Solve Equations: -20 = -4x - 6x Explained

by Alex Johnson 42 views

Welcome, math enthusiasts! Today, we're diving into a fundamental concept in algebra: solving linear equations. Our focus will be on a specific example that elegantly demonstrates how to isolate a variable. The equation we'll be dissecting is βˆ’20=βˆ’4xβˆ’6x-20 = -4x - 6x. Don't let the negative signs or the combined terms intimidate you; by the end of this article, you'll have a clear, step-by-step understanding of how to find the value of xx that makes this statement true. This skill is a cornerstone of mathematical problem-solving, appearing in everything from basic arithmetic to advanced calculus, and even in practical applications like physics and engineering. We'll break down the process into manageable steps, ensuring that the logic behind each move is transparent. Remember, the goal in solving any equation is to get the variable (in this case, xx) all by itself on one side of the equals sign. We achieve this by performing inverse operations – essentially, undoing what's being done to the variable. Think of it like a balancing scale; whatever you do to one side, you must do to the other to keep it balanced. So, let's roll up our sleeves and tackle this equation with confidence, transforming it from a puzzle into a solved problem. We'll cover combining like terms, which is a crucial first step in simplifying this particular equation, and then we'll move on to isolating the variable using division. Throughout this journey, we'll emphasize the importance of precision and attention to detail, as even a small slip can lead to an incorrect answer. Get ready to sharpen your algebraic skills and gain a deeper appreciation for the elegance of equation solving.

Understanding the Equation: βˆ’20=βˆ’4xβˆ’6x-20 = -4x - 6x

Before we embark on the journey to solve the equation βˆ’20=βˆ’4xβˆ’6x-20 = -4x - 6x, let's take a moment to truly understand what we're looking at. On the left side of the equals sign, we have a constant value, βˆ’20-20. This number stands alone and doesn't change. On the right side, we have βˆ’4xβˆ’6x-4x - 6x. This is where the variable xx comes into play. The terms βˆ’4x-4x and βˆ’6x-6x are called like terms because they both contain the same variable (xx) raised to the same power (which is 1, even though it's not explicitly written). The numbers βˆ’4-4 and βˆ’6-6 are the coefficients of these terms. The presence of the variable xx means that the value of the right side of the equation depends on what number xx represents. Our ultimate mission is to find that specific numerical value for xx that makes the left side (βˆ’20-20) exactly equal to the right side (βˆ’4xβˆ’6x-4x - 6x). This process of finding the value of the variable is what we mean by solving the equation. The equals sign is the central pivot; it asserts that the expression on the left is equivalent in value to the expression on the right. To solve for xx, we need to manipulate the equation using algebraic rules so that xx is isolated on one side. This involves a series of logical steps, each designed to simplify the equation without altering the fundamental truth it represents. We'll start by simplifying the right side, as this is a key characteristic of this particular problem. Combining like terms is a fundamental algebraic operation that makes equations easier to manage. It's like gathering similar items together before you start organizing them. For example, if you had 4 apples and then someone gave you 6 more apples, you'd have 10 apples in total. Algebraically, this translates to combining the coefficients of like terms. The distributive property also plays a role here, although it's often implicitly applied when combining like terms. The equation is a statement of equality, and our goal is to maintain that equality as we transform it. We're not just changing the numbers; we're rewriting the equation in a simpler form that still holds true. This foundational understanding is critical because it sets the stage for the systematic approach we'll employ. Keep in mind that the variable xx represents an unknown quantity, and our task is to uncover its identity.

Step 1: Combine Like Terms

The first crucial step in solving the equation βˆ’20=βˆ’4xβˆ’6x-20 = -4x - 6x involves simplifying the right-hand side by combining the like terms. As we identified, βˆ’4x-4x and βˆ’6x-6x are like terms because they both involve the variable xx raised to the power of one. Combining like terms is a fundamental operation in algebra that makes equations more manageable. To combine them, we simply add or subtract their coefficients – the numbers that precede the variable. In this case, the coefficients are βˆ’4-4 and βˆ’6-6. So, we perform the operation: βˆ’4+(βˆ’6)-4 + (-6). When adding two negative numbers, we add their absolute values and keep the negative sign. Therefore, βˆ’4+(βˆ’6)=βˆ’10-4 + (-6) = -10. This means that βˆ’4xβˆ’6x-4x - 6x is equivalent to βˆ’10x-10x. Our equation now transforms into a much simpler form: βˆ’20=βˆ’10x-20 = -10x. This step is vital because it reduces the complexity of the equation, making the subsequent steps more straightforward. It’s akin to clearing away clutter before you start building something. The property that allows us to do this is the distributive property in reverse. Mathematically, we can express this as: βˆ’4xβˆ’6x=(βˆ’4βˆ’6)x=βˆ’10x-4x - 6x = (-4 - 6)x = -10x. By combining these terms, we've effectively summarized the operations involving xx into a single term. This simplification is key to isolating xx. Always look for opportunities to combine like terms on each side of the equation first. This reduces the number of terms you have to work with, minimizing the chances of errors. It’s a principle of efficiency in algebraic manipulation. Remember, the goal is always to move towards a state where the variable is alone. This initial simplification is a significant stride in that direction. The equation βˆ’20=βˆ’10x-20 = -10x is now much easier to analyze and solve than the original βˆ’20=βˆ’4xβˆ’6x-20 = -4x - 6x. It clearly shows that βˆ’20-20 is equal to ten times the unknown value xx, with a negative sign in front.

Step 2: Isolate the Variable

Now that we have simplified our equation to βˆ’20=βˆ’10x-20 = -10x, the next logical step is to isolate the variable xx. Our current equation shows that βˆ’10-10 is being multiplied by xx. To undo multiplication, we use the inverse operation, which is division. We need to divide both sides of the equation by the coefficient of xx, which is βˆ’10-10, to maintain the balance of the equation. It's critical to perform the same operation on both sides to ensure the equality remains true. So, we divide the left side by βˆ’10-10 and the right side by βˆ’10-10:

-20 rac{}{-10} = -10x rac{}{-10}

Let's perform the division on each side. On the left side, βˆ’20-20 divided by βˆ’10-10 equals 22. Remember that a negative number divided by a negative number results in a positive number. On the right side, βˆ’10x-10x divided by βˆ’10-10 simplifies to xx, because any number or variable divided by itself equals 1 (and 1imesx1 imes x is just xx).

This leads us to the solution:

2=x2 = x

Alternatively, we can write this as x=2x = 2. This step is the heart of solving linear equations. By applying the inverse operation, we systematically peel away the numbers that are attached to the variable until the variable stands alone. It's like carefully unwrapping a present. The rule of performing the same operation on both sides is fundamental and ensures that the mathematical statement remains valid. If we only divided one side, the scale would tip, and the equality would be broken. Therefore, isolating the variable by using division (the inverse of multiplication) is the key to finding its value. This technique applies broadly across algebra; whenever a variable is multiplied by a coefficient, you divide both sides by that coefficient to solve for the variable.

Step 3: Verify the Solution

To ensure we have correctly solved the equation, it's always a good practice to verify the solution x=2x = 2 by substituting it back into the original equation: βˆ’20=βˆ’4xβˆ’6x-20 = -4x - 6x. This step acts as a self-check, confirming that our answer is accurate. If the left side of the equation equals the right side after substituting our value for xx, then our solution is correct. Let's substitute x=2x = 2 into the original equation:

βˆ’20=βˆ’4(2)βˆ’6(2)-20 = -4(2) - 6(2)

Now, perform the multiplications on the right side:

βˆ’4(2)=βˆ’8-4(2) = -8

βˆ’6(2)=βˆ’12-6(2) = -12

Substitute these values back into the equation:

βˆ’20=βˆ’8βˆ’12-20 = -8 - 12

Finally, perform the subtraction on the right side:

βˆ’8βˆ’12=βˆ’20-8 - 12 = -20

So, the equation becomes:

βˆ’20=βˆ’20-20 = -20

Since the left side equals the right side, our solution x=2x = 2 is indeed correct. This verification process is invaluable. It not only builds confidence in your answer but also reinforces the understanding that the variable represents a specific value that satisfies the equality. It's a way to prove to yourself that your algebraic manipulations were sound. Always take the time to verify your solutions, especially in more complex problems or during assessments. It's a small effort that can prevent significant errors and lead to a deeper mastery of algebraic concepts.

Conclusion

We have successfully navigated the process of solving the linear equation βˆ’20=βˆ’4xβˆ’6x-20 = -4x - 6x. By following a systematic approach, we first combined the like terms on the right side of the equation, simplifying it to βˆ’20=βˆ’10x-20 = -10x. Then, we isolated the variable xx by dividing both sides by βˆ’10-10, which yielded the solution x=2x = 2. Finally, we verified our answer by substituting x=2x = 2 back into the original equation, confirming that both sides were equal. This step-by-step method of combining like terms and then isolating the variable is a fundamental technique applicable to a vast array of algebraic problems. Mastering these basic skills is crucial for building a strong foundation in mathematics. Remember, the key principles are to maintain equality by performing the same operation on both sides and to use inverse operations to undo existing operations. Whether you're tackling homework assignments, preparing for exams, or applying mathematical concepts in real-world scenarios, the ability to confidently solve equations is an indispensable skill. Keep practicing, and don't hesitate to review these steps whenever you encounter a similar problem. For further exploration into algebraic concepts and equation solving, you can refer to resources like Khan Academy, which offers comprehensive lessons and practice exercises.