Testing A Solution In A System Of Equations

Welcome to our deep dive into the world of systems of equations! Today, we're tackling a common question: How do we determine if a specific point is a solution to a system of equations? This is a fundamental skill in algebra, and understanding it will unlock many doors in your mathematical journey. We'll be using the point $(-9, 4)$ and the system of equations:

$$\begin{array}{l} 8 x-11 y=14 \\ 15 x-3 y=-13 \end{array}$$

To find out if $(-9, 4)$ is indeed a solution, we need to check if this point satisfies both equations simultaneously. If it works in one but not the other, it's not a solution to the system. Think of it like a key fitting into two locks; it has to open both to be the right key for the whole mechanism.

The Core Concept: Substitution

The most straightforward way to test a point is through substitution. This involves taking the x-coordinate and the y-coordinate of the point and plugging them into each equation. If both equations hold true after the substitution, then the point is a valid solution. Let's break down the process for our specific problem. We have the point $(-9, 4)$, which means $x = -9$ and $y = 4$. We will now substitute these values into each of the equations provided.

Testing the First Equation: $8x - 11y = 14$

Our first equation is $8x - 11y = 14$. Let's substitute $x = -9$ and $y = 4$ into this equation. We perform the multiplication first, following the order of operations (PEMDAS/BODMAS). So, $8$ times $-9$ is $-72$, and $-11$ times $4$ is $-44$. The equation now becomes: $-72 - 44$. Now, we perform the subtraction. $-72$ minus $44$ equals $-116$. So, the left side of the equation evaluates to $-116$. The right side of the equation is $14$. We compare the two sides: $-116 = 14$. This statement is false. Since the point $(-9, 4)$ does not satisfy the first equation, we can already conclude that it is not a solution to the system of equations. However, for the sake of thoroughness and to demonstrate the complete process, let's go ahead and test the second equation as well. This reinforces the concept that both equations must be satisfied for a point to be a solution.

Testing the Second Equation: $15x - 3y = -13$

Now, let's move on to the second equation in our system: $15x - 3y = -13$. Again, we substitute $x = -9$ and $y = 4$. First, we calculate $15$ times $-9$, which gives us $-135$. Next, we calculate $-3$ times $4$, which results in $-12$. Plugging these into the equation, we get: $-135 - (-12)$. Remember that subtracting a negative is the same as adding a positive. So, $-135 - (-12)$ becomes $-135 + 12$. Performing this addition, we find that $-135 + 12 = -123$. The left side of the equation evaluates to $-123$. The right side of the equation is $-13$. We compare these two values: $-123 = -13$. This statement is also false. As we suspected, the point $(-9, 4)$ does not satisfy the second equation either. This further confirms our initial conclusion. A point must satisfy every equation in a system to be considered a solution to that system. If it fails even one, it's out.

Conclusion: A Point's True Identity

After performing the substitution and calculations for both equations, we found that the point $(-9, 4)$ results in false statements for both $8x - 11y = 14$ (where $-116 eq 14$) and $15x - 3y = -13$ (where $-123 eq -13$). Therefore, the point $(-9, 4)$ is not a solution to this system of equations. This process of verification is crucial when working with algebraic equations. It's not just about solving for variables; it's also about understanding the relationship between points and the lines or curves they represent. When a point satisfies a system of equations, it means that the point lies on all the graphs corresponding to those equations. For linear equations, this means the point is the intersection point of the lines.

Remember, the key takeaway here is that a solution to a system of equations must satisfy every single equation within that system. If a point doesn't work in even one of them, it can't be considered a solution to the whole group. Keep practicing these substitution methods, and you'll become a pro at identifying solutions in no time! For more on systems of equations and how to solve them, check out resources like Khan Academy.