Solving 3/x + 5 = X: A Step-by-Step Guide

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Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the world of algebra to tackle a particularly juicy equation: $\frac{3}{x}+5=x$. Don't let those fractions and variables intimidate you; we're going to break this down piece by piece, making it super clear how to solve it and get those answers rounded to three significant figures. This is a classic type of problem you'll often see in math exams, so mastering it is key. We're not just going to give you the answer; we're going to walk you through the why and the how, ensuring you understand the logic behind each step. So, grab your calculators, maybe a comfy seat, and let's get this equation solved!

Understanding the Equation

First off, let's get familiar with our equation: $\frac{3}{x}+5=x$. The first thing you probably notice is the $\frac{3}{x}$ term. This is what we call a rational expression, and it means our variable, $x$, is in the denominator. This is super important because it immediately tells us that $x$ cannot be zero. If $x$ were zero, we'd be dividing by zero, which is a big no-no in mathematics, leading to an undefined result. So, right from the start, we establish that $x \neq 0$. Understanding this constraint is fundamental to solving rational equations correctly. The goal here is to transform this rational equation into a more familiar form, likely a quadratic equation, which we know how to solve using various techniques. The presence of $x$ on both sides of the equation, with one instance in the denominator, suggests that multiplying through by $x$ will be a crucial step in simplifying things. This maneuver will clear the fraction and set us up for a standard polynomial equation. We'll be looking for values of $x$ that make the left side of the equation equal to the right side. It's like finding the magic numbers that balance the scales. We'll explore the different types of solutions we might encounter – will they be positive, negative, or perhaps a mix? And importantly, we'll double-check our solutions against our initial constraint that $x \neq 0$. This foundational understanding is what separates a quick guess from a solid mathematical solution.

Clearing the Fraction

Alright, team, the most effective way to deal with that pesky fraction $\frac{3}{x}$ is to eliminate it entirely. How do we do that? By multiplying every single term in the equation by $x$. Remember, whatever we do to one side of the equation, we must do to the other to maintain balance. So, let's see what happens when we multiply each term by $x$:

$x \times \left(\frac{3}{x}\right) + x \times 5 = x \times x$

When we multiply $x$ by $\frac{3}{x}$, the $x$ in the numerator cancels out the $x$ in the denominator, leaving us with just $3$. This is the magic of multiplying by the denominator! Next, we have $x \times 5$, which simply becomes $5x$. And on the right side, $x \times x$ gives us $x^2$.

So, after this step, our equation transforms from $\frac{3}{x}+5=x$ into $3 + 5x = x^2$.

See? No more fractions! This is a huge leap forward. This process of clearing fractions by multiplying by the least common denominator (in this case, just $x$) is a standard technique for solving rational equations. It converts a potentially tricky equation into a simpler polynomial form. It's important to be meticulous here; ensure you multiply every term, including the ones on the right-hand side. Missing even one term can lead to an incorrect final equation. This step sets the stage for solving what will likely be a quadratic equation, so let's make sure this transformation is solid before we move on. The goal is to isolate terms and arrange them in a standard format. This initial simplification is crucial for the subsequent steps.

Rearranging into a Quadratic Equation

Now that we've successfully banished the fraction, we're left with $3 + 5x = x^2$. This looks much friendlier, right? The next logical step is to rearrange this equation into the standard form of a quadratic equation, which is $ax^2 + bx + c = 0$. To do this, we need to move all the terms to one side of the equation, setting the other side to zero. It doesn't strictly matter which side you move them to, but it's common practice to have the $x^2$ term with a positive coefficient. In our case, $x^2$ is already positive on the right side, so let's move the $3$ and $5x$ from the left side over to the right.

To move the $5x$, we subtract $5x$ from both sides:

$3 + 5x - 5x = x^2 - 5x$

This simplifies to:

$3 = x^2 - 5x$

Now, to move the $3$, we subtract $3$ from both sides:

$3 - 3 = x^2 - 5x - 3$

This leaves us with:

$0 = x^2 - 5x - 3$

Or, written in the standard form we're used to:

$x^2 - 5x - 3 = 0$

And there you have it! Our original rational equation has been transformed into a clean, standard quadratic equation. This transformation is a critical part of solving these types of problems. By getting it into the $ax^2 + bx + c = 0$ format, we can now employ powerful tools like the quadratic formula or factoring (if possible) to find the values of $x$. Remember, the coefficients here are $a=1$, $b=-5$, and $c=-3$. These values will be essential for the next step, where we find the actual solutions for $x$. The ability to manipulate equations and change their form while preserving their equality is a cornerstone of algebraic problem-solving. This step highlights that seemingly different equations can often be reduced to a common, solvable structure.

Using the Quadratic Formula

We've successfully converted our equation into $x^2 - 5x - 3 = 0$. Now, we need to find the values of $x$ that satisfy this quadratic equation. Since this particular quadratic doesn't look easily factorable (we'd need two numbers that multiply to -3 and add to -5, which aren't straightforward integers), the most reliable method is to use the quadratic formula. The quadratic formula is a lifesaver for any quadratic equation in the form $ax^2 + bx + c = 0$. The formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $x^2 - 5x - 3 = 0$, we have $a=1$, $b=-5$, and $c=-3$. Let's substitute these values into the formula carefully:

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-3)}}{2(1)}$

Now, let's simplify this step by step. First, $-(-5)$ becomes $5$. Then, $(-5)^2$ is $25$. And $-4(1)(-3)$ is $+12$. So, the expression under the square root (the discriminant) becomes $25 + 12 = 37$.

$x = \frac{5 \pm \sqrt{37}}{2}$

This gives us two potential solutions because of the $\pm$ (plus-minus) sign. We have:

1. $x_1 = \frac{5 + \sqrt{37}}{2}$
2. $x_2 = \frac{5 - \sqrt{37}}{2}$

Now, we need to calculate these values and round them to three significant figures, as requested.

First, let's find the value of $\sqrt{37}$. Using a calculator, $\sqrt{37} \approx 6.08276...$

For $x_1$:

$x_1 = \frac{5 + 6.08276...}{2} = \frac{11.08276...}{2} \approx 5.54138...$

Rounding $5.54138...$ to three significant figures gives us $5.54$.

For $x_2$:

$x_2 = \frac{5 - 6.08276...}{2} = \frac{-1.08276...}{2} \approx -0.54138...$

Rounding $-0.54138...$ to three significant figures gives us $-0.541$.

So, our two solutions are approximately $5.54$ and $-0.541$. These are the values of $x$ that make the original equation true. The quadratic formula is incredibly powerful because it works for any quadratic equation, even those that are impossible to factor.

Checking Our Solutions

Alright, mathletes, we've done the heavy lifting and found our potential solutions: $x \approx 5.54$ and $x \approx -0.541$. But in math, especially when dealing with equations that had fractions or other complexities, it's always a smart move to check our answers. This means plugging these values back into the original equation ($\frac{3}{x}+5=x$) to see if both sides match up. This verification step helps catch any errors we might have made during the solving process. Plus, it solidifies our understanding.

Checking $x \approx 5.54$:

Let's substitute $5.54$ into the original equation:

Left side: $\frac{3}{5.54} + 5$

Using a calculator: $\frac{3}{5.54} \approx 0.541516...$

So, the left side is approximately $0.541516... + 5 = 5.541516...$

Right side: $x = 5.54$

Comparing the left side ($5.541516...$) and the right side ($5.54$), they are very close. The slight difference is due to rounding our solution to three significant figures. If we used the unrounded value $\frac{5 + \sqrt{37}}{2}$, the sides would be exactly equal. This looks good!

Checking $x \approx -0.541$:

Now, let's substitute $-0.541$ into the original equation:

Left side: $\frac{3}{-0.541} + 5$

Using a calculator: $\frac{3}{-0.541} \approx -5.545286...

So, the left side is approximately $-5.545286... + 5 = -0.545286...$

Right side: $x = -0.541$

Again, comparing the left side ($-0.545286...$) and the right side ($-0.541$), they are very close. The discrepancy arises from rounding. The unrounded value $\frac{5 - \sqrt{37}}{2}$ would yield an exact match. This confirms our second solution is also correct.

Remember our initial condition that $x \neq 0$? Both of our solutions, $5.54$ and $-0.541$, are not zero, so they are valid solutions. This checking process is super important for building confidence in your answers and for identifying potential mistakes. It's like double-checking your work before submitting an important assignment.

Final Answer

So, after all that hard work, guys, we have successfully solved the equation $\frac{3}{x}+5=x$. We transformed it into a quadratic equation, used the quadratic formula to find the roots, and then checked our answers. The solutions, rounded to three significant figures, are:

$x \approx 5.54$

and

$x \approx -0.541$

Keep practicing these types of problems, and you'll become algebra wizards in no time! Don't forget the key steps: clear the fraction, rearrange into a quadratic, use the appropriate formula, and always check your work. Math is all about building these skills step-by-step. Until next time, keep those equations balanced!