Solve For X: 0 = 2 - 4(ax - 3)

by Andrew McMorgan 31 views

Hey guys! Today, we're diving deep into the awesome world of algebra to tackle a common type of equation that pops up in math class. We've got the equation 0=2−4(ax−3)0 = 2 - 4(ax - 3), and our mission, should we choose to accept it, is to isolate that elusive variable 'xx'. This isn't just about getting the right answer; it's about understanding the process, the step-by-step logic that unlocks the solution. We'll be using some fundamental algebraic techniques – think distribution, combining like terms, and inverse operations – to peel back the layers of this equation and reveal 'xx'. So, grab your pencils, a fresh sheet of paper, and let's get ready to flex those math muscles! This equation might look a little intimidating with its parentheses and multiple terms, but trust me, with a methodical approach, it's totally manageable. We'll break it down piece by piece, ensuring that every move we make is mathematically sound. We’re aiming to get 'xx' all by itself on one side of the equals sign, and to do that, we need to get rid of everything else that’s hanging around with it. This involves a bit of strategic maneuvering, much like a game of chess, where each move has a purpose. We'll start by simplifying the expression within the parentheses, then work our way outwards, neutralizing each term until 'xx' is standing alone, victorious.

Let's get started by tackling the equation: 0=2−4(ax−3)0 = 2 - 4(ax - 3). Our primary goal here is to isolate 'xx'. The first thing we should do is distribute the '-4' to the terms inside the parentheses. This means multiplying -4 by 'axax' and then by '-3'. So, we get: −4imesax=−4ax-4 imes ax = -4ax and −4imes−3=+12-4 imes -3 = +12. Now, our equation transforms into 0=2−4ax+120 = 2 - 4ax + 12. See? We're already making progress! The next step is to combine the constant terms on the right side of the equation. We have a '2' and a '+12'. Adding them together gives us 2+12=142 + 12 = 14. So, the equation now looks like this: 0=14−4ax0 = 14 - 4ax. We're getting closer to isolating 'xx'. Remember, the goal is to get 'xx' by itself. To do this, we need to move the constant term ('14') to the other side of the equation. We can do this by subtracting 14 from both sides. Subtracting 14 from the left side gives us 0−14=−140 - 14 = -14, and subtracting 14 from the right side gives us 14−14−4ax=−4ax14 - 14 - 4ax = -4ax. So, our equation is now: −14=−4ax-14 = -4ax. Now, 'xx' is being multiplied by '-4a'. To isolate 'xx', we need to perform the inverse operation, which is division. We will divide both sides of the equation by '-4a'. On the left side, we have −14/−4a-14 / -4a. On the right side, we have −4ax/−4a-4ax / -4a, which simplifies to just 'xx'. So, we have x = rac{-14}{-4a}. Simplifying the fraction, we can divide both the numerator and the denominator by -2. This gives us x = rac{7}{2a}.

Wait a minute, something doesn't seem right! Let's retrace our steps. It's super common to make small errors when you're working through these problems, and that’s totally okay! It’s part of the learning process, guys. Let’s go back to the original equation: 0=2−4(ax−3)0 = 2 - 4(ax - 3). We correctly distributed the -4 to get 0=2−4ax+120 = 2 - 4ax + 12. Then we combined the constants to get 0=14−4ax0 = 14 - 4ax. Now, we want to get the term with 'xx' by itself. Instead of moving the 14, let's move the '-4ax' term to the left side. We can do this by adding 4ax4ax to both sides of the equation. On the left side, we get 0+4ax=4ax0 + 4ax = 4ax. On the right side, we have 14−4ax+4ax=1414 - 4ax + 4ax = 14. So, the equation becomes 4ax=144ax = 14. This looks much cleaner, right? Now, 'xx' is being multiplied by '4a4a'. To isolate 'xx', we need to divide both sides by '4a4a'. On the left side, we get 4ax/4a=x4ax / 4a = x. On the right side, we get 14/4a14 / 4a. So, we have x = rac{14}{4a}. We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2. This gives us x = rac{7}{2a}.

Okay, let's pause and reflect. We've arrived at x = rac{7}{2a}. Now, let's compare this to the answer choices provided: A. - rac{1}{e}, B. - rac{b}{e}, C. rac{5}{a}, D. rac{1}{a}. Uh oh! Our calculated answer, rac{7}{2a}, doesn't match any of the options. This usually means one of two things: either there was a mistake in our calculation, or there might be an issue with the provided options. Let's do one more careful check of our algebra. The equation is 0=2−4(ax−3)0 = 2 - 4(ax - 3). Distribute the -4: 0=2−4ax+120 = 2 - 4ax + 12. Combine constants: 0=14−4ax0 = 14 - 4ax. Add 4ax4ax to both sides: 4ax=144ax = 14. Divide by 4a4a: x = rac{14}{4a}. Simplify: x = rac{7}{2a}. The algebra seems solid, guys. It's possible the question or the options have a typo.

Let's consider if the equation was slightly different, to see if we could land on one of the answers. What if the equation was 0=2−4(ax−extbf5/4)0 = 2 - 4(ax - extbf{5/4})? Then 0=2−4ax+4(5/4)=2−4ax+5=7−4ax0 = 2 - 4ax + 4(5/4) = 2 - 4ax + 5 = 7 - 4ax. So, 4ax=74ax = 7, and x = rac{7}{4a}. Still not matching. What if the equation was 0=extbf5−4(ax−3)0 = extbf{5} - 4(ax - 3)? Then 0=5−4ax+12=17−4ax0 = 5 - 4ax + 12 = 17 - 4ax. So, 4ax=174ax = 17, and x = rac{17}{4a}. Nope. What if the equation was 0=2−extbf2(ax−3)0 = 2 - extbf{2}(ax - 3)? Then 0=2−2ax+6=8−2ax0 = 2 - 2ax + 6 = 8 - 2ax. So, 2ax=82ax = 8, and x = rac{8}{2a} = rac{4}{a}. Getting closer to rac{5}{a} and rac{1}{a}, but still not there.

Let's consider another possibility. What if the 'a' in the equation was supposed to be a '1'? Then 0=2−4(x−3)0 = 2 - 4(x - 3). 0=2−4x+120 = 2 - 4x + 12. 0=14−4x0 = 14 - 4x. 4x=144x = 14. x = rac{14}{4} = rac{7}{2}. This is a numerical answer, so it won't match the options with 'a' in them. Now, let's assume one of the options is correct and try to work backwards. If x = rac{5}{a} (Option C), then substituting this back into the original equation: 0 = 2 - 4(a( rac{5}{a}) - 3). 0=2−4(5−3)0 = 2 - 4(5 - 3). 0=2−4(2)0 = 2 - 4(2). 0=2−80 = 2 - 8. 0=−60 = -6. This is false. So, Option C is incorrect.

Let's try Option D, x = rac{1}{a}. Substituting this into the original equation: 0 = 2 - 4(a( rac{1}{a}) - 3). 0=2−4(1−3)0 = 2 - 4(1 - 3). 0=2−4(−2)0 = 2 - 4(-2). 0=2+80 = 2 + 8. 0=100 = 10. This is also false. It appears there might be an error in the question or the provided answer choices, as our derived solution x = rac{7}{2a} does not match any of the options, and testing options C and D shows they don't satisfy the equation. In a real test scenario, you'd want to double-check your work meticulously, and if you're still confident in your answer, flag it or make a note about the discrepancy. For the sake of this exercise, and assuming there might be a typo that would lead to one of the answers, let's imagine a scenario where the equation was meant to result in rac{5}{a}. For x = rac{5}{a} to be true, we would need 4ax=204ax = 20. Working backwards from 4ax=204ax=20 to the equation form 0=extconstant−4ax0 = ext{constant} - 4ax, we'd need the constant to be −20-20. This would mean the original equation looked something like 0=extsomeconstant−4ax0 = ext{some constant} - 4ax. If we had 0=extbf20−4(ax−3)0 = extbf{20} - 4(ax - 3), then 0=20−4ax+12=32−4ax0 = 20 - 4ax + 12 = 32 - 4ax, leading to 4ax=324ax = 32 and x = rac{32}{4a} = rac{8}{a}. Still not rac{5}{a}. Let's try to make 0=2−4(ax−extbfk)0 = 2 - 4(ax - extbf{k}) yield x = rac{5}{a}. We need 4ax=144ax = 14, so x = rac{14}{4a} = rac{7}{2a}. The only way to get rac{5}{a} is if x = rac{20}{4a}. This would imply 4ax=204ax = 20. If the equation was 0=extbfconstant−4ax0 = extbf{constant} - 4ax, then the constant would need to be 2020. So, if the equation was 0=20−4(ax−3)0 = 20 - 4(ax - 3), then 0=20−4ax+12=32−4ax0 = 20 - 4ax + 12 = 32 - 4ax, giving 4ax=324ax=32 and x= rac{8}{a}. This is proving tricky!

Let's assume the option rac{5}{a} is indeed the correct answer, and work backwards to see what the original equation should have been. If x = rac{5}{a}, then ax = a( rac{5}{a}) = 5. Substituting this into the structure 0=2−4(ax−3)0 = 2 - 4(ax - 3): 0=2−4(5−3)0 = 2 - 4(5 - 3). 0=2−4(2)0 = 2 - 4(2). 0=2−80 = 2 - 8. 0=−60 = -6. This is false. This confirms that with the given equation and options, none of the choices are correct. However, if the question intended to have a solution like rac{5}{a}, a possible intended equation might have been derived from the result 4ax=204ax = 20. For example, if the equation was 0=extbf20−4ax0 = extbf{20} - 4ax, then 4ax=204ax = 20 and x = rac{20}{4a} = rac{5}{a}. To get 0=20−4ax0 = 20 - 4ax from the form 0=C−4(ax−D)0 = C - 4(ax - D), we would need C+4D=20C + 4D = 20. If D=3D=3, then C+12=20C + 12 = 20, so C=8C=8. So, if the equation was 0=extbf8−4(ax−3)0 = extbf{8} - 4(ax - 3), then 0=8−4ax+12=20−4ax0 = 8 - 4ax + 12 = 20 - 4ax, which leads to 4ax=204ax = 20 and x = rac{5}{a}. This shows how a slight change in the constant term can shift the answer significantly. But based strictly on the equation provided, 0=2−4(ax−3)0 = 2 - 4(ax - 3), our derived answer is x = rac{7}{2a}. Since this isn't an option, and testing the options shows they don't work, it's safe to conclude there's an error in the question's setup. It's a good lesson in carefully checking your work and being aware that sometimes questions can have mistakes! The process of solving for xx involved distributing, combining like terms, and using inverse operations, which are fundamental skills.

In conclusion, guys, while we meticulously followed the steps to solve the equation 0=2−4(ax−3)0 = 2 - 4(ax - 3) for xx, our derived solution is x = rac{7}{2a}. Unfortunately, this result does not align with any of the provided answer choices (A. - rac{1}{e}, B. - rac{b}{e}, C. rac{5}{a}, D. rac{1}{a}). We even went the extra mile to test options C and D by substituting them back into the original equation, and they proved to be incorrect. This strongly suggests there may be a typo in the question itself or in the answer options provided. It's a valuable reminder in mathematics that accuracy is key, and sometimes you'll encounter problems that require a critical eye to identify potential errors. The algebraic techniques we employed – distribution, simplification, and isolating the variable using inverse operations – are robust and correctly applied. The process is what matters, and understanding how to manipulate equations is a crucial skill. If this were an exam, the best course of action would be to double-check your calculations one last time and then perhaps make a note of the discrepancy. For now, we stand by our derived answer of x = rac{7}{2a} based on the equation as written. Keep practicing, keep questioning, and keep those math skills sharp!

Final Answer Derivation: Starting equation: 0=2−4(ax−3)0 = 2 - 4(ax - 3)

  1. Distribute the -4: 0=2−4ax+120 = 2 - 4ax + 12
  2. Combine constant terms: 0=14−4ax0 = 14 - 4ax
  3. Move the term with xx to the left side by adding 4ax4ax to both sides: 4ax=144ax = 14
  4. Isolate xx by dividing both sides by 4a4a: x = rac{14}{4a}
  5. Simplify the fraction: x = rac{7}{2a}

Since rac{7}{2a} is not among the options, and testing the options shows they don't satisfy the equation, we acknowledge the likely error in the question or options.