Solving Equations: Finding The Number Of Solutions

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Solving Equations: Finding the Number of Solutions

Hey guys! Let's dive into the world of equations and figure out how to determine the number of solutions they have. Specifically, we'll tackle the equation 12(x+12)=4xβˆ’1\frac{1}{2}(x+12)=4x-1. Understanding how many solutions an equation possesses is a fundamental skill in algebra, and it's super useful for all sorts of problem-solving. It's like having a roadmap to understand the behavior of the equation. We'll explore different scenarios and learn how to identify whether an equation has no solution, one solution, two solutions, or even an infinite number of solutions. This is not just about memorizing rules, it is about understanding the core concepts of algebra that let you explore the relationship between variables and equations. So, let’s get started, shall we?

First things first, let's talk about what a 'solution' actually means. A solution to an equation is a value (or values) of the variable that makes the equation true. For example, if we have the equation x+2=5x + 2 = 5, the solution is x=3x = 3 because when we substitute 3 for x, the equation holds true (3+2=53 + 2 = 5). An equation can have one solution, no solutions, or infinitely many solutions. Understanding what kind of solutions is crucial in math. Knowing how to determine the number of solutions will help you solve more complex problems in the future. Now, let’s get into the specifics of our example and see how to find the answer. The equation we are working with today is 12(x+12)=4xβˆ’1\frac{1}{2}(x+12)=4x-1. Let’s get into how to solve it and determine how many solutions it has. Remember to pay close attention to each step because it’s important to understand the concept.

Unveiling the Solution: Step-by-Step Approach

Alright, let's break down how to solve the equation 12(x+12)=4xβˆ’1\frac{1}{2}(x+12)=4x-1 step by step. This is where we put our detective hats on and carefully work through each step to find the value(s) of x that make the equation true. Don’t worry; it's a piece of cake if you follow along with me. We are going to isolate x and then be able to determine the number of solutions. This process of isolating x will help you when you’re facing different algebraic equations. It will help you find the different values of x. It's like a game where the goal is to get x all by itself on one side of the equation. Trust me, it’s going to be much easier than you think! Are you ready to dive in, guys?

  1. Distribute the 12\frac{1}{2}: The first step involves distributing the 12\frac{1}{2} across the terms inside the parentheses. So, 12(x+12)\frac{1}{2}(x + 12) becomes 12x+6\frac{1}{2}x + 6. Our equation now looks like this: 12x+6=4xβˆ’1\frac{1}{2}x + 6 = 4x - 1. This is the first step in simplifying the equation and getting closer to solving for x. Remember, distributing is like giving each term inside the parenthesis a little treat of 12\frac{1}{2}. It is important that you do this step carefully. It ensures that you don’t make any mistakes when solving the equation.

  2. Isolate the x terms: Next, we want to get all the x terms on one side of the equation. To do this, let's subtract 12x\frac{1}{2}x from both sides. This gives us: 6=4xβˆ’12xβˆ’16 = 4x - \frac{1}{2}x - 1. Let’s simplify that: 4xβˆ’12x4x - \frac{1}{2}x equals 3.5x3.5x or 72x\frac{7}{2}x. So, we now have 6=72xβˆ’16 = \frac{7}{2}x - 1. Basically, we're grouping all the terms that have x together. This helps us see how x behaves and gets us closer to finding its value. Remember, what you do on one side, you have to do on the other. That is the rule for any algebraic equation.

  3. Isolate the constant terms: Now, let's get all the constant terms (the numbers without x) on the other side. Add 1 to both sides: 6+1=72x6 + 1 = \frac{7}{2}x. This simplifies to 7=72x7 = \frac{7}{2}x. We are trying to isolate x. This is the main point of what we are doing. Remember that. Each step is moving us closer to the solution.

  4. Solve for x: Finally, to isolate x, we need to get rid of the fraction. You can do this by multiplying both sides by 27\frac{2}{7} (the reciprocal of 72\frac{7}{2}). So, we have 7βˆ—27=x7 * \frac{2}{7} = x. Which simplifies to x=2x = 2. We have successfully found the value of x! Now we have a solution that makes the equation correct! This is what it’s all about, figuring out the value of x.

Decoding the Number of Solutions

Now that we have successfully solved for x and found that x=2x = 2, it's time to figure out how many solutions our equation has. Given our result, 12(x+12)=4xβˆ’1\frac{1}{2}(x+12)=4x-1 has only one solution, which is x equals 2. The solution is unique and there are no other values of x that will make the original equation true. This tells us the equation represents a situation where there is a single point of intersection. Imagine two lines on a graph: this equation describes a scenario where those two lines cross each other at just one point. The value of x equals 2 represents the x-coordinate of that intersection point. Getting this conclusion requires us to understand what a solution to an equation means and how to find it. This highlights how essential it is to have a good understanding of algebra! It is like the foundation of solving more complex mathematical problems. Understanding how to find the number of solutions is a key concept that will help you solve more complex equations. And it's not just about getting the right answer; it's about understanding the 'why' behind it.

When you solve an equation and find a single value for the variable, as we did here with x=2x=2, you have one solution. This indicates a very specific point where the equation is satisfied. Equations can also have no solutions or infinitely many solutions, but in our case, we have one and only one. The ability to identify the number of solutions is a fundamental skill in algebra, as it helps interpret the nature of the relationship described by the equation. Recognizing the number of solutions helps us analyze and interpret the problem. This skill is useful in various areas of mathematics and related fields.

Exploring Other Possibilities: No Solutions and Infinite Solutions

While our equation had one solution, let's quickly discuss the other possibilities: no solutions and infinitely many solutions. It’s like we are adding more tools to our toolbox. These possibilities can show up in different equations, so it's good to know what to look for. Are you ready?

  • No Solution: An equation has no solution if there's no value of the variable that makes the equation true. This often happens when you end up with a contradiction, such as 2=52 = 5. This happens when you have a false statement as a result of your solving process. Think about it like two parallel lines. They never intersect. They never have a common point. An equation with no solution means there is no value of x that makes both sides of the equation equal. It is like the equation is impossible.

  • Infinitely Many Solutions: An equation has infinitely many solutions if any value of the variable makes the equation true. This occurs when both sides of the equation are essentially the same. For example, if you simplify an equation and end up with something like x+2=x+2x + 2 = x + 2, this means any value of x will make the equation true. Think about it like having two identical lines. They overlap at every point. This means that any value of x is a solution. This situation is less common, but understanding it is super important. It highlights when an equation is always true, no matter what value we assign to the variable.

Understanding these different types of solutions is an important part of solving algebraic problems. They help us analyze and interpret what the equation is trying to tell us. Now you are well-equipped to tackle different kinds of equations and determine how many solutions they have. It's like having a superpower. You can quickly see the behavior of the equation! Don't you think it's cool?

Final Thoughts: Wrapping It Up

We have explored how to determine the number of solutions for a given equation. We solved 12(x+12)=4xβˆ’1\frac{1}{2}(x+12)=4x-1 and found out that it has one solution, where x=2x = 2. Along the way, we touched on the concepts of no solutions and infinitely many solutions, giving us a complete understanding. Knowing how to find the number of solutions helps us understand the nature of the equation and its implications. It's an essential skill for algebra and further studies in mathematics. Remember, practice is key! The more you solve equations, the more familiar you will become with these concepts. Keep practicing! Are you ready to solve another equation? I am sure you can do it!

So, the correct answer is B. one.