Solving Systems: Which Equation Works?

Hey mathletes! Let's dive into a classic algebra problem. We're given the equation $3x + 5y = 59$ and asked which of the following equations, when substituted into a system with another equation, would still produce the same solution. Essentially, we're looking for an equation that is mathematically equivalent to the original equation, or at least shares the same solution set in relation to another equation in the system. Let's break down the options and see what we can find. We'll explore each choice, combining our algebra skills with some good old-fashioned logic to unravel this math mystery. Keep in mind that when we talk about a 'solution,' we're talking about the values of x and y that make both equations in the system true. So, let's get started and see if we can crack this equation code! Understanding this problem is key to mastering systems of equations and knowing how to manipulate them to find the right answers. It's like having a superpower to simplify complex problems! This knowledge is useful for all sorts of situations. Knowing how to solve systems of equations is a fundamental skill in mathematics, useful for understanding more advanced concepts and in a variety of real-world applications. Being able to quickly evaluate and manipulate equations allows us to solve for unknowns in a more efficient way.

A. $10x - 5y = -20$

Alright, first up, we have $10x - 5y = -20$. Now, this equation looks a bit different from our original $3x + 5y = 59$, right? The key here is to realize that this equation is not mathematically equivalent to the original. Because the equation is not equivalent, we can use it to create a completely different solution set in a system. To check it out, let's see if there is any way we can manipulate $3x + 5y = 59$ to obtain $10x - 5y = -20$. A quick inspection does not give us an easy transformation. This immediately tells us that the solution will be different. While this may be a valid equation, it doesn't give us the same solution as our original setup. Think of it like this: this equation defines a completely different line on a graph. The point where that line intersects with another line (defined by the second equation in the system) will be different from where the original equation's line intersects. So, the chances of it sharing a solution with the original are slim to none. This is an important concept in algebra. Being able to understand how equations change when you manipulate them can help you with your algebra homework and any real-world problem-solving you may encounter in the future. Remember that the solutions to a system of equations are always the points where the equations' lines intersect. Keep that in mind, and you will do great.

Now, let's try to manipulate this equation just to see if we can glean any insight. If we divide the equation by 5, we get $2x - y = -4$. That result might look somewhat familiar because it uses similar values to option C! That’s a good example of how you can use the result of these equations for the process of elimination. Since it is not the same equation as the original, the solution set is going to be completely different when placed in a system of equations. So, this option is likely incorrect. We can immediately disregard this option. The lines will not intersect at the same point, so the solution to the system will not be the same. The way we would solve this is by using substitution or elimination, as we'll do in the next sections. Stay tuned, because the next options will be a little more involved.

B. $7x = 39$

Let's move on to option B: $7x = 39$. This one throws a bit of a curveball. It only has one variable! With just one variable, it is going to be incredibly difficult to make the solution to the system match up with the original. When we look at this equation, we see that we can easily solve for x: $x = 39/7$. This is not going to get us anywhere near a solution set that matches up with the original equation, as it doesn't take into consideration y. A key concept here is that for a system to have the same solution, the new equation needs to represent the same relationship between x and y (or be able to be manipulated to represent that relationship). Think of it like this: If the system contains two lines, for the solution to be the same, the lines would have to intersect at the same point. Since the new equation would only contain one variable, it means that the line would be vertical. This is not the case with our original equation. So, the chances of this option working are slim. Also, this equation completely ignores the y variable. This means, this equation does not provide the same solution as the original one, which has both x and y. So, we can eliminate this option, too. The other thing to keep in mind is that the original equation provides us with two variables. The solution is always going to be the same for the system. Remember, the solution to a system of equations requires the values of both variables, x and y, to be the same. This would provide a single value for x, but not y. Therefore, this is not the right choice for our equation.

C. $2x - y = -4$

Alright, let's turn our attention to option C: $2x - y = -4$. Hmm, this one's interesting, because it also has both x and y variables. Let's see if we can manipulate this equation to get something resembling our original. If we take our original equation, $3x + 5y = 59$, we can't directly transform it into $2x - y = -4$ with simple algebraic manipulations. This indicates that their solutions won't match, unless there's some trickery involved. However, because we cannot directly manipulate the equation, we can safely assume this is not the right answer. We can see that there's no easy way to get from $3x + 5y$ to $2x - y$. This would involve multiplication and division by different numbers, which would lead to a completely different equation. We can manipulate an equation by adding or subtracting the same value, but we cannot divide by different values. Keep this in mind when you are solving problems. Remember that the correct answer needs to have the same solution as the original equation. It would need to represent the same relationship between x and y. So, since this option does not, we can eliminate it. This option will be incorrect. If you were to plug the numbers into this equation, you would not get the same answer. Keep this in mind! The solution is unique to the system of equations. Since this option is not the right answer, we can eliminate it.

D. $13x = 39$

Finally, we've arrived at option D: $13x = 39$. Wait a second! This looks promising. Let's solve for x: $x = 39 / 13$, which gives us $x = 3$. This is a great starting point, but it's not the complete solution, because we still don't know y. If we were to substitute this value into the original equation, we would get: $3(3) + 5y = 59$. This would give us $9 + 5y = 59$. Subtracting 9 from both sides, we get $5y = 50$, and finally, $y = 10$. So, if we know x = 3, and y = 10, then we know this is the only solution to the system. Since we can determine a unique value for both x and y, we can see that this is a great option. However, let's consider this. If we were to solve for x and y, the answer would be the same. The question asks which equation can replace the original one and still give us the same solution. This option lets us solve for both x and y. The answer to the equation is $x=3$ and $y=10$, which satisfies the original system. Therefore, this equation can absolutely replace the original, as it doesn't change the ultimate solution to the system.

Conclusion

After a thorough review of the options, it's clear that D. $13x = 39$ is the one. While other options might look similar or have some connection to the original equation, only D provides a relationship that, when considered with another equation, preserves the solution set of the original system of equations. Remember, the key is to ensure that the new equation, in combination with the second equation in the system, still yields the same values for x and y as the original equations. Keep practicing, and you'll be solving these equation puzzles in no time! So, keep up the great work, and keep those math muscles flexing!