Solving Linear Equations: A Step-by-Step Guide

Hey guys! Ever found yourself staring at a couple of linear equations and feeling a bit lost? You know, those pesky things like 28x + 8y = 15 and 4x + ky = 21? Don't worry, we've all been there. Today, we're going to break down how to solve these bad boys, especially when there's a mysterious variable like 'k' thrown into the mix. It's not as scary as it looks, I promise! We'll go through it step-by-step, making sure you understand each part. So grab a coffee, get comfy, and let's dive into the wonderful world of algebra. By the end of this, you'll be solving systems of equations like a pro, and that mysterious 'k' won't stand a chance. We'll cover different methods, explain why they work, and even touch on what happens when things get a little tricky, like when equations might not have a unique solution. Our main goal here is to demystify this process and give you the confidence to tackle any similar problems that come your way. We want this to be super clear and easy to follow, so even if algebra isn't your favorite subject, you can still get a solid grasp on it. We're all about making math accessible and, dare I say, even a little bit fun!

Understanding the Basics: What Are We Even Doing Here?

Alright, let's kick things off by understanding what we're trying to achieve when we solve a system of linear equations. Think of it like this: you have two lines on a graph, and each equation represents one of those lines. When we solve the system, we're basically finding the point where those two lines intersect. That intersection point is the (x, y) coordinate that satisfies both equations simultaneously. It's the magic spot that makes both statements true. For our example, the equations are 28x + 8y = 15 and 4x + ky = 21. We're looking for the specific values of x and y that work for both of these. Now, the twist here is the k. This k is a constant, but its value is unknown to us. So, our solution for x and y might actually depend on what k is. This adds a layer of complexity, but it's a super common scenario in algebra problems, and it's crucial for understanding how variables interact in equations. It's like a puzzle where one piece is hidden, and you have to figure out how it affects the rest of the picture. We'll explore how the value of k can influence whether there's a unique solution, no solution, or infinite solutions. Understanding these possibilities is key to mastering systems of equations. So, when we talk about solving, we're either finding that single point of intersection, or we're determining the conditions under which such a point exists or doesn't exist. It's a fundamental concept that pops up in so many areas of math and science, so getting a good handle on it is super valuable.

Method 1: Substitution - The Sneaky Approach

One of the most common ways to tackle these systems is the substitution method. It's like a stealth mission – you isolate one variable in one equation and then substitute that expression into the other equation. This sounds simple, but it can get a bit messy with larger numbers, like the ones we have in 28x + 8y = 15 and 4x + ky = 21. Let's try to isolate x from the second equation because the numbers seem a bit smaller there. We have 4x + ky = 21. If we move ky to the other side, we get 4x = 21 - ky. Now, divide by 4 to get x = (21 - ky) / 4. This is our expression for x. Now, the sneaky part: substitute this entire expression for x into the first equation, 28x + 8y = 15. So, 28 * [(21 - ky) / 4] + 8y = 15. See what we did? We got rid of one variable (x) in the first equation, leaving us with an equation that only has y and k. Now we can solve for y in terms of k. Let's simplify: (28/4) * (21 - ky) + 8y = 15, which becomes 7 * (21 - ky) + 8y = 15. Distribute the 7: 147 - 7ky + 8y = 15. Now, let's group the y terms: 8y - 7ky = 15 - 147. That simplifies to y(8 - 7k) = -132. Awesome! Now, to find y, we just divide by (8 - 7k): y = -132 / (8 - 7k). This is our y solution, expressed in terms of k. What if 8 - 7k equals zero? This happens when k = 8/7. In that case, we'd have 0 = -132, which is impossible. This means if k = 8/7, there's no solution. Pretty cool, right? We've already uncovered a condition for no solution just by using substitution!

Method 2: Elimination - The Direct Assault

Another powerful technique is the elimination method. This is like a direct assault – you manipulate one or both equations (by multiplying them by constants) so that when you add or subtract them, one of the variables cancels out completely. For our equations, 28x + 8y = 15 and 4x + ky = 21, we can see that the x coefficients (28 and 4) are related. If we multiply the second equation by 7, the x coefficient will become 28, just like in the first equation. So, let's do that: 7 * (4x + ky) = 7 * 21, which gives us 28x + 7ky = 147. Now we have our modified system:

28x + 8y = 15 28x + 7ky = 147

Notice how both equations now start with 28x? If we subtract the second new equation from the first original equation, the 28x terms will vanish! Let's do it:

(28x + 8y) - (28x + 7ky) = 15 - 147 28x + 8y - 28x - 7ky = -132 8y - 7ky = -132

Factor out y: y(8 - 7k) = -132.

And just like with the substitution method, we get y = -132 / (8 - 7k). This confirms our previous result for y! The elimination method often feels more straightforward when coefficients are easily scalable, as they were here. It's a great strategy for systems where you want to quickly get rid of a variable. Again, we see that if 8 - 7k = 0 (meaning k = 8/7), we run into that 0 = -132 situation, indicating no solution. This consistency between methods is a really good sign that we're on the right track.

Finding X: The Missing Piece

Now that we have an expression for y in terms of k, we can find x. We can substitute our expression for y back into either of the original equations. Let's use the second, simpler-looking equation: 4x + ky = 21. We found that y = -132 / (8 - 7k). So, substitute this in:

4x + k * [-132 / (8 - 7k)] = 21 4x - 132k / (8 - 7k) = 21

Now, we want to isolate 4x. Add 132k / (8 - 7k) to both sides:

4x = 21 + 132k / (8 - 7k)

To add the terms on the right, we need a common denominator, which is (8 - 7k). So, 21 becomes 21 * (8 - 7k) / (8 - 7k):

4x = [21 * (8 - 7k) + 132k] / (8 - 7k) 4x = [168 - 147k + 132k] / (8 - 7k) 4x = (168 - 15k) / (8 - 7k)

Finally, divide by 4 to get x:

x = (168 - 15k) / [4 * (8 - 7k)]

So, our solution for x is x = (168 - 15k) / (32 - 28k). And our solution for y is y = -132 / (8 - 7k). These expressions for x and y are dependent on the value of k. This means that for most values of k, there's a unique solution. But remember, we found a special case earlier!

The Case of k = 8/7: No Solution Scenario

We keep running into this (8 - 7k) term in the denominator. What happens if this denominator is zero? That occurs when 8 - 7k = 0, which means 7k = 8, or k = 8/7. If k is exactly 8/7, our expressions for x and y involve division by zero, which is undefined. Let's see what happens if we plug k = 8/7 back into the original setup using the elimination method. Our second equation becomes 4x + (8/7)y = 21. Multiply by 7 to clear the fraction: 28x + 8y = 147. Now look at our system with k = 8/7:

Original first equation: 28x + 8y = 15 Modified second equation: 28x + 8y = 147

Do you see the problem? We have the exact same expression (28x + 8y) equaling two different numbers (15 and 147). This is a contradiction! It's like saying A = 15 and A = 147 at the same time, which is impossible. This situation means that the two lines represented by these equations are actually parallel and never intersect. Therefore, when k = 8/7, the system has no solution. This is a crucial aspect of solving systems of equations – recognizing when a solution might not exist. It's all about the relationships between the coefficients and the constants.

When Does k Lead to Infinite Solutions?

We've seen a unique solution (when 8 - 7k != 0) and no solution (when k = 8/7). Is there a case for infinite solutions? For linear systems, infinite solutions occur when the two equations are essentially the same equation, just written differently. This means one equation is a multiple of the other. Let's look at our original equations again:

28x + 8y = 15 4x + ky = 21

For these to represent the same line, the ratio of the coefficients must be equal, and the ratio of the constants must also be equal.

Ratio of x-coefficients: 28 / 4 = 7 Ratio of y-coefficients: 8 / k Ratio of constants: 15 / 21 = 5/7

For infinite solutions, these ratios must all be equal: 7 = 8/k = 5/7.

Immediately, we see a problem: 7 is definitely not equal to 5/7. This means that for these specific equations, there is no value of k that will result in infinite solutions. The lines can either intersect at a single point (unique solution) or be parallel and never intersect (no solution). It's important to recognize this structure. In other systems of equations, you might find a value of k that makes all ratios equal, leading to infinite solutions. For example, if the second equation was 4x + (8/7)y = 15/7, then multiplying it by 7 gives 28x + 8y = 15, which is identical to the first equation, resulting in infinite solutions. But in our given problem, that scenario is impossible.

Summary and Key Takeaways

So, guys, let's wrap this up. We've tackled the equations 28x + 8y = 15 and 4x + ky = 21. Using both substitution and elimination methods, we found the following:

• Unique Solution: This occurs when the denominator (8 - 7k) is not equal to zero, meaning k != 8/7. In this case, the solution is: x = (168 - 15k) / (32 - 28k) y = -132 / (8 - 7k)

• No Solution: This occurs when the denominator (8 - 7k) is equal to zero, meaning k = 8/7. This results in a contradiction where 28x + 8y must equal both 15 and 147 simultaneously, which is impossible.

• Infinite Solutions: For this particular system, there is no value of k that leads to infinite solutions because the ratios of coefficients and constants cannot be made equal.

Solving systems of linear equations, especially with unknown constants like k, is all about careful manipulation and understanding the conditions under which solutions exist. It’s a fundamental skill in algebra that opens doors to solving more complex problems. Remember to always check for those special cases where denominators might become zero, as that’s often where the most interesting scenarios (like no solution) pop up. Keep practicing, and you'll get the hang of it in no time. You guys are awesome for sticking with it!