Solving Equations: Find All Solutions!

Hey guys! Let's dive into a cool math problem where we're gonna find all the solutions to an equation. It's like a treasure hunt, but instead of gold, we're looking for the values of 'x' that make the equation true. We'll break down the problem step-by-step so it's super easy to follow. So, grab your calculators (or your brains!) and let's get started. We're going to solve the equation: $\frac{x+4}{x-1}=\frac{-x^2+3 x+12}{(x+3)(x-1)}$.

Understanding the Problem: The Equation Unveiled

Alright, first things first, let's understand what we're dealing with. The equation $\frac{x+4}{x-1}=\frac{-x^2+3 x+12}{(x+3)(x-1)}$ looks a little intimidating at first glance, but trust me, it's not as scary as it seems. We have two fractions set equal to each other. Our main goal is to find the values of 'x' that satisfy this equation, meaning when you plug those values into the equation, both sides are equal. One important thing to keep in mind is that we need to be careful about any values of 'x' that might cause the denominator (the bottom part of the fraction) to equal zero. Why? Because dividing by zero is a big no-no in math. It's undefined, which means it doesn't make sense. So, we'll need to check for those values later. This step is super important when we're dealing with rational expressions (fractions with variables). We're essentially trying to find the 'x' values that make the equation work. It's like solving a puzzle; we need to find the pieces (the values of 'x') that fit perfectly to make both sides of the equation equal.

So, before we start solving, let's take a quick look at the denominators: we have (x-1) and (x+3)(x-1). This tells us that 'x' cannot be equal to 1 or -3, because those values would make the denominators zero, and we can't have that. This is something we must remember, because at the end of the calculations we need to exclude these answers. Now that we know what to look for, and what to avoid, we can start our calculations.

Now, let's clear those fractions and solve for the unknown 'x'.

Solving the Equation: Step-by-Step Guide

Now comes the fun part: solving the equation! We'll go through the steps carefully to make sure we don't miss anything. Our goal is to isolate 'x' on one side of the equation. To do this, we'll need to get rid of the fractions. The first step to solve is to get rid of fractions. To do this, we'll multiply both sides of the equation by the common denominator, which is (x+3)(x-1). This is a really important move because it simplifies the equation and makes it much easier to solve. When we multiply the left side of the equation $\frac{x+4}{x-1}$ by $(x+3)(x-1)$, the $(x-1)$ terms cancel out, leaving us with $(x+4)(x+3)$. On the right side of the equation, the entire denominator $(x+3)(x-1)$ cancels out, and we are left with just the numerator, which is $-x^2+3x+12$.

So now our equation looks like this: $(x+4)(x+3) = -x^2+3x+12$. Now it's time to expand the left side of the equation by multiplying out the terms.

Expanding $(x+4)(x+3)$ means multiplying each term in the first parenthesis by each term in the second parenthesis. So, x times x is $x^2$, x times 3 is 3x, 4 times x is 4x, and 4 times 3 is 12. So, we get $x^2 + 3x + 4x + 12$. Combining the like terms, 3x and 4x, we get $7x$. So, the left side of the equation simplifies to $x^2 + 7x + 12$.

Now our equation looks like this: $x^2 + 7x + 12 = -x^2 + 3x + 12$. Now, let's move all the terms to one side of the equation to set it equal to zero. To do this, we'll add $x^2$ to both sides, subtract $3x$ from both sides, and subtract 12 from both sides. When we do this, we get $2x^2 + 4x = 0$. Now we have a simpler equation, but we are not finished yet. Now we have an easier equation to work with! To find the values of 'x' that satisfy the equation, we can factor out a common term, which in this case is 2x. This gives us $2x(x + 2) = 0$. For this equation to be true, either $2x = 0$ or $(x+2) = 0$.

Finding the Solutions: Identifying the Values of 'x'

Alright, we're in the home stretch now! We've simplified the equation, and now we need to find the actual solutions, the values of 'x' that make everything work. From the equation $2x(x + 2) = 0$, we know that either $2x = 0$ or $(x + 2) = 0$. Let's solve each of these separately. If $2x = 0$, then dividing both sides by 2 gives us $x = 0$. This is one of our solutions! Easy peasy. If $(x + 2) = 0$, then subtracting 2 from both sides gives us $x = -2$. This is our second solution. So, we've found two possible solutions: $x = 0$ and $x = -2$. But wait! Remember how we talked about those pesky values that make the denominator zero? Let's go back and check our initial restrictions: x cannot be 1 or -3. Our solutions, 0 and -2, are both valid because they don't violate these restrictions. We need to be vigilant about the restrictions, as those can invalidate one of our possible solutions.

So, our potential solutions are $x = 0$ and $x = -2$. Now, let's consider the multiple-choice options. We have to choose all answers that apply, so let's check them. Option A is -5, which is not one of our solutions. Option B is -3, but we said that $x$ cannot be -3. Option C is -2, which is one of our solutions! Option D is 0, which is also one of our solutions. Option E is 1, but we said that $x$ cannot be 1. Option F is 4, which is not one of our solutions.

So, the correct answers are C and D.

Checking the Solutions: Making Sure They Work

It's always a good idea to check your solutions. Let's plug each solution back into the original equation to make sure it works. For $x = 0$, the left side of the equation is $\frac{0+4}{0-1} = \frac{4}{-1} = -4$. The right side of the equation is $\frac{-0^2+3(0)+12}{(0+3)(0-1)} = \frac{12}{(3)(-1)} = \frac{12}{-3} = -4$. Since both sides are equal, $x = 0$ is indeed a solution. Now, let's check $x = -2$. The left side of the equation is $\frac{-2+4}{-2-1} = \frac{2}{-3} = -\frac{2}{3}$. The right side of the equation is $\frac{-(-2)^2+3(-2)+12}{(-2+3)(-2-1)} = \frac{-4-6+12}{(1)(-3)} = \frac{2}{-3} = -\frac{2}{3}$. Since both sides are equal, $x = -2$ is also a solution. This step is super important because it confirms that our solutions are correct. It's always a good practice to go back and check your work to minimize errors. By plugging the solutions back into the original equation, we can ensure that we haven't made any mistakes along the way. Seeing both sides of the equation match up gives you a great sense of confidence that you've solved the problem correctly.

We did it, guys! We successfully found all the solutions to the equation. Isn't that awesome? We started with a seemingly complicated equation, simplified it step-by-step, and found the values of 'x' that make it true. Keep practicing, and you'll become a pro at solving equations in no time! Remember to always check your solutions, and pay attention to those restrictions! Great job!