Solving For K In Homogeneous System 3x + Y = K^2 - 9 And X - 2y = K + 3

Aug 3, 2025 by Mei Lin 72 views

Solving for k in a Homogeneous System of Equations

Hey guys! Let's dive into solving a system of equations to find the value of 'k' that makes it homogeneous. This is a super cool topic in linear algebra, and I'm excited to break it down for you step by step. We'll make sure everything is crystal clear so you can tackle similar problems with confidence. So, let's get started and unravel this mathematical puzzle together!

Understanding Homogeneous Systems

Before we jump into the calculations, let's quickly recap what a homogeneous system of equations actually means. A system of linear equations is said to be homogeneous if all the constant terms are zero. In other words, when you write the equations in the form:

a₁x + b₁y = c₁
a₂x + b₂y = c₂

For a homogeneous system, both c₁ and c₂ must be equal to zero. This is a crucial point because homogeneous systems have some special properties. For example, they always have at least one solution – the trivial solution, where all variables are zero (x = 0, y = 0). The interesting part is figuring out when they have non-trivial solutions as well.

In our case, the system is given as:

3x + y = k² - 9
x - 2y = k + 3

To make this system homogeneous, we need to find the values of k that make the right-hand sides of both equations equal to zero. That means we need to solve the following equations:

k² - 9 = 0
k + 3 = 0

These equations are our key to unlocking the value(s) of k that satisfy the condition for a homogeneous system. So, let’s break down how to solve them!

Solving the Equations for k

Okay, let’s tackle these equations one by one to find the values of k. The first equation we have is:

k² - 9 = 0

This is a quadratic equation, and we can solve it in a couple of ways. One method is to factor it. Notice that this is a difference of squares, which can be factored as:

(k - 3)(k + 3) = 0

This gives us two possible solutions for k:

k - 3 = 0 => k = 3
k + 3 = 0 => k = -3

So, from the first equation, we have two potential values for k: 3 and -3. Now, let's move on to the second equation:

k + 3 = 0

This one is much simpler to solve. Just subtract 3 from both sides:

k = -3

So, from the second equation, we have one value for k: -3. Now, we need to find the value(s) of k that satisfy both equations simultaneously. This is where the magic happens!

Finding the Common Solution

We've found the solutions for k from each equation separately. Now, we need to find the value(s) that make both equations true at the same time. This is like finding the intersection point of two conditions. From the first equation, k² - 9 = 0, we found that k could be 3 or -3. From the second equation, k + 3 = 0, we found that k must be -3.

So, which value(s) do they have in common? Well, the only value that appears in both sets of solutions is k = -3. This is the value that makes both equations equal to zero, thus making the system homogeneous. If we plug k = -3 back into our original system, we get:

3x + y = (-3)² - 9  =>  3x + y = 9 - 9  =>  3x + y = 0
x - 2y = -3 + 3  =>  x - 2y = 0

As you can see, both equations now have a zero on the right-hand side, which confirms that the system is indeed homogeneous when k = -3. This is a fantastic check to ensure our solution is correct!

Verifying the Solution

To be absolutely sure, it's always a good idea to verify our solution. We found that k = -3 makes the system homogeneous. Let's plug this value back into the original system and see what happens:

3x + y = k² - 9
x - 2y = k + 3

Substituting k = -3:

3x + y = (-3)² - 9
x - 2y = -3 + 3

Simplifying:

3x + y = 9 - 9  =>  3x + y = 0
x - 2y = 0

Now we have a homogeneous system:

3x + y = 0
x - 2y = 0

To solve this system, we can use various methods, such as substitution or elimination. Let's use the substitution method. From the second equation, we can express x in terms of y:

x = 2y

Now, substitute this into the first equation:

3(2y) + y = 0
6y + y = 0
7y = 0

This gives us:

y = 0

Now, substitute y = 0 back into x = 2y:

x = 2(0) = 0

So, the solution to the homogeneous system is x = 0 and y = 0. This is the trivial solution, which is expected for a homogeneous system. The fact that we obtained this solution further confirms that our value of k = -3 is correct. It’s always satisfying when the math checks out, right?

Conclusion

Awesome! We've successfully found the value of k that makes the given system of equations homogeneous. By setting the constant terms to zero, we transformed the problem into solving for k in the equations k² - 9 = 0 and k + 3 = 0. We found that k = -3 is the common solution that satisfies both equations. We even verified our solution by plugging it back into the original system and confirming that it results in a homogeneous system with the trivial solution (x = 0, y = 0).

I hope this breakdown has been helpful and that you now feel more confident in tackling similar problems. Remember, the key is to understand the properties of homogeneous systems and then carefully solve the resulting equations. Keep practicing, and you'll become a pro at these in no time! If you have any questions or want to explore more examples, just let me know. Happy solving, guys!