Find The System Of Linear Equations With Solution (-3,-3)

Hey math whizzes and problem solvers! Ever feel like you're trying to find the perfect match for something? Well, in the world of mathematics, we often need to find the perfect match for a system of linear equations. This perfect match is called the solution, and today, guys, we're diving deep into how to find it. Our specific challenge is to figure out which system of linear equations has the point $(-3,-3)$ as its solution. This means that when we plug in $x = -3$ and $y = -3$ into both equations of a system, both equations will hold true. It's like a secret handshake that only the correct pair of equations will share with our given point. We'll be exploring different systems and testing them out, one by one, to see which one satisfies this condition. Get ready to flex those algebraic muscles because we're about to break down how to identify the correct system. We'll be looking at four different options, and it's our job to plug in the values and see which pair of equations clicks. It's a process of elimination and verification, ensuring that our chosen system truly represents the intersection point $(-3,-3)$. So, grab your notebooks, sharpen your pencils, and let's get to the bottom of this mathematical mystery!

Understanding the Concept of a Solution

Alright, let's really get our heads around what it means for a point to be the solution to a system of linear equations. Think of each linear equation as a line on a graph. When we have a system of two linear equations, we're essentially looking at two lines. The solution to this system is the exact point where these two lines intersect. If they intersect at a single point, that point $(x, y)$ is the unique solution. If the lines are parallel and never meet, there's no solution. If the lines are actually the same line, then there are infinitely many solutions because every point on the line is a solution. In our case, we are given a specific point, $(-3,-3)$, and we're told that this is the solution. This means that if we were to graph these lines, they would cross precisely at $x = -3$ and $y = -3$. The task at hand is to identify which pair of equations, when graphed, would indeed intersect at this very spot. To do this without graphing, which can be time-consuming and sometimes inaccurate, we use substitution. We take the proposed solution point, $(-3,-3)$, and substitute its $x$ and $y$ values into each equation within each system. If both equations in a system are satisfied by these values, then that system has $(-3,-3)$ as its solution. It's a rigorous check that confirms the intersection point. This method is straightforward but requires careful calculation to avoid errors. We need to be meticulous with our arithmetic, especially when dealing with negative numbers. So, the core idea is: a solution point must make every equation in the system true simultaneously. This is the fundamental principle we'll use to solve our problem.

Testing the Options: A Step-by-Step Approach

Now, let's get down to business and test each of the given systems to see which one has $(-3,-3)$ as its solution. Remember, we need to substitute $x = -3$ and $y = -3$ into both equations of each system. If both equations work out, we've found our winner!

Option A: Equation 1: $x - 5y = -12$ Substitute: $(-3) - 5(-3) = -3 + 15 = 12$. This does not equal $-12$. Since the first equation is not satisfied, we can stop here for Option A. This is not our system.

Option B: Equation 1: $x - 5y = -12$ Substitute: $(-3) - 5(-3) = -3 + 15 = 12$. This also does not equal $-12$. Wait a minute, guys, let's recheck our arithmetic. Ah, I see the mistake! $(-3) - 5(-3) = -3 + 15 = 12$. The equation states it should be $-12$. So, $12 eq -12$. Therefore, Option B is also incorrect because the first equation is not satisfied. It's crucial to double-check these steps!

Let's re-evaluate Option A and B carefully. It seems I made a small slip in the initial check. Let's re-do it with absolute precision.

Re-testing Option A: Equation 1: $x - 5y = -12$ Substitute $x = -3$ and $y = -3$: $(-3) - 5(-3) = -3 + 15 = 12$. The equation requires this to be $-12$. Since $12 eq -12$, the first equation is not satisfied. Thus, Option A is incorrect.

Re-testing Option B: Equation 1: $x - 5y = -12$ Substitute $x = -3$ and $y = -3$: $(-3) - 5(-3) = -3 + 15 = 12$. Again, this result is $12$, but the equation requires $-12$. So, $12 eq -12$. The first equation is not satisfied. Thus, Option B is incorrect.

It appears there might be a slight issue with how the options were presented or my initial interpretation. Let me re-examine the structure of these options and the goal. The goal is to find the system where both equations are true for $(-3,-3)$. If the first equation in multiple options yields an incorrect result, we need to be extra careful. Let's assume the original prompt and options are correct and proceed with thorough checking.

Let's restart the testing process, being extra diligent.

Option A: Equation 1: $x - 5y = -12$ Substitute $x = -3, y = -3$: $(-3) - 5(-3) = -3 + 15 = 12$. Target is $-12$. $12 eq -12$. Option A fails on the first equation.

Option B: Equation 1: $x - 5y = -12$ Substitute $x = -3, y = -3$: $(-3) - 5(-3) = -3 + 15 = 12$. Target is $-12$. $12 eq -12$. Option B fails on the first equation.

This is unusual. Let me check the possibility of a typo in my understanding or transcription. If the first equation was $x + 5y$ or the constant was $+12$, it might work. However, I must work with what's given. Let's assume there might be a mistake in my calculation or interpretation and proceed with the remaining options, but with extreme caution.

Let's reconsider the first equation: $x - 5y$. If $x=-3$ and $y=-3$, then $x - 5y = (-3) - 5(-3) = -3 + 15 = 12$. So, for the first equation to be satisfied, the right-hand side must be $12$, not $-12$. This means options A and B, which have $x-5y = -12$, cannot be the correct answer if our point $(-3,-3)$ is indeed the solution.

This leads me to believe that the correct option must have $x - 5y = 12$ as its first equation. This narrows down our choices to Option C and Option D.

Let's test these:

Option C: Equation 1: $x - 5y = 12$ Substitute $x = -3, y = -3$: $(-3) - 5(-3) = -3 + 15 = 12$. This equation is satisfied! Great news, guys. Now we need to check the second equation in Option C.

Equation 2: $3x + 2y = -15$ Substitute $x = -3, y = -3$: $3(-3) + 2(-3) = -9 + (-6) = -9 - 6 = -15$. This equation is also satisfied!

Since both equations in Option C are satisfied by the point $(-3,-3)$, Option C is the correct answer.

Let's quickly check Option D just to be absolutely sure and to illustrate the process fully.

Option D: Equation 1: $x - 5y = 12$ Substitute $x = -3, y = -3$: $(-3) - 5(-3) = -3 + 15 = 12$. This equation is satisfied. So far, so good.

Equation 2: $3x + 2y = 15$ Substitute $x = -3, y = -3$: $3(-3) + 2(-3) = -9 + (-6) = -9 - 6 = -15$. The equation requires this to be $15$. Since $-15 eq 15$, the second equation is not satisfied.

Therefore, Option D is incorrect.

Conclusion: The Verified Solution

After meticulously testing each option by substituting the given solution point $(-3,-3)$ into both equations of each system, we have found our match. Option C is the only system where both equations, $x - 5y = 12$ and $3x + 2y = -15$, are simultaneously true when $x = -3$ and $y = -3$. This confirms that the point $(-3,-3)$ is indeed the solution to the system presented in Option C. It's always essential to perform these checks carefully, paying close attention to signs and arithmetic, especially when dealing with negative numbers. This process of verification is fundamental in algebra and ensures we arrive at the correct conclusions. Keep practicing these steps, and you'll become a master at solving systems of equations in no time! Remember, every problem solved is a step towards greater mathematical understanding and confidence. You've got this!