Solve X/4 = 9/(4x) - Easy Math Guide

Welcome, math enthusiasts! Today, we're diving into a common type of algebraic equation that often pops up in math classes and standardized tests: solving rational equations. Specifically, we'll tackle the problem: rac{x}{4}=rac{9}{4 x}. This equation might look a little intimidating at first glance with variables in the denominator, but don't worry! By following a few straightforward steps, we can break it down and find the solution(s) for x. Our goal is to isolate x and determine what value(s) it can take to make this equation a true statement. Remember, when dealing with equations containing fractions, especially those with variables in the denominator, it's crucial to be mindful of any values that would make those denominators zero, as division by zero is undefined. We'll keep this in mind as we progress through the solution. The beauty of mathematics lies in its systematic approach, and rational equations are no exception. We'll employ methods like cross-multiplication and checking for extraneous solutions to ensure our final answer is accurate and valid. So, grab your notebooks, and let's get started on unraveling this algebraic puzzle!

Understanding the Equation: rac{x}{4}=rac{9}{4 x}

Before we jump into solving, let's take a moment to truly understand the equation rac{x}{4}=rac{9}{4 x}. This is a rational equation because it involves fractions where the variable, x, appears in the denominator of at least one of the fractions. The equation states that the ratio of x to 4 is equal to the ratio of 9 to 4x. Our primary objective is to find the value or values of x that satisfy this equality. A key consideration with rational equations is the domain of the variable. In this specific equation, the term rac{9}{4 x} has x in the denominator. This means that x cannot be zero, because division by zero is undefined. So, from the outset, we establish that any solution we find must not be x = 0. If we arrive at x = 0 as a potential solution during our calculations, we must discard it because it is an extraneous solution – a solution that arises from the algebraic manipulation but does not satisfy the original equation. We will use this restriction later when we check our answers. The structure of this equation, with a proportion (one fraction equal to another), often lends itself well to a technique called cross-multiplication. This method simplifies the equation by eliminating the denominators, transforming it into a polynomial equation that is generally easier to solve. Keep these fundamental concepts in mind as we move forward to the step-by-step solution process. The ability to identify potential pitfalls, like division by zero, is a hallmark of a proficient problem-solver in mathematics.

Step-by-Step Solution: Cracking the Code

Now, let's get down to business and solve the equation rac{x}{4}=rac{9}{4 x} step by step. The most efficient way to handle an equation like this, where we have a proportion (one fraction equals another), is through cross-multiplication. This technique involves multiplying the numerator of the first fraction by the denominator of the second fraction and setting it equal to the product of the denominator of the first fraction and the numerator of the second fraction. So, for our equation, we will multiply x by 4x and set it equal to 4 multiplied by 9.

Step 1: Cross-Multiply

$$x imes (4x) = 4 imes 9$$

This simplifies to:

$$4x^2 = 36$$

Step 2: Isolate the variable term

Our goal now is to get the x term by itself. To do this, we need to isolate $x^2$. We can achieve this by dividing both sides of the equation by 4:

$$\frac{4x^2}{4} = \frac{36}{4}$$

This gives us:

$$x^2 = 9$$

Step 3: Solve for x

Now we have $x^2 = 9$. To find the value of x, we need to take the square root of both sides of the equation. It's extremely important to remember that when you take the square root of a number to solve an equation like this, there are two possible solutions: a positive one and a negative one. This is because both a positive number squared and its negative counterpart squared will result in a positive number.

\sqrt{x^2} = \pm\sqrt{9}

Therefore, we get:

x = \pm 3

This means our potential solutions are $x = 3$ and $x = -3$.

Checking for Extraneous Solutions: The Final Verification

We've arrived at our potential solutions: $x = 3$ and $x = -3$. However, in mathematics, especially when dealing with rational equations, it's crucial to check our answers to ensure they are valid and not extraneous. Remember our initial observation? We noted that x cannot be zero because it appears in the denominator of the original equation rac{x}{4}=rac{9}{4 x}. Division by zero is undefined, so any solution that makes a denominator zero must be rejected.

Let's test our first potential solution, $x = 3$:

Substitute x = 3 into the original equation:

\frac{3}{4} = \frac{9}{4 imes 3}

\frac{3}{4} = \frac{9}{12}

Now, simplify the right side of the equation:

\frac{9}{12} = \frac{3 imes 3}{3 imes 4} = \frac{3}{4}

So, we have rac{3}{4} = rac{3}{4}. This is a true statement, which means $x = 3$ is a valid solution.

Now, let's test our second potential solution, $x = -3$:

Substitute x = -3 into the original equation:

\frac{-3}{4} = \frac{9}{4 imes (-3)}

\frac{-3}{4} = \frac{9}{-12}

Simplify the right side of the equation:

\frac{9}{-12} = \frac{-9}{12} = \frac{-3 imes 3}{3 imes 4} = \frac{-3}{4}

So, we have rac{-3}{4} = rac{-3}{4}. This is also a true statement, which means $x = -3$ is also a valid solution.

Since neither of our solutions resulted in a division by zero in the original equation, and both satisfy the equation, we can confidently say that both $x = 3$ and $x = -3$ are the correct solutions to the equation rac{x}{4}=rac{9}{4 x}. This rigorous checking process is what separates a good mathematician from a great one!

Conclusion: Mastering Rational Equations

We've successfully navigated the process of solving the rational equation rac{x}{4}=rac{9}{4 x}. By employing the powerful technique of cross-multiplication, we transformed a fractional equation into a simpler quadratic form. We then carefully solved for x, remembering to consider both the positive and negative roots when taking the square root. Most importantly, we underscored the critical step of checking for extraneous solutions. This vigilance is paramount in rational equations, where variables in denominators can impose restrictions on the possible values of x. In this case, our restriction was that x could not equal zero. Fortunately, both of our calculated solutions, $x = 3$ and $x = -3$, were valid and did not violate this condition. This systematic approach—understanding the equation, executing the algebraic steps, and diligently verifying the results—is the key to mastering rational equations and many other algebraic challenges. Remember, practice makes perfect! The more you encounter and solve problems like this, the more intuitive the process will become.

For further exploration and practice with algebraic equations and concepts, you can refer to reputable sources like Khan Academy which offers a vast library of free resources, tutorials, and practice exercises on a wide range of mathematical topics. Their comprehensive approach can help solidify your understanding and build your confidence in tackling more complex problems. Exploring resources like mathsisfun.com can also provide alternative explanations and interactive tools that make learning mathematics engaging and accessible.