Solve For X: 6(x-5) = -24

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Solve for x: Understanding Algebraic Equations

When we talk about solving for x, we're diving into the fascinating world of algebra, where letters like 'x' stand in for unknown numbers. The equation 6(xβˆ’5)=βˆ’246(x-5)=-24 is a classic example of a linear equation that we can unravel to find the value of 'x'. At its core, solving for 'x' means isolating it on one side of the equation. Think of an equation like a perfectly balanced scale; whatever you do to one side, you must do to the other to keep it balanced. Our goal is to carefully manipulate the equation using mathematical operations – addition, subtraction, multiplication, and division – to get 'x' all by itself. This process is fundamental to understanding more complex mathematical concepts and is a building block for problem-solving in many areas of science, engineering, and even everyday life. Let's break down how we can conquer this particular equation, step by step, making it clear and accessible for anyone curious about algebra.

The First Step: Simplifying the Equation

The first thing we notice in the equation 6(xβˆ’5)=βˆ’246(x-5)=-24 is the presence of parentheses. The number '6' is multiplying the entire expression inside the parentheses, which is '(x-5)'. To make things simpler, we can use the distributive property. This property tells us that we can multiply the '6' by each term inside the parentheses. So, 6 multiplied by 'x' is 6x, and 6 multiplied by '-5' is -30. Applying this, our equation transforms from 6(xβˆ’5)=βˆ’246(x-5)=-24 to 6xβˆ’30=βˆ’246x - 30 = -24. This simplification is crucial because it removes the parentheses and presents us with a more straightforward equation to work with. It’s like clearing away clutter before you can see the path forward. This initial step is all about making the equation more manageable and easier to manipulate as we proceed to isolate 'x'. Remember, the goal is to simplify and prepare for the next stages of our algebraic journey.

Isolating the 'x' Term: Getting Closer to the Solution

Now that we have our simplified equation, 6xβˆ’30=βˆ’246x - 30 = -24, our next major objective is to get the term containing 'x' – that's '6x' – by itself on one side of the equation. To do this, we need to deal with the '-30' that's currently on the same side as '6x'. Since it's being subtracted, the inverse operation is addition. We'll add '30' to both sides of the equation. This is where the balance of the equation comes into play. If we add 30 to the left side (6xβˆ’30+306x - 30 + 30), the '-30' and '+30' cancel each other out, leaving us with just '6x'. Now, to maintain the balance, we must also add 30 to the right side of the equation: βˆ’24+30-24 + 30. Performing this addition, we find that βˆ’24+30-24 + 30 equals 6. So, our equation now looks like 6x=66x = 6. We've successfully isolated the term with 'x', which is a significant milestone in our quest to solve for 'x'. This step demonstrates the power of inverse operations in moving terms around an equation.

The Final Step: Finding the Value of 'x'

We've reached the final stage in solving for x in the equation 6x=66x = 6. Here, '6x' means '6 multiplied by x'. To get 'x' completely by itself, we need to undo this multiplication. The inverse operation of multiplication is division. Therefore, we will divide both sides of the equation by '6'. On the left side, 6x6x divided by 6 leaves us with just 'x'. On the right side, we perform the same operation: 6 divided by 6. And what do we get? 6Γ·6=16 \div 6 = 1. So, the solution to our equation is x=1x = 1. We have successfully found the value of 'x' that makes the original equation 6(xβˆ’5)=βˆ’246(x-5)=-24 true! This entire process, from simplifying with the distributive property to using inverse operations to isolate 'x', is the essence of solving linear algebraic equations. It's a logical sequence of steps that, when followed correctly, guarantees the right answer. Keep practicing these steps, and you'll become a pro at solving for 'x' in no time!

Understanding the Solution

Once we've found that x=1x=1, it's always a good practice to check our answer by plugging this value back into the original equation. This verification step ensures that our solution is correct and that we haven't made any mistakes along the way. Let's substitute '1' for 'x' in the equation 6(xβˆ’5)=βˆ’246(x-5)=-24. So, we have 6(1βˆ’5)6(1-5). Inside the parentheses, 1βˆ’51-5 equals -4. Now, we multiply this result by 6: 6imes(βˆ’4)6 imes (-4). This gives us -24. And indeed, the original equation stated that 6(xβˆ’5)6(x-5) should equal -24. Since our calculation resulted in -24, our solution x=1x=1 is correct! This process of checking reinforces our understanding and builds confidence in our algebraic abilities. It's a small but vital part of the problem-solving journey.

Why is Solving for 'x' Important?

Solving for x is more than just an academic exercise; it's a fundamental skill with wide-ranging applications. In mathematics, it's the gateway to understanding more advanced concepts like calculus, statistics, and physics. In science and engineering, equations are used to model everything from the trajectory of a rocket to the flow of electricity. Being able to solve for an unknown variable like 'x' allows scientists and engineers to predict outcomes, design systems, and solve complex problems. For instance, if you're trying to figure out how much fertilizer you need for a garden based on a specific yield formula, you might need to solve for a variable representing the amount of fertilizer. In economics, 'x' could represent the price of a product that maximizes profit. Even in everyday situations, like budgeting or planning a trip, we often implicitly solve for unknowns. Understanding how to manipulate equations to find 'x' empowers us with analytical skills applicable to countless real-world scenarios. It's a core component of critical thinking and logical reasoning.

Further Exploration in Algebra

As you become more comfortable with solving simple linear equations like 6(xβˆ’5)=βˆ’246(x-5)=-24, you'll discover a vast and exciting landscape of algebraic concepts. This includes dealing with equations that have more than one variable, such as 2x+3y=102x + 3y = 10, which introduces the idea of systems of equations. You'll also encounter quadratic equations, which involve a variable raised to the power of two (like x2x^2), leading to solutions that might be found by factoring, completing the square, or using the quadratic formula. Inequalities, where we compare expressions using symbols like '<' (less than) or '>' (greater than), are another important area. Furthermore, functions, which describe relationships between variables, play a central role in advanced mathematics. Each of these topics builds upon the foundational skills of manipulating equations and solving for unknowns. The journey through algebra is progressive, with each new concept making previous ones clearer and opening doors to more complex problem-solving.

For those interested in delving deeper into the world of mathematics and algebra, exploring resources from Khan Academy can be incredibly beneficial. They offer a comprehensive range of free courses and exercises that cover everything from basic arithmetic to advanced calculus, with a strong emphasis on building a solid understanding of algebraic principles. You can find great explanations and practice problems on their website.