Solve For S: 2s + 4 = 5(-4 - 2s)

In the realm of mathematics, solving algebraic equations is a fundamental skill that forms the bedrock of more complex problem-solving. Today, we're going to tackle a specific linear equation: $2s + 4 = 5(-4 - 2s)$. Our goal is to find the value of the variable '$s$' that makes this equation true. Don't worry if algebra sometimes feels like a puzzle; we'll break it down step-by-step, making it as clear and accessible as possible. This equation involves basic arithmetic operations, the distributive property, and the process of isolating a variable. By the end of this article, you'll not only have the solution but also a clearer understanding of the techniques used to arrive at it. So, grab a pen and paper, and let's dive into the fascinating world of algebraic manipulation!

Understanding the Equation: $2s + 4 = 5(-4 - 2s)$

Before we jump into solving, let's take a moment to understand the equation we're working with: $2s + 4 = 5(-4 - 2s)$. This is a linear equation because the highest power of the variable '$s$' is 1. Our objective is to find the value of '$s$' that satisfies this equality. To do this, we need to simplify both sides of the equation and then isolate '$s$' on one side. The left side, $2s + 4$, is already quite simple. The right side, $5(-4 - 2s)$, requires a bit more attention. Here, we see a number (5) multiplied by an expression in parentheses $(-4 - 2s)$. This immediately signals the need to apply the distributive property. The distributive property states that for any numbers $a$, $b$, and $c$, $a(b + c) = ab + ac$. In our case, '$a$' is 5, '$b$' is -4, and '$c$' is -2s. This property is crucial for simplifying expressions where a number is multiplying a sum or difference within parentheses. Mastering the distributive property is key to unraveling many algebraic equations, as it allows us to remove parentheses and combine terms effectively. It's like opening up a box to see all the individual items inside, making it easier to manage.

Applying the Distributive Property

Now, let's apply the distributive property to the right side of our equation: $5(-4 - 2s)$. We multiply 5 by each term inside the parentheses. So, we have:

$5 \times (-4) = -20$

and

$5 \times (-2s) = -10s$

Combining these results, the right side of the equation simplifies to $-20 - 10s$. So, our original equation, $2s + 4 = 5(-4 - 2s)$, now becomes:

$2s + 4 = -20 - 10s$

This is a significant step because we've eliminated the parentheses, making the equation easier to work with. Remember, the distributive property is your best friend when you encounter numbers multiplying expressions in parentheses. It's a fundamental rule in algebra that helps simplify expressions and move towards isolating the variable. Think of it as distributing the 'influence' of the number outside the parentheses to every term within. Without this step, we'd be stuck with the parentheses, making it much harder to combine like terms and solve for '$s$'. It's all about simplifying and making the equation more manageable, step by step.

Gathering Like Terms

With the equation now in its simplified form, $2s + 4 = -20 - 10s$, our next task is to bring all the terms containing the variable '$s$' to one side of the equation and all the constant terms (numbers without variables) to the other side. This process is often referred to as 'gathering like terms'. To achieve this, we can use addition and subtraction, remembering that whatever operation we perform on one side of the equation, we must perform the exact same operation on the other side to maintain the balance of the equality.

Let's start by moving the '$s$' terms. We have '$2s$' on the left and '$-10s$' on the right. To get all '$s$' terms to the left side, we can add $10s$ to both sides of the equation. This is because adding $10s$ to $-10s$ on the right side will cancel out the '$s$' term there ($-10s + 10s = 0$).

$2s + 4 + 10s = -20 - 10s + 10s$

Combining the '$s$' terms on the left side ($2s + 10s = 12s$), the equation becomes:

$12s + 4 = -20$

Now, we need to move the constant terms to the right side. We have '+4' on the left side. To eliminate it, we subtract 4 from both sides of the equation. This will cancel out the '+4' on the left side ($+4 - 4 = 0$).

$12s + 4 - 4 = -20 - 4$

This simplifies to:

$12s = -24$

This step of gathering like terms is crucial. It brings order to the equation, grouping all the knowns and all the unknowns separately. It's like tidying up your workspace before starting a detailed task; it makes the subsequent steps much smoother and less prone to error. By systematically moving terms across the equals sign using inverse operations, we're systematically isolating the variable we're trying to find.

Isolating the Variable '$s$'

We've reached a point where our equation is $12s = -24$. Now, the variable '$s$' is being multiplied by 12. To isolate '$s$', we need to perform the inverse operation of multiplication, which is division. We will divide both sides of the equation by 12. This will leave '$s$' by itself on the left side.

rac{12s}{12} = rac{-24}{12}

Performing the division:

$s = -2$

And there we have it! We have successfully solved for '$s$'. The value of '$s$' that satisfies the original equation is -2. This final step of isolating the variable is the culmination of all the previous simplification and rearranging. It's the moment where we reveal the unknown value. Remember, in algebra, the goal is often to get the variable alone on one side of the equation. To do this, we use inverse operations: if a variable is multiplied by a number, we divide by that number; if it's divided, we multiply; if a number is added, we subtract; if it's subtracted, we add. Each step is designed to undo operations and bring us closer to the solution. It’s like carefully taking apart a complex mechanism to understand how each part contributes to the whole.

Verification: Checking Our Solution

It's always a good practice in mathematics to verify our solution to ensure we haven't made any errors. We found that $s = -2$. Let's substitute this value back into the original equation: $2s + 4 = 5(-4 - 2s)$.

First, let's evaluate the left side (LHS) of the equation with $s = -2$:

LHS = $2s + 4$

LHS = $2(-2) + 4$

LHS = $-4 + 4$

LHS = $0$

Now, let's evaluate the right side (RHS) of the equation with $s = -2$:

RHS = $5(-4 - 2s)$

RHS = $5(-4 - 2(-2))$

RHS = $5(-4 - (-4))$

RHS = $5(-4 + 4)$

RHS = $5(0)$

RHS = $0$

Since the Left Hand Side (LHS) equals the Right Hand Side (RHS) ($0 = 0$), our solution $s = -2$ is correct! This verification step is incredibly important. It confirms that our algebraic manipulations were accurate and that the value we found for '$s$' indeed makes the original equation true. It’s the final check that gives us confidence in our answer. Think of it as double-checking your work before submitting an important assignment – it prevents silly mistakes and ensures accuracy. Always take the time to plug your solution back into the original problem; it's a habit that will serve you well in all your mathematical endeavors.

Conclusion

We have successfully navigated the process of solving the linear equation $2s + 4 = 5(-4 - 2s)$. By applying the distributive property to simplify the right side, gathering like terms to one side, and then isolating the variable '$s$', we arrived at the solution $s = -2$. The verification step confirmed that this value indeed satisfies the original equation. This problem highlights the core principles of algebraic manipulation: maintaining equality by performing operations on both sides, understanding the distributive property, and systematically isolating the unknown variable. These skills are foundational for tackling more intricate mathematical challenges across various fields, from science and engineering to economics and computer science. Remember, consistent practice is key to mastering these concepts.

For further exploration into the fascinating world of algebra and equation solving, you can visit Khan Academy, a fantastic resource offering free courses and practice exercises on a wide range of mathematical topics. Another excellent resource for understanding algebraic concepts is Math is Fun, which provides clear explanations and interactive tools to make learning enjoyable.