Solving 3(-2r-5)=2w+2(w+1): A Step-by-Step Guide

by Rajiv Sharma 49 views

Hey guys! Math problems can sometimes feel like a puzzle, right? Today, we're going to break down the equation 3(-2r-5)=2w+2(w+1) step by step, so you can understand exactly how to solve it. Whether you're prepping for a test, helping with homework, or just want to brush up on your algebra skills, this guide is here to help. We'll go through each stage of the process in detail, making sure you grasp the fundamental concepts and can tackle similar problems with confidence. So, grab your pencils and notebooks, and let's dive in!

Understanding the Equation

Before we jump into the solution, let's first understand the equation: 3(-2r-5)=2w+2(w+1). This equation might look intimidating at first glance, but don't worry! It's essentially an algebraic expression with variables (r and w) and numbers. Our goal is to simplify this equation and see if we can solve for either r or w. However, there's a catch! Notice that we have two variables in a single equation. This means we can't find a single numerical value for either r or w. Instead, we'll simplify the equation and express one variable in terms of the other. Think of it like translating a sentence from one language to another – we're not finding a single "answer," but rather re-writing the equation in a simpler, more understandable form.

To make things clearer, let's break down the key components of the equation:

  • Constants: These are the numbers in the equation, like 3, -5, 2, and 1. They have fixed values and don't change.
  • Variables: These are the letters r and w. They represent unknown values that we're trying to relate to each other.
  • Coefficients: These are the numbers that multiply the variables. For example, -2 is the coefficient of r, and 2 is the coefficient of w in certain terms.
  • Terms: These are the individual parts of the equation separated by plus or minus signs. For example, -2r, -5, 2w, and 2(w+1) are all terms.
  • Parentheses: These indicate that we need to perform certain operations first, following the order of operations (PEMDAS/BODMAS).

Understanding these components is crucial because it helps us navigate the steps we'll take to simplify the equation. We'll be using the distributive property, combining like terms, and isolating variables – all of which rely on recognizing these fundamental elements. By dissecting the equation into its basic parts, we can approach the solution with a clear strategy and avoid common mistakes. Remember, math isn't about memorizing formulas; it's about understanding the underlying principles and applying them logically.

Step 1: Distribute the Constants

The first step in simplifying our equation, 3(-2r-5)=2w+2(w+1), is to distribute the constants outside the parentheses. This means we'll multiply the number outside the parentheses by each term inside. Remember the distributive property: a(b + c) = ab + ac. This is a fundamental rule in algebra, and it's essential for expanding expressions and simplifying equations. Think of it like this: you're "distributing" the multiplication across all the terms inside the parentheses, ensuring that everything gets multiplied correctly.

Let's start with the left side of the equation: 3(-2r-5). We need to distribute the 3 to both -2r and -5.

  • 3 * (-2r) = -6r
  • 3 * (-5) = -15

So, the left side becomes -6r - 15.

Now, let's move to the right side of the equation: 2w+2(w+1). Here, we only need to distribute the 2 in the term 2(w+1).

  • 2 * (w) = 2w
  • 2 * (1) = 2

So, the term 2(w+1) becomes 2w + 2. Adding this to the existing 2w, the right side now looks like 2w + 2w + 2.

After distributing the constants, our equation now looks much simpler: -6r - 15 = 2w + 2w + 2. We've eliminated the parentheses, which makes it easier to work with the equation. Distributing the constants is a crucial step because it allows us to combine like terms and further simplify the equation. It's like clearing away the clutter in a room before you start organizing – you need to get rid of the initial obstacles to see the bigger picture.

Step 2: Combine Like Terms

Now that we've distributed the constants, the next step is to combine like terms. This means identifying terms that have the same variable or are constants and adding or subtracting them together. Combining like terms helps to further simplify the equation and makes it easier to isolate variables later on. Think of it like sorting your laundry – you group similar items together to make them easier to manage.

Let's look at our equation: -6r - 15 = 2w + 2w + 2. On the left side, we have -6r and -15. These are not like terms because -6r has the variable r, while -15 is a constant. So, we can't combine them.

On the right side, we have 2w, 2w, and 2. Notice that we have two terms with the variable w: 2w and 2w. These are like terms, and we can combine them by adding their coefficients.

  • 2w + 2w = 4w

So, the right side simplifies to 4w + 2.

Our equation now looks like this: -6r - 15 = 4w + 2. Notice how much simpler it looks after combining like terms! We've reduced the number of terms, which makes the equation easier to analyze and manipulate. Combining like terms is a fundamental skill in algebra, and it's used extensively in solving equations, simplifying expressions, and working with polynomials. It's like streamlining a process – you eliminate unnecessary steps to make the overall task more efficient.

Step 3: Isolate One Variable

Since our equation, -6r - 15 = 4w + 2, has two variables (r and w), we can't solve for a single numerical value for either one. Instead, we need to isolate one variable in terms of the other. This means we'll rewrite the equation so that one variable is alone on one side of the equation, and the other variable and constants are on the other side. Think of it like rearranging furniture in a room – you're moving things around to create a specific arrangement.

Let's choose to isolate r. This means we want to get r by itself on the left side of the equation. To do this, we need to get rid of the -15 that's being subtracted from -6r. We can do this by adding 15 to both sides of the equation. Remember, whatever we do to one side of the equation, we must do to the other side to keep it balanced.

  • -6r - 15 + 15 = 4w + 2 + 15

This simplifies to:

  • -6r = 4w + 17

Now, we have -6r on the left side, but we want just r. To get rid of the -6, we need to divide both sides of the equation by -6.

  • -6r / -6 = (4w + 17) / -6

This simplifies to:

  • r = (4w + 17) / -6

We can also rewrite this as:

  • r = -(4w + 17) / 6

Or, we can distribute the negative sign in the numerator:

  • r = (-4w - 17) / 6

Now, we have r isolated on the left side of the equation, and it's expressed in terms of w. This means that the value of r depends on the value of w. We've successfully isolated one variable! Isolating a variable is a crucial skill in algebra, and it's used extensively in solving equations, graphing functions, and working with systems of equations. It's like focusing a lens – you adjust the settings to bring a specific element into sharp focus.

Step 4: Express w in Terms of r (Optional)

We've successfully isolated r in terms of w. But just for practice, and to fully understand the relationship between the variables, let's also express w in terms of r. This is an optional step, but it's a great way to reinforce your algebra skills and see the equation from a different perspective. Think of it like looking at a painting from different angles – you gain a deeper appreciation for the artwork.

Let's go back to our equation before we divided by -6: -6r - 15 = 4w + 2. This time, we want to isolate w on one side of the equation.

First, let's get rid of the +2 on the right side by subtracting 2 from both sides:

  • -6r - 15 - 2 = 4w + 2 - 2

This simplifies to:

  • -6r - 17 = 4w

Now, we have 4w on the right side, but we want just w. To get rid of the 4, we need to divide both sides of the equation by 4.

  • (-6r - 17) / 4 = 4w / 4

This simplifies to:

  • w = (-6r - 17) / 4

Now, we have w isolated on the left side of the equation, and it's expressed in terms of r. This means that the value of w depends on the value of r. We've successfully expressed w in terms of r! This step demonstrates the flexibility of algebraic manipulation and how we can rewrite equations to highlight different relationships between variables. It's like having a conversation with someone – you can express the same idea in different ways to make sure you're understood.

Final Thoughts

So there you have it! We've successfully taken the equation 3(-2r-5)=2w+2(w+1) and simplified it, isolated r in terms of w, and even expressed w in terms of r. Remember, the key to solving algebraic equations is to break them down into smaller, manageable steps. Distribute constants, combine like terms, and isolate variables – these are the fundamental techniques that will help you tackle even the most challenging problems.

Math might seem daunting at times, but with practice and a clear understanding of the underlying principles, you can conquer any equation. Keep practicing, keep asking questions, and most importantly, keep believing in yourself. You've got this!