Solving For 'w': A Step-by-Step Guide

by Andrew McMorgan 38 views

Hey Plastik Magazine readers! Let's dive into a math problem and break it down step-by-step. Today, we're going to solve for 'w' in the equation โˆ’32โˆ’73w=โˆ’45-\frac{3}{2}-\frac{7}{3} w=-\frac{4}{5}. Don't worry if it looks intimidating at first; we'll simplify this together, making sure everyone understands each part. This guide is designed to be super clear, so even if math isn't your favorite, you'll feel confident by the end. We'll use simple language, and the explanations are tailored for everyone, from those who are new to algebra to those who just need a refresher. So, grab your pencils and let's get started. We aim to not only find the value of w but also to demonstrate a clear method that can be applied to similar problems. This method involves isolating the variable, which is a fundamental skill in algebra and is used extensively in different scientific and engineering applications. Understanding this process will significantly boost your problem-solving abilities. Ready to get started, guys?

So, the equation is โˆ’32โˆ’73w=โˆ’45-\frac{3}{2}-\frac{7}{3} w=-\frac{4}{5}. Our main goal here is to isolate w on one side of the equation, meaning we want to get w by itself. To do this, we'll need to perform a series of operations that will systematically remove everything else that's with w. The first step is to deal with the constant term, which in this case is โˆ’32-\frac{3}{2}. We will get rid of this by adding 32\frac{3}{2} to both sides of the equation. This is because adding the same value to both sides keeps the equation balanced. This fundamental principle ensures the equality remains true throughout all our manipulations. It is essential when solving any equation. When you add 32\frac{3}{2} to both sides, the โˆ’32-\frac{3}{2} on the left side cancels out, which leaves us with only the term involving w. On the right side, we'll need to combine โˆ’45-\frac{4}{5} and 32\frac{3}{2}.

Step 1: Isolating the 'w' term

Our first step is to isolate the term containing 'w'. This involves getting rid of that pesky โˆ’32-\frac{3}{2} on the left side. Here's how we'll do it. Let's add 32\frac{3}{2} to both sides of the equation. Why? Because adding 32\frac{3}{2} to โˆ’32-\frac{3}{2} will make it zero, and anything plus zero is just itself. This will leave us with a term involving w on the left side. It's like a balancing act. Whatever we do on one side, we have to do the same on the other to maintain equality. So, our equation โˆ’32โˆ’73w=โˆ’45-\frac{3}{2}-\frac{7}{3} w=-\frac{4}{5} turns into:

โˆ’32+32โˆ’73w=โˆ’45+32-\frac{3}{2} + \frac{3}{2} - \frac{7}{3} w = -\frac{4}{5} + \frac{3}{2}

This simplifies to:

โˆ’73w=โˆ’45+32- \frac{7}{3} w = -\frac{4}{5} + \frac{3}{2}

See? The โˆ’32-\frac{3}{2} disappeared on the left side, and now we only have terms with w on one side. Now we need to figure out what is on the right side. Before we get into solving for w, we need to simplify the fractions on the right side. To do that, we need a common denominator. The lowest common denominator for 5 and 2 is 10.

Step 2: Combining Fractions

Okay, now that we've isolated the term with w, we need to combine the fractions on the right side of the equation. Remember, our equation now looks like this: โˆ’73w=โˆ’45+32- \frac{7}{3} w = -\frac{4}{5} + \frac{3}{2}. Combining the fractions means we need to find a common denominator. The least common multiple (LCM) of 5 and 2 is 10, so we'll convert both fractions to have a denominator of 10. Hereโ€™s how:

For โˆ’45-\frac{4}{5}: Multiply the numerator and denominator by 2:

โˆ’4ร—25ร—2=โˆ’810\frac{-4 \times 2}{5 \times 2} = -\frac{8}{10}

For 32\frac{3}{2}: Multiply the numerator and denominator by 5:

3ร—52ร—5=1510\frac{3 \times 5}{2 \times 5} = \frac{15}{10}

Now, substitute these new fractions back into the equation: โˆ’73w=โˆ’810+1510- \frac{7}{3} w = -\frac{8}{10} + \frac{15}{10}. Add the fractions on the right side: โˆ’810+1510=710- \frac{8}{10} + \frac{15}{10} = \frac{7}{10}. Now our equation looks like this: โˆ’73w=710- \frac{7}{3} w = \frac{7}{10}. Pretty neat, huh? We've managed to simplify the right side into a single fraction. We're getting closer to solving for w!

Step 3: Solving for 'w'

Alright, guys, let's solve for w! Our simplified equation is now โˆ’73w=710- \frac{7}{3} w = \frac{7}{10}. To isolate w, we need to get rid of the โˆ’73-\frac{7}{3} that's multiplying it. The best way to do this is to multiply both sides of the equation by the reciprocal of โˆ’73-\frac{7}{3}, which is โˆ’37-\frac{3}{7}. Remember, multiplying by the reciprocal effectively cancels out the fraction on the side with w. Letโ€™s write it out so it's clear:

โˆ’37ร—โˆ’73w=710ร—โˆ’37-\frac{3}{7} \times -\frac{7}{3} w = \frac{7}{10} \times -\frac{3}{7}

On the left side, โˆ’37ร—โˆ’73-\frac{3}{7} \times -\frac{7}{3} equals 1, so we're left with just w. On the right side, we multiply the fractions: 710ร—โˆ’37\frac{7}{10} \times -\frac{3}{7}. The 7 in the numerator and denominator cancel out, which leaves us with โˆ’310-\frac{3}{10}.

So now we have:

w=โˆ’310w = -\frac{3}{10}

That's it! We solved for w. The value of w that satisfies the original equation is โˆ’310-\frac{3}{10}. Congratulations, you've successfully solved for w! This final step involves the use of reciprocals to isolate the variable. This is a crucial technique in algebra. The method we used ensures the equation stays balanced. This final result is the solution, and it is a fundamental aspect of understanding and solving linear equations. Remember, the core idea is to perform operations that simplify the equation without changing its fundamental truth.

Verification and Conclusion

Verifying our solution is an important step, because it allows us to ensure we got our answer correct. To do this, we plug our value for w back into the original equation and see if it holds true. Remember the original equation: โˆ’32โˆ’73w=โˆ’45-\frac{3}{2}-\frac{7}{3} w=-\frac{4}{5}. We'll substitute โˆ’310-\frac{3}{10} for w:-

โˆ’32โˆ’73ร—(โˆ’310)=โˆ’45-\frac{3}{2} - \frac{7}{3} \times (-\frac{3}{10}) = -\frac{4}{5}

Let's simplify:

โˆ’32+2130=โˆ’45-\frac{3}{2} + \frac{21}{30} = -\frac{4}{5}

Simplify 2130\frac{21}{30} to 710\frac{7}{10}:-

โˆ’32+710=โˆ’45-\frac{3}{2} + \frac{7}{10} = -\frac{4}{5}

Convert โˆ’32-\frac{3}{2} to a fraction with a denominator of 10, which is โˆ’1510-\frac{15}{10}:-

โˆ’1510+710=โˆ’45-\frac{15}{10} + \frac{7}{10} = -\frac{4}{5}

โˆ’810=โˆ’45-\frac{8}{10} = -\frac{4}{5}

Simplify โˆ’810-\frac{8}{10} to โˆ’45-\frac{4}{5}:-

โˆ’45=โˆ’45-\frac{4}{5} = -\frac{4}{5}

Since both sides are equal, our solution is correct! This confirms that our solution is accurate and satisfies the original equation. In conclusion, the value of w that solves the equation โˆ’32โˆ’73w=โˆ’45-\frac{3}{2}-\frac{7}{3} w=-\frac{4}{5} is โˆ’310-\frac{3}{10}. We successfully isolated the variable by systematically applying inverse operations to both sides of the equation, always ensuring the balance and maintaining the truth of the equation. This process is fundamental in algebra and is essential for solving many mathematical and real-world problems. Keep practicing these steps, and you'll become a pro at solving equations like this in no time! Keep experimenting with different equations and practice the steps until it becomes second nature. Great job, everyone! And remember, practice makes perfect! Stay curious, and keep exploring the amazing world of mathematics!