Solve 6(y-3) = 4y-28: A Step-by-Step Guide
Hey guys! Let's dive into solving a common type of algebraic equation. Today, we're going to break down the equation 6(y-3) = 4y - 28. Don't worry if it looks intimidating at first. We'll go through each step together, making it super clear and easy to understand. This equation falls under the category of linear equations, which are fundamental in algebra and often pop up in various real-world applications. Understanding how to solve them is a crucial skill for anyone studying mathematics or related fields. So, grab your pencils, and let's get started!
Understanding the Basics of Algebraic Equations
Before we jump into the nitty-gritty, let's quickly recap what an algebraic equation actually is. At its heart, an equation is a mathematical statement that asserts the equality of two expressions. These expressions are connected by an equals sign (=). The goal when solving an equation is to find the value(s) of the variable(s) that make the equation true. In our case, the variable is 'y,' and we're aiming to find the value of 'y' that satisfies the equation 6(y-3) = 4y - 28. Think of it like a puzzle where we need to figure out what 'y' should be to make both sides of the equation balance perfectly.
Linear equations, like the one we're tackling, are a specific type of equation where the highest power of the variable is 1. This means we won't see any terms like y² or y³ in our equation. Linear equations are relatively straightforward to solve, and they form the building blocks for more complex algebraic problems. The techniques we learn here will be invaluable as we progress in our mathematical journey. Remember, practice makes perfect, so the more we solve these equations, the more comfortable and confident we'll become. We need to understand the basics to effectively solve the equation. Now, let's get our hands dirty and start cracking this equation!
Step 1: Distribute the 6 on the Left Side
Our first move is to simplify the equation by dealing with the parentheses. Remember the distributive property? It's our best friend here! The distributive property states that a(b + c) = ab + ac. This means we need to multiply the 6 outside the parentheses by each term inside the parentheses. So, we multiply 6 by 'y' and 6 by '-3'. This gives us:
6 * y = 6y 6 * -3 = -18
So, the left side of our equation becomes 6y - 18. Now, our equation looks like this:
6y - 18 = 4y - 28
See? We've already made progress! By applying the distributive property, we've eliminated the parentheses and made the equation a bit easier to handle. This step is crucial because it allows us to combine like terms later on. Without distributing, we'd be stuck with those parentheses, and the equation would be much harder to solve. It's like unlocking a door that leads us closer to the solution. Don't underestimate the power of the distributive property; it's a fundamental tool in algebra. This step simplifies the equation, and we are now ready to proceed to the next step.
Step 2: Move the 'y' Terms to One Side
Now, let's gather all the terms containing 'y' on one side of the equation. It doesn't matter which side we choose, but let's aim to keep the coefficient of 'y' positive to avoid dealing with negative signs unnecessarily. In our case, we have 6y on the left side and 4y on the right side. To move the 4y from the right side to the left side, we need to subtract 4y from both sides of the equation. Remember, whatever we do to one side, we must do to the other to maintain the balance.
So, we subtract 4y from both sides:
6y - 18 - 4y = 4y - 28 - 4y
This simplifies to:
2y - 18 = -28
Notice how the 4y on the right side disappears because 4y - 4y = 0. We've successfully moved the 'y' term to the left side, and now we have a simpler equation to work with. This step is crucial because it isolates the variable term, bringing us closer to finding the value of 'y'. It's like separating the ingredients we need from the ones we don't, making it easier to bake our mathematical cake. Next, we will move on to isolating the 'y' by moving the constant terms to the other side.
Step 3: Move the Constant Terms to the Other Side
We're making great progress! Now that we have all the 'y' terms on the left side, let's move the constant terms (the numbers without 'y') to the right side. In our equation, 2y - 18 = -28, we have -18 on the left side and -28 on the right side. To move the -18 to the right side, we need to add 18 to both sides of the equation. Again, we must maintain the balance by performing the same operation on both sides.
So, we add 18 to both sides:
2y - 18 + 18 = -28 + 18
This simplifies to:
2y = -10
The -18 on the left side disappears because -18 + 18 = 0. We've successfully moved the constant term to the right side, and we're one step closer to solving for 'y'. It's like clearing away the clutter to reveal the treasure hidden underneath. By isolating the 'y' term, we've set the stage for the final step: dividing to find the value of 'y'. We are in the penultimate step towards the answer. Just one more step and we will have the value of 'y'.
Step 4: Isolate 'y' by Dividing
We're almost there! We now have the equation 2y = -10. This means 2 times 'y' equals -10. To isolate 'y' and find its value, we need to divide both sides of the equation by the coefficient of 'y,' which is 2. Remember, division is the inverse operation of multiplication, so it will undo the multiplication and leave 'y' by itself.
So, we divide both sides by 2:
(2y) / 2 = (-10) / 2
This simplifies to:
y = -5
And there we have it! We've successfully solved for 'y,' and we found that y = -5. It's like reaching the summit after a challenging climb. We've gone through all the steps, applied the necessary operations, and arrived at the solution. This final step of dividing isolates the variable and reveals its value. Remember, it's always a good idea to check our answer to make sure it's correct.
Step 5: Check Your Solution
To be absolutely sure we've got the right answer, let's check our solution by plugging y = -5 back into the original equation. This step is crucial because it helps us catch any errors we might have made along the way. It's like proofreading our work to ensure it's perfect.
Our original equation was:
6(y - 3) = 4y - 28
Now, substitute y = -5:
6(-5 - 3) = 4(-5) - 28
Simplify the expressions inside the parentheses:
6(-8) = -20 - 28
Multiply:
-48 = -48
We see that both sides of the equation are equal! This confirms that our solution, y = -5, is indeed correct. It's like the final piece of the puzzle clicking into place. Checking our solution gives us confidence that we've solved the equation accurately. Remember, always check your work to ensure you've arrived at the correct answer. This also reinforces our understanding of the entire solving process.
Conclusion: Mastering the Art of Solving Equations
Woohoo! We did it! We successfully solved the equation 6(y-3) = 4y - 28 and found that y = -5. We tackled each step methodically, from distributing to isolating the variable, and we even checked our solution to be sure. Solving equations like this is a fundamental skill in algebra, and it's something you'll use time and time again.
The key takeaways from this exercise are:
- The distributive property is crucial for eliminating parentheses.
- We need to maintain balance by performing the same operation on both sides of the equation.
- Isolating the variable is the ultimate goal.
- Checking our solution is always a good idea.
Remember, practice makes perfect! The more equations you solve, the more comfortable you'll become with the process. Don't be afraid to make mistakes; they're part of the learning journey. Each time you solve an equation, you're strengthening your problem-solving skills and building a solid foundation for more advanced math concepts.
So, keep practicing, keep exploring, and most importantly, keep having fun with math! You've got this!
