Extraneous Solutions In Equations A Detailed Explanation
Hey everyone! Today, we're diving into a fascinating problem in mathematics that involves solving an equation and identifying extraneous solutions. Extraneous solutions, guys, are like those unwanted guests at a party – they show up in our calculations but don't actually belong in the final answer. We're going to break down the equation $\frac{x}{2} + \frac{2x-9}{x-7} = \frac{5}{x-7}$ and figure out if the solutions $x = -4$ and $x = 7$ are legit or just crashers.
Before we jump into the nitty-gritty, let's make sure we're all on the same page about what extraneous solutions really are. In simple terms, an extraneous solution is a value that we get when solving an equation, but it doesn't actually satisfy the original equation. This usually happens when we perform operations that aren't reversible, like squaring both sides of an equation or, in our case, dealing with fractions that have variables in the denominator. When we have variables in the denominator, certain values can make the denominator equal to zero, which is a big no-no in math because division by zero is undefined. So, we need to be extra careful and check our solutions to make sure they don't cause any mathematical mayhem.
Why do these extraneous solutions pop up? Well, when we solve equations, we often manipulate them to make them easier to handle. For example, we might multiply both sides by an expression to get rid of fractions. This is a perfectly valid technique, but it can sometimes introduce solutions that weren't there in the first place. It's like adding extra ingredients to a recipe – you might end up with something that looks delicious, but it doesn't quite taste right. So, always remember: verification is key! Once we find potential solutions, we've got to plug them back into the original equation to see if they hold up. If a solution makes the equation true, it's a keeper. If it leads to an undefined expression or a contradiction, we've got an extraneous solution on our hands, and we need to kick it to the curb. This careful checking process is what ensures we get the correct and meaningful solutions to our mathematical puzzles.
Okay, let's get down to business and solve the equation. We've got $\frac{x}{2} + \frac{2x-9}{x-7} = \frac{5}{x-7}$. The first thing we want to do is get rid of those fractions, because honestly, who likes dealing with fractions? To do that, we need to find the least common denominator (LCD) of all the fractions in the equation. In this case, our denominators are 2 and (x-7), so the LCD is simply 2(x-7). Now, we're going to multiply both sides of the equation by this LCD. This is a crucial step because it will clear out the fractions and make the equation much easier to work with. When we multiply each term by 2(x-7), we get:
Now, let's simplify this beast. The 2 in the first term cancels out, and the (x-7) in the second and third terms cancel out, leaving us with:
See? Much cleaner already! Now, let's distribute and expand everything:
Combine like terms to get a quadratic equation:
Subtract 10 from both sides to set the equation to zero:
Now, we've got a quadratic equation in the standard form. We can solve this by factoring, using the quadratic formula, or completing the square. Factoring is often the quickest method if we can find factors easily. In this case, we're looking for two numbers that multiply to -28 and add up to -3. Those numbers are -7 and 4. So, we can factor the quadratic equation as:
This gives us two potential solutions: $x = 7$ and $x = -4$. But hold on! We're not done yet. We need to check these solutions to see if they actually work in the original equation or if they're those sneaky extraneous ones.
Alright, we've got our potential solutions: $x = 7$ and $x = -4$. Now comes the crucial step: checking for extraneous solutions. Remember, we need to plug each of these values back into the original equation to see if they make it true. This is super important, guys, because if we skip this step, we might end up with the wrong answer. Let's start with $x = 7$. Plugging this into our original equation, $\frac{x}{2} + \frac{2x-9}{x-7} = \frac{5}{x-7}$, we get:
Simplifying, we have:
Uh oh! We've got division by zero, which is a major red flag. Division by zero is undefined in mathematics, meaning that this expression is meaningless. So, $x = 7$ makes the denominators in the original equation zero, which tells us that it is indeed an extraneous solution. It's a no-go!
Now, let's check the other solution, $x = -4$. Plugging this into the original equation, we get:
Simplifying each term, we have:
Now, let's convert -2 to a fraction with a denominator of 11:
Hooray! This equation holds true. So, $x = -4$ is a valid solution. It passes the test and earns its place in our final answer.
So, what have we learned today, guys? We started with the equation $\frac{x}{2} + \frac{2x-9}{x-7} = \frac{5}{x-7}$, found two potential solutions ($x = 7$ and $x = -4$), and then did the crucial step of checking for extraneous solutions. We discovered that $x = 7$ is an extraneous solution because it leads to division by zero in the original equation. On the other hand, $x = -4$ is a valid solution because it satisfies the original equation. Therefore, the only valid solution to the equation is $x = -4$.
Remember, in mathematics, it's not enough to just find potential solutions. We always need to verify them, especially when dealing with equations that have variables in the denominator or radicals. Checking for extraneous solutions is like double-checking your work – it ensures that you get the correct and meaningful answers. So, keep practicing, keep checking, and keep conquering those mathematical challenges!
Therefore, the solution $x = 7$ is extraneous.
