Solving For A And B In The Equation X - (1/2)y = -4

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Hey guys! Let's dive into a fun math problem where we're given an equation and a table, and our mission is to find the missing values. This is a classic algebra challenge, and we’re going to break it down step by step so it's super clear. Our main goal here is to find the values of a and b that fit perfectly into our equation. So, grab your thinking caps, and let's get started!

Understanding the Equation and the Table

Okay, first things first, let's take a good look at what we have. We've got the equation xβˆ’12y=βˆ’4x - \frac{1}{2}y = -4. This is a linear equation, which means it represents a straight line when graphed. The equation tells us how x and y are related to each other. For every value of x, there's a corresponding value of y that makes the equation true, and vice versa. Understanding this relationship is key to solving our problem.

Then, we have a table with some x and y values. Some of these values are given, and some are missing. Our job is to fill in the blanks. The table looks like this:

x y
0 a
-2 4
b 0

See those a and b? Those are the mystery values we need to uncover. We’re given some pairs of x and y values, and we're going to use the equation xβˆ’12y=βˆ’4x - \frac{1}{2}y = -4 to find the missing ones. This is where the fun begins!

Now, why is this important? Well, these kinds of problems pop up all the time in algebra and beyond. They help us understand how equations work and how to solve for unknowns. Plus, they're a great way to sharpen our problem-solving skills. So, let's jump into solving for a and b.

Solving for a

Let's kick things off by finding the value of a. Remember, a is the y value when x is 0. So, we're going to plug x = 0 into our equation and solve for y. This is a classic substitution method, a fundamental technique in algebra. By substituting known values into an equation, we simplify the problem and make it solvable.

Our equation is xβˆ’12y=βˆ’4x - \frac{1}{2}y = -4. We're replacing x with 0, so the equation becomes:

0βˆ’12y=βˆ’40 - \frac{1}{2}y = -4

This simplifies to:

βˆ’12y=βˆ’4-\frac{1}{2}y = -4

Now, we need to isolate y. To do this, we can multiply both sides of the equation by -2. Why -2? Because multiplying βˆ’12-\frac{1}{2} by -2 will give us 1, leaving y by itself. This is a crucial step in solving for any variable: perform the inverse operation to undo the operation affecting the variable.

So, we have:

(βˆ’2)βˆ—(βˆ’12y)=(βˆ’2)βˆ—(βˆ’4)(-2) * (-\frac{1}{2}y) = (-2) * (-4)

This simplifies to:

y=8y = 8

Woohoo! We found a! Since y is 8 when x is 0, we know that a = 8. That wasn't so bad, right? We just used a little substitution and some basic algebra to crack the code. Now, let’s jot this down in our table. The first row is complete now: when x is 0, y is 8. Knowing one piece of the puzzle helps us visualize the whole picture better.

But we're not done yet. We still have b to find. Are you ready for the next challenge? Let's move on to solving for b.

Solving for b

Alright, let's tackle b! This time, we're looking for the x value when y is 0. So, just like before, we’re going to use substitution. But this time, we're plugging in y = 0 into our trusty equation xβˆ’12y=βˆ’4x - \frac{1}{2}y = -4. Substitution is such a powerful tool in algebra; it's like having a key that unlocks the solution.

Our equation becomes:

xβˆ’12(0)=βˆ’4x - \frac{1}{2}(0) = -4

Simplifying this, we get:

xβˆ’0=βˆ’4x - 0 = -4

Which is just:

x=βˆ’4x = -4

Boom! We found b! When y is 0, x is -4. So, b = -4. See how straightforward that was? By substituting y = 0, we immediately found the value of x. This highlights the beauty of algebraic equations: they provide a clear relationship between variables, making it possible to solve for unknowns.

Now, let's update our table with this new information. We now know that when y is 0, x is -4. This completes the third row of our table. We’re making great progress! Filling in these missing pieces not only solves the problem but also gives us a better understanding of the line represented by the equation.

So, we've found both a and b. We’re on a roll! But let's not stop here. Let's take a moment to recap what we've done and then verify our solutions. This is a crucial step in problem-solving: checking your work to ensure accuracy.

Verifying the Solutions

Okay, so we've found that a = 8 and b = -4. Now, let's make sure these values actually work in our equation. We're going to plug them back into xβˆ’12y=βˆ’4x - \frac{1}{2}y = -4 to see if everything checks out. This is like the detective work of math – we’ve got our suspects (the values of a and b), and now we need to make sure they fit the crime (the equation).

First, let's check the x = 0, y = 8 pair. Plugging these values into our equation, we get:

0βˆ’12(8)=βˆ’40 - \frac{1}{2}(8) = -4

Simplifying, we have:

0βˆ’4=βˆ’40 - 4 = -4

Which is:

βˆ’4=βˆ’4-4 = -4

Yay! It checks out! This confirms that our value for a is correct. We’ve got one down, one to go. The satisfaction of seeing the equation balance out is one of the best parts of math, isn’t it?

Now, let's verify the x = -4, y = 0 pair. Plugging these values into our equation, we get:

βˆ’4βˆ’12(0)=βˆ’4-4 - \frac{1}{2}(0) = -4

Simplifying, we have:

βˆ’4βˆ’0=βˆ’4-4 - 0 = -4

Which is:

βˆ’4=βˆ’4-4 = -4

Double yay! This also checks out! Our value for b is correct too. We’ve successfully verified both solutions. This step is so important because it gives us confidence that we’ve solved the problem correctly. It's like the final piece of the puzzle clicking into place.

So, we’ve not only found the values of a and b but also confirmed that they are correct. Great job, guys! Now, let's wrap things up with a neat little summary.

Conclusion

Alright, we did it! We successfully found the values of a and b for the equation xβˆ’12y=βˆ’4x - \frac{1}{2}y = -4. We discovered that a = 8 and b = -4. We used the power of substitution to plug in known values and solve for the unknowns. And, just to be sure, we verified our solutions by plugging them back into the equation. This entire process highlights the importance of careful steps and checking your work in mathematics.

But more than just finding the right answers, we also practiced some super important skills. We worked with linear equations, used substitution, and honed our problem-solving abilities. These are skills that will come in handy in all sorts of math problems and even in everyday life. Thinking logically, breaking down a problem, and finding solutions – these are things we use all the time.

So, what's the big takeaway here? Equations are like puzzles, and with the right tools and a little bit of brainpower, we can solve them. Whether it's finding missing values or understanding how variables relate, math gives us a framework for making sense of the world around us. Keep practicing, keep exploring, and remember, every problem is just a chance to learn something new. You guys rock!