Simple Algebra: Solving 4z + 3 = 6 + 2z

Hey guys, let's dive into a super common type of math problem you'll bump into: solving algebraic equations. Today, we're tackling a specific one: what is the answer of 4z + 3 = 6 + 2z? This might look a little intimidating at first, especially if you're new to algebra, but trust me, it's totally manageable once you break it down. We're going to go through it step-by-step, so by the end of this, you'll feel confident in solving this kind of equation, and many more like it. Algebra is all about figuring out the unknown, represented by letters like 'z' here, and we do that by using a set of logical rules to isolate that unknown variable. Think of it like a puzzle; we're given clues (the numbers and operations) and we need to find the missing piece (the value of 'z'). So, grab a drink, maybe a snack, and let's get started on unraveling this equation together. We'll cover the basic principles, show you the exact steps, and even touch on why these steps work. By the time we're done, you'll see that solving for 'z' in this equation is not just possible, but pretty straightforward.

Understanding the Equation: 4z + 3 = 6 + 2z

Alright, let's really look at the equation: 4z + 3 = 6 + 2z. What we're trying to do here is find the value of 'z' that makes both sides of the equals sign true. That's the golden rule of equations – whatever you do to one side, you must do to the other to keep it balanced. Our goal is to get 'z' all by itself on one side of the equation. To do that, we need to move all the terms with 'z' to one side and all the constant numbers (the ones without 'z') to the other side. Think about it like sorting laundry; you want all the socks in one pile and all the t-shirts in another. We'll be using inverse operations to achieve this sorting. For example, if a term is being added, we'll subtract it from both sides. If a term is being multiplied, we'll divide both sides by it. It's all about undoing what's being done to 'z' so we can see its true value. The equation has 'z' terms on both the left (4z) and the right (2z) sides, and constants on both sides (+3 and +6). This means we've got a bit of rearranging to do, but nothing we can't handle. We'll start by tackling the 'z' terms, and then we'll move on to the constants.

Step-by-Step Solution: Isolating 'z'

So, how do we actually solve 4z + 3 = 6 + 2z? Let's break it down into clear, actionable steps. First things first, we want to gather all the 'z' terms on one side. It's usually easiest to move the smaller 'z' term to avoid dealing with negative numbers if possible. In our equation, we have 4z on the left and 2z on the right. Since 2z is smaller than 4z, let's move the 2z over to the left side. How do we do that? Well, 2z is currently being added on the right side (even though there's no plus sign, it's implied as positive). To move it, we do the opposite: we subtract 2z from both sides of the equation. So, the equation becomes:

4z + 3 - 2z = 6 + 2z - 2z

On the left, 4z minus 2z gives us 2z. On the right, +2z minus 2z cancels each other out, leaving just 6. So, our equation is now simplified to:

2z + 3 = 6

See? We've already made progress! Now, we need to get the constant terms together. We have +3 on the left side with our 'z' term, and we want to move it to the right side. Since 3 is being added on the left, we do the opposite: we subtract 3 from both sides of the equation.

2z + 3 - 3 = 6 - 3

On the left, +3 minus 3 cancels out, leaving us with just 2z. On the right, 6 minus 3 equals 3. So now we have:

2z = 3

We're in the home stretch, guys! The last step is to get 'z' completely by itself. Right now, 'z' is being multiplied by 2. To undo multiplication, we do the opposite: we divide both sides of the equation by 2.

2z / 2 = 3 / 2

On the left, 2 divided by 2 is 1, leaving us with 1z, or just 'z'. On the right, 3 divided by 2 is 1.5 or 3/2.

And there you have it!

z = 3/2 or z = 1.5

So, the answer to what is the answer of 4z + 3 = 6 + 2z is z = 3/2 or 1.5. Pretty cool, right? We took a slightly confusing equation and, with a few simple, logical steps, found our unknown value.

Verification: Checking Our Answer

Now, the beauty of algebra is that you can always check your work. This is super important, especially in tests, to make sure you haven't made any silly mistakes. To verify our answer, z = 3/2, we need to plug this value back into the original equation: 4z + 3 = 6 + 2z. If our value for 'z' is correct, then both sides of the equation should be equal. Let's substitute 3/2 for every 'z' we see.

• Left side: 4z + 3 becomes 4*(3/2) + 3.

  • First, multiply 4 by 3/2: (4/1) * (3/2) = 12/2 = 6.
  • So, the left side is 6 + 3 = 9.

• Right side: 6 + 2z becomes 6 + 2*(3/2).

  • First, multiply 2 by 3/2: (2/1) * (3/2) = 6/2 = 3.
  • So, the right side is 6 + 3 = 9.

Since the left side (9) equals the right side (9), our solution, z = 3/2, is absolutely correct! This verification step is your best friend in algebra. It gives you that confidence that you've nailed the problem. If the two sides didn't match, we'd go back and re-examine our steps to find where we might have gone wrong. But in this case, everything checks out perfectly. It's like a secret handshake between your answer and the original problem, confirming they belong together. Always take that extra minute to plug your answer back in; it's a small effort for a huge payoff in accuracy.

Why Does This Work? The Logic Behind the Steps

So, why are we allowed to just add, subtract, multiply, and divide on both sides of the equation? It all comes down to the fundamental property of equality. Think of an equation like a perfectly balanced scale. The equals sign (=) is the pivot point. Whatever is on the left side has the exact same weight (value) as whatever is on the right side. If you add weight to one side of a scale, you have to add the same amount of weight to the other side to keep it balanced. The same applies when you subtract, multiply, or divide.

• Adding/Subtracting: When we subtracted 2z from both sides of 4z + 3 = 6 + 2z, we were essentially removing the same