How To Solve For W In -6/w = -8

by Andrew McMorgan 32 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of algebra, specifically tackling a super common type of problem: solving for a variable. We've got a cool equation here, βˆ’6w=βˆ’8-\frac{6}{w}=-8, and our mission, should we choose to accept it (and we totally will!), is to figure out what number ww represents. This might seem a little tricky at first because our variable, ww, is chilling in the denominator. But don't sweat it! With a few smart algebraic moves, we'll have this solved in no time. So, grab your calculators, a piece of paper, and let's get this math party started!

Understanding the Equation: βˆ’6w=βˆ’8-\frac{6}{w}=-8

Alright, let's break down this equation, βˆ’6w=βˆ’8-\frac{6}{w}=-8. At its core, an equation is like a balanced scale. Whatever you do to one side, you must do to the other to keep it balanced. Our goal is to isolate ww, meaning we want to get ww all by itself on one side of the equals sign. Right now, ww is part of a fraction, and it's in the denominator. This is a key point, guys. Fractions can sometimes throw us off, but they're just numbers waiting to be manipulated. The negative sign in front of the fraction means the entire fraction is negative. We can think of βˆ’6w-\frac{6}{w} as βˆ’1Γ—6w-1 \times \frac{6}{w}. We also have βˆ’8-8 on the other side. So, we're looking for a number ww such that when you divide -6 by it, you get -8. Seems simple enough when you put it that way, right? We'll be using inverse operations to get ww out of that denominator and then isolate it. Remember, inverse operations are like opposite actions that undo each other – addition undoes subtraction, multiplication undoes division, and vice versa. We'll be using these superpowers to solve for ww.

Step 1: Getting ww Out of the Denominator

The first hurdle we need to overcome is that ww is stuck in the denominator. To get it out, we need to multiply both sides of the equation by ww. Why ww? Because multiplying by ww will cancel out the ww in the denominator. Think about it: if you have ab\frac{a}{b} and you multiply it by bb, you get aa. So, let's apply this to our equation: βˆ’6w=βˆ’8-\frac{6}{w}=-8. We multiply both sides by ww:

wΓ—(βˆ’6w)=wΓ—(βˆ’8)w \times \left(-\frac{6}{w}\right) = w \times (-8)

On the left side, the ww in the numerator and the ww in the denominator cancel each other out, leaving us with just βˆ’6-6. On the right side, we have ww multiplied by βˆ’8-8, which we can write as βˆ’8w-8w. So, our equation now looks much simpler:

βˆ’6=βˆ’8w-6 = -8w

See? We've successfully moved ww out of the denominator and into a much more manageable position. This is a huge step towards solving for ww. This technique of multiplying by the variable in the denominator is a fundamental skill in algebra, especially when dealing with rational expressions. It's all about using multiplication to undo division and simplify the expression. We’re making great progress, guys, and this is just the beginning!

Step 2: Isolating ww

Now that we have the equation βˆ’6w=βˆ’8-\frac{6}{w}=-8 simplified to βˆ’6=βˆ’8w-6 = -8w, our next goal is to get ww all by itself. Right now, ww is being multiplied by βˆ’8-8. To undo multiplication, we use its inverse operation: division. So, we need to divide both sides of the equation by βˆ’8-8. This will isolate ww and give us its value. Remember, whatever we do to one side, we must do to the other to maintain the balance of the equation.

Let's divide both sides by βˆ’8-8:

βˆ’6βˆ’8=βˆ’8wβˆ’8\frac{-6}{-8} = \frac{-8w}{-8}

On the left side, we have βˆ’6βˆ’8\frac{-6}{-8}. A negative number divided by a negative number results in a positive number. Also, we can simplify the fraction 68\frac{6}{8} by dividing both the numerator and the denominator by their greatest common divisor, which is 2. So, 68\frac{6}{8} simplifies to 34\frac{3}{4}.

On the right side, the βˆ’8-8 in the numerator and the βˆ’8-8 in the denominator cancel each other out, leaving us with just ww. So, our equation becomes:

34=w\frac{3}{4} = w

And there you have it! We've successfully isolated ww. The solution is w=34w = \frac{3}{4}. This step is all about using the inverse operation of multiplication, which is division, to peel away the coefficient and leave the variable standing alone. It’s a classic move in algebra, and once you get the hang of it, you’ll be solving equations like this in your sleep!

Step 3: Checking Your Answer

Whenever you solve an equation, especially in math class or on a test, it's always a good idea to check your answer. This means plugging the value you found for ww back into the original equation to see if it makes the equation true. It's like giving your answer a little confidence boost! Our original equation was βˆ’6w=βˆ’8-\frac{6}{w}=-8, and we found that w=34w = \frac{3}{4}. Let's substitute 34\frac{3}{4} for ww and see if it works:

βˆ’634=βˆ’8-\frac{6}{\frac{3}{4}} = -8

Now, we need to simplify the left side. Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of 34\frac{3}{4} is 43\frac{4}{3}. So, we can rewrite the left side as:

βˆ’6Γ—43-6 \times \frac{4}{3}

Let's multiply βˆ’6-6 by 43\frac{4}{3}. We can think of βˆ’6-6 as βˆ’61-\frac{6}{1}. So, we have:

βˆ’61Γ—43=βˆ’6Γ—41Γ—3=βˆ’243-\frac{6}{1} \times \frac{4}{3} = -\frac{6 \times 4}{1 \times 3} = -\frac{24}{3}

Now, we simplify βˆ’243-\frac{24}{3}. Twenty-four divided by 3 is 8. Since we have a negative sign, the result is βˆ’8-8:

βˆ’243=βˆ’8-\frac{24}{3} = -8

So, the left side of our equation equals βˆ’8-8, and the right side of our original equation is also βˆ’8-8. Since βˆ’8=βˆ’8-8 = -8, our solution is correct! This step is crucial for confirming your work and building confidence in your algebraic skills. It’s a small step, but it makes a big difference in ensuring accuracy. Always double-check your work, guys!

Conclusion: You've Mastered Solving for ww!

And there you have it, folks! We successfully solved the equation βˆ’6w=βˆ’8-\frac{6}{w}=-8 for ww. We learned how to handle variables in the denominator by using multiplication to get them out, and then we used division to isolate the variable and find its value. Finally, we verified our answer by plugging it back into the original equation. This process of solving equations is a fundamental building block in mathematics, and mastering it opens doors to more complex problems. Remember these key steps: identify the variable, use inverse operations to isolate it, and always check your work. You guys did an amazing job following along. Keep practicing, and soon you'll be solving all sorts of algebraic puzzles. Until next time on Plastik Magazine, stay curious and keep those brains buzzing!