Solve 4x - 6 = 10x - 3: Step-by-Step Guide

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Hey everyone! Today, we're diving into a classic algebra problem that might seem a little daunting at first glance, but trust me, it's totally manageable. We're going to tackle the equation 4xβˆ’6=10xβˆ’34x - 6 = 10x - 3 and find out what the value of xx is. Whether you're a math whiz or just trying to get a handle on your algebra homework, this guide is for you. We'll break it down step-by-step, making sure you understand each part of the process. So, grab a pen and paper, and let's get started on solving this equation!

Understanding the Goal: Isolating 'x'

Alright guys, the main goal when we're faced with an equation like 4xβˆ’6=10xβˆ’34x - 6 = 10x - 3 is to isolate the variable xx. Think of it like a puzzle where you need to get xx all by itself on one side of the equals sign. To do this, we'll use a few basic algebraic rules that are super important. These rules essentially say that whatever you do to one side of the equation, you have to do the exact same thing to the other side. This keeps the equation balanced, like a perfectly calibrated scale. We'll be moving terms around, combining like terms, and ultimately figuring out the specific numerical value that xx represents to make this equation true.

It's crucial to remember that the equals sign isn't just a separator; it signifies balance. If you have 55 on the left and 55 on the right, they're equal. If you add 22 to the left, making it 77, you must add 22 to the right, making it 77 as well, to maintain that equality. We use this principle to move numbers and variables around. For example, if you have a term you want to get rid of on one side, say a '+5', you can subtract 55 from that side. But remember, you must also subtract 55 from the other side. This might seem like a lot of steps, but each one moves us closer to our final answer – the value of xx. So, let's get our hands dirty and start applying these principles to our specific equation!

Step 1: Gathering 'x' Terms

Okay, so the first tactical move in solving 4xβˆ’6=10xβˆ’34x - 6 = 10x - 3 is to get all the terms containing xx onto one side of the equation. It doesn't really matter which side you choose, but often it's easier to move the smaller xx term to the side with the larger xx term. This helps avoid dealing with negative coefficients for xx right away, though it's not strictly necessary. In our case, we have 4x4x on the left and 10x10x on the right. Since 10x10x is bigger than 4x4x, let's move the 4x4x over to the right side.

To move the 4x4x from the left side, we need to perform the opposite operation. Since it's currently 4x4x (which is 4x4x plus nothing else on that side), we'll subtract 4x4x. But remember the golden rule: whatever we do to one side, we must do to the other. So, we subtract 4x4x from both sides of the equation:

4xβˆ’6βˆ’4x=10xβˆ’3βˆ’4x4x - 6 - 4x = 10x - 3 - 4x

Now, let's simplify both sides. On the left side, 4xβˆ’4x4x - 4x cancels out, leaving us with just βˆ’6-6. On the right side, 10xβˆ’4x10x - 4x gives us 6x6x. So, our equation now looks much cleaner:

βˆ’6=6xβˆ’3-6 = 6x - 3

See? We've successfully gathered all the xx terms onto one side. This is a huge step towards finding our solution!

Step 2: Isolating the 'x' Term

Now that we have our xx terms consolidated on one side (the right side, in this case, giving us 6x6x), our next mission is to isolate this 6x6x term. This means we need to get rid of any other numbers that are hanging out with it. Looking at our current equation, βˆ’6=6xβˆ’3-6 = 6x - 3, we see that the 6x6x term has a βˆ’3-3 next to it.

To get 6x6x by itself, we need to undo the subtraction of 33. The opposite of subtracting 33 is adding 33. So, we're going to add 33 to both sides of the equation to keep things balanced:

βˆ’6+3=6xβˆ’3+3-6 + 3 = 6x - 3 + 3

Let's simplify again. On the left side, βˆ’6+3-6 + 3 equals βˆ’3-3. On the right side, βˆ’3+3-3 + 3 cancels out, leaving us with just 6x6x. Our equation is now:

βˆ’3=6x-3 = 6x

We are so close, guys! The 6x6x term is almost isolated. We just have one more step to get xx completely by itself.

Step 3: Solving for 'x'

The final frontier in solving βˆ’3=6x-3 = 6x is to get xx completely alone. Right now, xx is being multiplied by 66. To reverse multiplication, we use division. So, we need to divide both sides of the equation by 66.

rac{-3}{6} = rac{6x}{6}

On the left side, rac{-3}{6} simplifies. Both 33 and 66 are divisible by 33. So, -3 rf 3 = -1 and 6 rf 3 = 2. This gives us - rac{1}{2}.

On the right side, rac{6x}{6} simplifies to just xx, because the 66's cancel out.

So, our final answer is:

x = - rac{1}{2}

And there you have it! We've successfully solved the equation.

Step 4: Checking Your Answer (The Cool Down Lap)

Now, a really smart move in algebra, especially when you're learning, is to check your answer. This is like a cool-down lap after a race – it makes sure you didn't miss any steps or make a silly calculation error. It gives you confidence that your solution is indeed correct.

Our solution is x = - rac{1}{2}. To check it, we substitute this value back into the original equation: 4xβˆ’6=10xβˆ’34x - 6 = 10x - 3. Let's see if both sides end up being equal.

First, let's evaluate the left side (LS): 4xβˆ’64x - 6

Substitute x = - rac{1}{2}:

LS = 4 imes (- rac{1}{2}) - 6

LS = - rac{4}{2} - 6

LS = βˆ’2βˆ’6-2 - 6

LS = βˆ’8-8

Now, let's evaluate the right side (RS): 10xβˆ’310x - 3

Substitute x = - rac{1}{2}:

RS = 10 imes (- rac{1}{2}) - 3

RS = - rac{10}{2} - 3

RS = βˆ’5βˆ’3-5 - 3

RS = βˆ’8-8

Look at that! The left side equals βˆ’8-8 and the right side also equals βˆ’8-8. Since LS = RS (βˆ’8=βˆ’8-8 = -8), our solution x = - rac{1}{2} is absolutely correct! This verification step is super valuable for confirming your work.

Conclusion: You've Got This!

So there you have it, guys! We took the equation 4xβˆ’6=10xβˆ’34x - 6 = 10x - 3 and, by following a clear, step-by-step process, we arrived at the solution x = - rac{1}{2}. We gathered the xx terms, isolated the xx term, and finally solved for xx. And to top it off, we even checked our answer to make sure it was spot on. Algebra might seem tricky sometimes, but with a little practice and by understanding the core principles of balancing equations, you can tackle any problem thrown your way. Keep practicing, and you'll become a pro in no time!

The key takeaways here are to always perform the same operation on both sides of the equation and to use inverse operations to isolate your variable. Whether it's adding to subtract, multiplying to divide, or vice versa, these are your tools for success. Remember that checking your answer isn't just a suggestion; it's a powerful way to build confidence in your mathematical abilities. So next time you're faced with a linear equation, remember these steps, stay calm, and solve it systematically. You've totally got this!