Simple Algebra: Solve For X

Welcome back to the exciting world of algebra! Today, we're diving into a fundamental skill: solving linear equations. Don't let the "algebra" label scare you; it's really just like a puzzle where you need to find the value of an unknown number, often represented by a letter like 'x'. Think of it as being a detective, and 'x' is the mystery you're trying to crack. Our focus today is on a specific type of equation, and we'll be using a classic example to guide us through the process. Let's get ready to unravel the mystery of 'x'!

The Basics of Equation Solving

At its heart, solving an equation is all about maintaining balance. Imagine a perfectly balanced scale. Whatever you do to one side, you must do the exact same thing to the other side to keep it from tipping. This principle is the golden rule of algebra. When we have an equation like -13s = -15s - 1, our goal is to isolate the variable, which is 's' in this case. This means getting 's' all by itself on one side of the equals sign. To do this, we use inverse operations. Addition is the inverse of subtraction, and multiplication is the inverse of division. If you see a number being added to your variable, you subtract it from both sides. If you see a number being multiplied by your variable, you divide both sides by that number. It's like a dance: whatever move you make on the left, you mirror it on the right. For our specific equation, -13s = -15s - 1, we notice that 's' appears on both sides. This is a common scenario in linear equations, and our first step is usually to gather all the terms with the variable onto one side. We can achieve this by adding 15s to both sides. Why 15s? Because it's the opposite of -15s. When we add 15s to -15s, they cancel each other out, leaving us with zero on that side. So, on the left side, we'll have -13s + 15s, which simplifies to 2s. On the right side, the -15s and +15s disappear, leaving us with just -1. Now our equation looks much simpler: 2s = -1. See? We're one step closer to finding out what 's' is. This process of moving terms is crucial, and understanding the concept of inverse operations is key. Remember, consistency is vital; keep that scale balanced!

Step-by-Step Solution

Let's meticulously work through our equation, -13s = -15s - 1, to find the value of 's'. We've already established the fundamental principle: maintain balance by performing the same operation on both sides. Our first move was to gather the 's' terms. We decided to add 15s to both sides of the equation. This is because we want to eliminate the -15s on the right side. So, the operation looks like this:

-13s + 15s = -15s - 1 + 15s

On the left side, -13s + 15s combines to 2s. On the right side, -15s + 15s cancels out to 0, leaving us with just -1. So, our simplified equation is:

2s = -1

Now, we're very close! The variable 's' is being multiplied by 2. To isolate 's', we need to perform the inverse operation of multiplication, which is division. We will divide both sides of the equation by 2:

2s / 2 = -1 / 2

On the left side, 2s / 2 simplifies to just s. On the right side, -1 / 2 is -1/2 or -0.5.

Therefore, the solution is:

s = -1/2 or s = -0.5

To be absolutely sure our answer is correct, we can perform a check. This involves substituting the value we found for 's' back into the original equation. Let's use s = -0.5.

Original equation: -13s = -15s - 1

Substitute s = -0.5:

-13 * (-0.5) = -15 * (-0.5) - 1

Calculate the left side: -13 * (-0.5) = 6.5

Calculate the right side: -15 * (-0.5) - 1 = 7.5 - 1 = 6.5

Since the left side (6.5) equals the right side (6.5), our solution s = -0.5 is indeed correct. This checking process is a vital part of solving equations; it confirms your work and builds confidence in your algebraic abilities.

Understanding the Concepts

Beyond just following steps, it's crucial to grasp the underlying concepts that make equation solving work. The core idea, as we've discussed, is maintaining equality. Think of the equals sign (=) as the pivot of a perfectly balanced seesaw. If you add weight to one side, you must add the same weight to the other to keep it level. In algebra, these