Solve Linear Equation: -1/3x + 12 = -2x + 2

Hey math whizzes! Today, we're diving into a classic algebra problem, tackling the linear equation $-\frac{1}{3} x+12=-2 x+2$. You might look at that fraction and feel a little bit of dread, but trust me, guys, it's totally manageable. We're going to break it down step-by-step, so by the end of this, you'll feel like an equation-solving ninja. Remember, the goal is to isolate the variable '$x$' on one side of the equation. We'll use inverse operations to move terms around until '$x$' is all by itself. So, grab your notebooks, maybe a lucky pen, and let's get this solved!

Understanding the Equation and Goal

So, what are we actually trying to do here? We've got the equation $-\frac{1}{3} x+12=-2 x+2$, and our mission, should we choose to accept it, is to find the value of '$x$' that makes this statement true. Think of it like a balancing scale; whatever we do to one side, we must do to the other to keep it balanced. The ultimate aim is to get '$x$' alone. To do this, we'll need to combine like terms and use inverse operations. For instance, if we have a '+12', we'll subtract 12 from both sides. If we have a '$-2x$', we'll add '$2x$' to both sides. The trickiest part here is definitely the fraction, $-\frac{1}{3} x$. We'll deal with that by either multiplying by its reciprocal or by getting rid of all the fractions initially. We're going to explore a few ways to tackle this, ensuring you have a solid understanding of the process. The options provided are A. $x=6$, B. $x=18$, C. $x=-18$, and D. $x=-6$. One of these has to be the correct answer, and we're going to find out which one it is through logical algebraic manipulation. So, let's get started with the first strategy!

Strategy 1: Clearing the Fraction First

Alright guys, let's tackle that pesky fraction first! The equation is $-\frac{1}{3} x+12=-2 x+2$. The easiest way to deal with the $-\frac{1}{3}$ is to multiply every single term in the equation by the denominator, which is 3. This will clear out the fraction like magic! Here’s how it goes:

• Multiply each term by 3: $3 \times (-\frac{1}{3} x) + 3 \times 12 = 3 \times (-2 x) + 3 \times 2$

• Simplify: This simplifies to: $-1x + 36 = -6x + 6$

See? No more fractions! Now the equation looks much friendlier. We have $-x + 36 = -6x + 6$. Our next step is to get all the '$x$' terms on one side and all the constant terms (the numbers without '$x$') on the other side. I usually like to get the '$x$' terms on the side that will result in a positive '$x$' coefficient, but either way works.

• Move the '$x$' terms: Let's add '$6x$' to both sides of the equation to get all the '$x$' terms on the left. $-x + 6x + 36 = -6x + 6x + 6$ $5x + 36 = 6$

• Move the constant terms: Now, we need to get the constants on the right side. We'll subtract 36 from both sides. $5x + 36 - 36 = 6 - 36$ $5x = -30$

• Isolate '$x$': The final step is to isolate '$x$' by dividing both sides by 5. $\frac{5x}{5} = \frac{-30}{5}$ $x = -6$

Boom! We found our answer using this method. The value of '$x$' is -6. This matches option D. Pretty neat, right? Let's quickly try another method to confirm our answer.

Strategy 2: Isolating '$x$' Without Clearing Fractions First

Okay guys, let's try solving this without getting rid of the fraction right away. It’s a good way to double-check our work and see if we get the same result. The equation is still $-\frac{1}{3} x+12=-2 x+2$. Remember, the goal is to get all the '$x$' terms on one side and the constants on the other.

• Combine '$x$' terms: Let's move the '$x$' terms to the left side. To do this, we'll add $2x$ to both sides. We need to find a common denominator to combine $-\frac{1}{3} x$ and $2x$. Remember that $2x$ is the same as $\frac{6}{3}x$. $-\frac{1}{3} x + 2x + 12 = -2x + 2x + 2$ $-\frac{1}{3} x + \frac{6}{3} x + 12 = 2$ $\frac{5}{3} x + 12 = 2$

• Combine constant terms: Now, let's move the constant term (12) to the right side by subtracting 12 from both sides. $\frac{5}{3} x + 12 - 12 = 2 - 12$ $\frac{5}{3} x = -10$

• Isolate '$x$': To get '$x$' by itself, we need to get rid of the $\frac{5}{3}$ coefficient. We can do this by multiplying both sides by the reciprocal of $\frac{5}{3}$, which is $\frac{3}{5}$. $\frac{3}{5} \times \frac{5}{3} x = -10 \times \frac{3}{5}$ $1x = -\frac{10 \times 3}{5}$ $x = -\frac{30}{5}$ $x = -6$

And there you have it! We got the exact same answer, $x = -6$. This confirms our previous result and shows that both methods are effective. It's great to have multiple ways to solve a problem, isn't it? This really solidifies that $-6$ is indeed the correct value for '$x$' in this equation.

Checking Our Answer

Guys, it's super important to check our work, especially in math! It's like proofreading your essay before you hand it in. If we found that $x = -6$, we should plug that value back into the original equation to see if both sides are equal. The original equation is $-\frac{1}{3} x+12=-2 x+2$.

• Substitute $x = -6$ into the left side: $-\frac{1}{3} (-6) + 12$ $= \frac{-1 \times -6}{3} + 12$ $= \frac{6}{3} + 12$ $= 2 + 12$ $= 14$

• Substitute $x = -6$ into the right side: $-2 (-6) + 2$ $= (-2 \times -6) + 2$ $= 12 + 2$ $= 14$

Since the left side (14) equals the right side (14), our solution $x = -6$ is absolutely correct! This step is crucial because it guarantees that we haven't made any calculation errors along the way. It gives us confidence in our final answer. So, when you're solving equations, always, always, always take a minute to plug your answer back in. It's a lifesaver!

Final Answer and Conclusion

So, after going through two different methods and a thorough check, we've definitively concluded that the solution to the linear equation $-\frac{1}{3} x+12=-2 x+2$ is $x = -6$. This matches option D. It's awesome how algebra works, right? By using inverse operations and a bit of strategic manipulation, we can find the unknown value that makes an equation true. Remember the key steps: get all '$x$' terms on one side, all constant terms on the other, and then isolate '$x$'. Don't be afraid of fractions; they can be cleared by multiplying by the denominator, or you can work with them directly using common denominators. The most important thing is to be consistent and perform the same operation on both sides of the equation to maintain balance. Keep practicing these types of problems, and soon you'll be solving them in your sleep! Great job today, everyone!