Solving The Equation: $x-rac{3x+5}{x-10}=5$

Nov 9, 2025 by Admin 45 views

Hey guys! Let's dive into solving the equation $x- rac{3x+5}{x-10}=5$ . This is a classic algebra problem that involves fractions, so we'll need to use some algebraic manipulation to get to the solution. Don't worry, it's not as scary as it looks! We'll break it down step by step, making sure you understand each move. The core idea here is to get rid of that pesky fraction and isolate x. Ready? Let's go!

Understanding the Problem and Initial Steps

Alright, so the equation we're tackling is $x- rac{3x+5}{x-10}=5$ . Our goal is to find the value(s) of x that make this equation true. The main challenge? That fraction! To get rid of it, we're going to use a tried-and-true method: multiplying both sides of the equation by the denominator, which in this case is $(x-10)$ . But before we jump in, let's just make sure we understand the potential pitfalls. See, we can't let $x$ equal 10, because that would make the denominator zero, and dividing by zero is a big no-no in math. So, keep that in the back of your mind; we'll need to check our answer later to make sure it doesn't cause any problems. Now, the first step is to multiply both sides of the equation by $(x-10)$ . This will eliminate the fraction and simplify the equation. This is a super important step in solving equations involving fractions, ensuring we get a clearer path toward finding the solution to the problem, and is a key technique for anyone looking to build a solid foundation in algebra. Think of it like this: if you have a recipe and some ingredients are hard to work with, simplifying is the key!

Multiplying by the Denominator and Simplifying

Okay, let's get to work! We're multiplying both sides of $x- rac{3x+5}{x-10}=5$ by $(x-10)$ . Here's how it looks:

$(x-10) * (x- rac{3x+5}{x-10}) = 5 * (x-10)$

Now, let's distribute on both sides. On the left side, we multiply each term by $(x-10)$ :

$x*(x-10) - rac{3x+5}{x-10}*(x-10) = 5x - 50$

This simplifies to:

$x^2 - 10x - (3x + 5) = 5x - 50$

See how the $(x-10)$ in the fraction cancels out? That's the magic of this method! Now, let's clean this up by distributing the negative sign and combining like terms:

$x^2 - 10x - 3x - 5 = 5x - 50$

This gives us:

$x^2 - 13x - 5 = 5x - 50$

We are making progress! Remember how we mentioned not letting $x$ equal 10? We'll need to make sure our final answer doesn't break this rule. This is a common and important practice when solving equations with fractions, ensuring the validity of our solution by avoiding any undefined expressions. It's like a quality check for your math problem-solving skills! So, we're slowly moving closer to a point where we can easily solve for x.

Rearranging and Solving the Quadratic Equation

Alright, so the next step is to get everything on one side of the equation and set it equal to zero. This will give us a quadratic equation, which we can then solve. Let's move those terms around! We'll subtract $5x$ and add $50$ to both sides of the equation $x^2 - 13x - 5 = 5x - 50$ :

$x^2 - 13x - 5 - 5x + 50 = 0$

Combining like terms, we get:

$x^2 - 18x + 45 = 0$

Great! Now we have a standard quadratic equation. There are a few ways to solve this: factoring, completing the square, or using the quadratic formula. Let's try factoring, as it's often the quickest method when it works. We're looking for two numbers that multiply to 45 and add up to -18. Those numbers are -3 and -15. So, we can factor the quadratic equation as follows:

$(x - 3)(x - 15) = 0$

For this to be true, either $(x - 3) = 0$ or $(x - 15) = 0$ . This gives us two potential solutions:

$x - 3 = 0 => x = 3$
$x - 15 = 0 => x = 15$

We've got two possible answers, guys! But remember, we need to check if they work in the original equation and don't make the denominator zero. Let's go to the next section and verify if our answers are really the solutions.

Checking the Solutions

Now, let's see if our solutions are valid. We found two possible values for $x$ : 3 and 15. We have to plug them back into the original equation to see if they hold true. Let's begin with $x = 3$ . We plug it into the original equation $x- rac{3x+5}{x-10}=5$ and get:

$3 - rac{3(3) + 5}{3 - 10} = 5$

$3 - rac{9 + 5}{-7} = 5$

$3 - rac{14}{-7} = 5$

$3 - (-2) = 5$

$3 + 2 = 5$

$5 = 5$

This checks out! $x = 3$ is a valid solution. Now, let's check $x = 15$ .

$15 - rac{3(15) + 5}{15 - 10} = 5$

$15 - rac{45 + 5}{5} = 5$

$15 - rac{50}{5} = 5$

$15 - 10 = 5$

$5 = 5$

This also checks out! $x = 15$ is also a valid solution. Both solutions are good and don't cause the denominator to be zero, which is excellent. Remember how we said that $x$ cannot be 10? Neither of our answers breaks that rule. Always be cautious when solving for fractional equations! It is good practice.

Conclusion and Final Thoughts

Awesome, guys! We've successfully solved the equation $x- rac{3x+5}{x-10}=5$ . We found two valid solutions: $x = 3$ and $x = 15$ . We did this by first eliminating the fraction, then rearranging the equation into a quadratic form, solving it and finally, checking the solutions to ensure they were valid. Solving equations with fractions can be tricky, but by following a clear, step-by-step approach, we were able to get to the answer. Remember to always check your solutions, especially when dealing with fractions, to avoid any division-by-zero errors. This process is a fundamental skill in algebra and is essential for tackling more complex mathematical problems. So, keep practicing, keep learning, and you'll become a pro at solving these types of equations! Now you should be well-equipped to tackle similar problems. Keep up the good work!