Equivalent System Of Equations: A Detailed Solution

Hey guys! Today, we're diving into the fascinating world of systems of equations. Specifically, we're tackling a problem where we need to find a system equivalent to a given one. This type of question often pops up in algebra, and understanding how to manipulate equations is super useful. Let's break it down step-by-step!

The Original System

First, let's lay out the original system of equations that we need to work with:

$\begin{cases} 5x^2 + 6y^2 = 50 \\ 7x^2 + 2y^2 = 10 \end{cases}$

Our mission, should we choose to accept it (and we do!), is to find an equivalent system. That means we're looking for a new set of equations that has the exact same solutions as this one. There are several ways we can manipulate systems of equations to achieve this. Common methods include substitution, elimination, and multiplying equations by constants.

Understanding Equivalent Systems

Before we dive into the nitty-gritty, let's make sure we're all on the same page about what makes two systems equivalent. Two systems of equations are considered equivalent if they have the same solution set. In other words, any pair of (x, y) values that satisfy the first system must also satisfy the second system, and vice versa. We achieve equivalent systems by performing operations that don't change the solution set, like adding a multiple of one equation to another or multiplying an equation by a non-zero constant.

For example, consider a simple system:

$\begin{cases} x + y = 5 \\ x - y = 1 \end{cases}$

The solution to this system is x = 3 and y = 2. An equivalent system could be obtained by adding the two equations:

$\begin{cases} x + y = 5 \\ 2x = 6 \end{cases}$

The solution remains x = 3 and y = 2. This illustrates the principle that certain operations maintain the solution set, thus creating an equivalent system. Now, let's get back to the original problem and find an equivalent system.

Strategy: Elimination Method

One of the most straightforward approaches here is the elimination method. The goal is to manipulate one or both equations so that, when we add or subtract them, one of the variables gets eliminated. Looking at our system:

$\begin{cases} 5x^2 + 6y^2 = 50 \\ 7x^2 + 2y^2 = 10 \end{cases}$

Notice that we can eliminate the $y^2$ term if we multiply the second equation by -3. This will give us a $-6y^2$ term, which will cancel out the $+6y^2$ in the first equation. Let's do that!

Performing the Multiplication

Multiply the second equation by -3:

$-3 * (7x^2 + 2y^2) = -3 * 10$

This simplifies to:

$-21x^2 - 6y^2 = -30$

Now we have a new system:

$\begin{cases} 5x^2 + 6y^2 = 50 \\ -21x^2 - 6y^2 = -30 \end{cases}$

Comparing with the Options

Now, let's compare this new system with the options provided. We are looking for the system that matches this:

A. $\begin{cases} 5x^2 + 6y^2 = 50 \\ -21x^2 - 6y^2 = 10 \end{cases}$

B. $\begin{cases} 5x^2 + 6y^2 = 50 \\ -21x^2 - 6y^2 = -30 \end{cases}$

The Correct Answer

By comparing our derived system with the options, we can clearly see that Option B is the correct one:

$\begin{cases} 5x^2 + 6y^2 = 50 \\ -21x^2 - 6y^2 = -30 \end{cases}$

This system is equivalent to the original system because we obtained it through valid algebraic manipulations (specifically, multiplying an equation by a constant). The solution set for this system will be identical to the solution set for the original system.

Why Other Options Are Incorrect

Let's quickly discuss why the other options are incorrect:

Option A: The second equation is $-21x^2 - 6y^2 = 10$, but after multiplying the original second equation by -3, we obtained $-21x^2 - 6y^2 = -30$. Therefore, Option A is incorrect.

Additional Insights and Tips

• Always Double-Check: When manipulating equations, always double-check your work to ensure you haven't made any arithmetic errors. A small mistake can lead to an incorrect answer.
• Multiple Approaches: Remember that there can be multiple ways to solve a system of equations. In this case, we used the elimination method, but substitution could also be used (though it might be more complex).
• Understanding Equivalence: Keep in mind that equivalent systems have the same solution set. Any operation you perform should maintain this property.
• Practice Makes Perfect: The more you practice solving systems of equations, the better you'll become at recognizing patterns and choosing the most efficient solution method.

Common Mistakes to Avoid

• Arithmetic Errors: As mentioned earlier, arithmetic errors are a common pitfall. Be careful when multiplying, adding, or subtracting equations.
• Incorrect Multiplication: Make sure you distribute the multiplication correctly across all terms in the equation.
• Forgetting the Constant: When multiplying an equation by a constant, don't forget to multiply the constant term on the right-hand side of the equation as well.
• Misinterpreting Equivalence: Understand that equivalent systems must have the exact same solution set. If your manipulations change the solution set, you've made an error.

Conclusion

So, there you have it! Finding an equivalent system of equations involves manipulating the original equations using valid algebraic operations. In this case, multiplying the second equation by -3 and then comparing the result with the given options led us to the correct answer. Remember to always double-check your work and understand the underlying principles to avoid common mistakes. Keep practicing, and you'll become a pro at solving systems of equations in no time! Keep rocking it, Plastik Magazine readers!