the coefficients have a convenient relationship like an integer multiple, elimination tends to be the more efficient option.

Conclusion

So, to wrap things up, the valid reason for using the elimination method to solve the given system of equations is that a coefficient in Equation I (4) is an integer multiple of a coefficient in Equation II (2). This relationship allows for easy manipulation to eliminate a variable and simplify the solution process. Choosing the right method can save you time and reduce the risk of errors, making problem-solving a whole lot smoother. Keep practicing, and you'll become a pro at solving systems of equations in no time! Keep it real!