Solve Tricky Equations: A Step-by-Step Guide

Aug 3, 2025 by Rajiv Sharma 45 views

Solving the Tricky Equation: $x ullet 3(3x-9)^2(3) + (3x-9)^3 = 27(x-3)^2(4x-3)$

Hey everyone! Today, we're diving into a really interesting mathematical problem. We're going to break down the steps to solve this equation: $x ullet 3(3x-9)^2(3) + (3x-9)^3 = 27(x-3)^2(4x-3)$ . It might look intimidating at first glance, but don't worry, we'll tackle it together step by step. So, grab your thinking caps, and let's get started!

Initial Assessment and Simplification

When we first look at the equation, it’s crucial to identify common factors and simplify where possible. Our equation is $x ullet 3(3x-9)^2(3) + (3x-9)^3 = 27(x-3)^2(4x-3)$ . The term $(3x - 9)$ appears multiple times, and we can factor out a 3 from it, rewriting it as $3(x - 3)$ . This is a common technique in algebra: identifying and factoring out common terms to simplify the expression.

By factoring out the 3, we make the equation much cleaner and easier to handle. This initial simplification is not just about making the numbers smaller; it’s about restructuring the equation to reveal its underlying form. It’s like decluttering a room before you start organizing—you need to clear the excess to see the structure. So, this step sets the stage for more complex manipulations and, ultimately, the solution. It is important to be able to notice the possibility of simplification, because it will make all subsequent steps easier.

Let's rewrite the equation using this simplification:

$x ullet 3[3(x-3)]^2(3) + [3(x-3)]^3 = 27(x-3)^2(4x-3)$

Now, we can simplify the terms inside the brackets. Remember, when you square or cube a product, you apply the exponent to each factor. This means $[3(x-3)]^2$ becomes $3^2(x-3)^2$ , and $[3(x-3)]^3$ becomes $3^3(x-3)^3$ . Applying these simplifications, the equation transforms into:

$x ullet 3 ullet 9(x-3)^2 ullet 3 + 27(x-3)^3 = 27(x-3)^2(4x-3)$

Notice how much cleaner the equation is already! We've gone from a complex-looking expression to something more manageable. This is the power of simplification. The numerical coefficients are now clearly visible, and the common factor of $(x-3)$ is more prominent.

Factoring and Further Simplification

After the initial simplification, the next crucial step is factoring out common terms. Looking at our equation, $x ullet 3 ullet 9(x-3)^2 ullet 3 + 27(x-3)^3 = 27(x-3)^2(4x-3)$ , we can see that $27(x-3)^2$ is a common factor on both sides. Factoring this out will help us reduce the equation to a more manageable form.

Let's rewrite the equation to explicitly show this common factor:

$27x(x-3)^2 + 27(x-3)^3 = 27(x-3)^2(4x-3)$

Now, we can factor out $27(x-3)^2$ from the left side:

$27(x-3)^2[x + (x-3)] = 27(x-3)^2(4x-3)$

This step is like isolating a key ingredient in a recipe. By factoring out the common term, we're highlighting a part of the equation that behaves in a predictable way. It allows us to see the remaining terms more clearly and how they interact with each other. This is a powerful technique in algebra: by identifying and factoring out common elements, we can often simplify complex expressions and reveal underlying structures.

Now that we've factored out the common term, we can see that $27(x-3)^2$ appears on both sides of the equation. If $(x-3) e 0$ , we can divide both sides by this factor. But before we do that, we need to consider the case where $(x-3) = 0$ . This is a critical point: whenever we divide by an expression, we must check if that expression could be zero, as division by zero is undefined. So, we'll come back to this case later. For now, let's assume $(x-3) e 0$ and divide both sides by $27(x-3)^2$ :

$x + (x-3) = 4x - 3$

Notice how much simpler the equation has become! We've gone from a complex expression with squares and cubes to a simple linear equation. This is the beauty of factoring: it allows us to peel away layers of complexity and get to the heart of the problem.

Solving the Simplified Equation

After factoring and simplifying, we've arrived at a much more manageable equation: $x + (x - 3) = 4x - 3$ . This is a linear equation, and solving it involves combining like terms and isolating the variable. Let's break down the steps.

First, combine the terms on the left side of the equation:

$2x - 3 = 4x - 3$

Now, we want to isolate the variable x. We can do this by subtracting $2x$ from both sides:

$-3 = 2x - 3$

Next, add 3 to both sides of the equation:

$0 = 2x$

Finally, divide both sides by 2 to solve for x:

$x = 0$

So, we have found one potential solution: $x = 0$ . This is a critical step in the problem-solving process. We've taken a complex equation and, through careful manipulation, reduced it to a simple form that we can solve directly. But, as we discussed earlier, we need to remember a crucial point: we made an assumption along the way.

Checking for Special Cases

Remember when we divided both sides of the equation by $27(x-3)^2$ ? We made the assumption that $(x-3) e 0$ . This is a critical point in solving equations: whenever we divide by an expression, we must consider the case where that expression might be zero. If we don't, we risk losing solutions.

So, let's investigate what happens when $(x-3) = 0$ . This occurs when $x = 3$ . We need to go back to our original equation and see if $x = 3$ is a solution.

Our original equation was:

$x ullet 3(3x-9)^2(3) + (3x-9)^3 = 27(x-3)^2(4x-3)$

Substitute $x = 3$ into the equation:

$3 ullet 3(3(3)-9)^2(3) + (3(3)-9)^3 = 27(3-3)^2(4(3)-3)$

Simplify the terms:

$3 ullet 3(9-9)^2(3) + (9-9)^3 = 27(0)^2(12-3)$

$3 ullet 3(0)^2(3) + (0)^3 = 27(0)(9)$

$0 + 0 = 0$

So, the equation holds true when $x = 3$ . This means that $x = 3$ is also a solution to the original equation. Guys, it’s very important to check these special cases, because they can sneakily be solutions that we might miss if we're not careful.

Final Solutions and Conclusion

Alright, we've reached the end of our mathematical journey! We started with a complex equation, $x ullet 3(3x-9)^2(3) + (3x-9)^3 = 27(x-3)^2(4x-3)$ , and through careful simplification, factoring, and problem-solving, we've arrived at the solutions.

We found two solutions for this equation:

$x = 0$ (from solving the simplified linear equation)
$x = 3$ (from considering the special case where $(x-3) = 0$ )

Therefore, the solutions to the equation are $x = 0$ and $x = 3$ . It's always a good idea to plug these solutions back into the original equation to double-check our work. This is like proofreading an essay: it helps catch any errors we might have made along the way.

In conclusion, this problem highlights the importance of several key mathematical techniques:

Simplifying expressions by factoring out common terms.
Recognizing and handling special cases, such as division by zero.
Reducing complex equations to simpler forms.
Carefully solving linear equations.

By mastering these techniques, you'll be well-equipped to tackle a wide range of mathematical problems. Keep practicing, and you'll become a math whiz in no time! Remember, mathematics is like a puzzle: it might seem daunting at first, but with the right tools and approach, you can solve it. Keep your mind sharp, and happy problem-solving!

Key Takeaways

Always simplify: Look for common factors and simplify the equation as much as possible before diving into more complex manipulations.
Factor wisely: Factoring out common terms can significantly reduce the complexity of the equation.
Check for special cases: Whenever you divide by an expression, make sure to check if that expression could be zero.
Verify your solutions: Plug your solutions back into the original equation to ensure they are correct.

By following these steps, you'll be able to solve even the trickiest equations with confidence. Keep practicing, and happy solving!