Solving Equations How Many Solutions Exist?
Hey guys! Today, we're diving into a fun math problem that involves figuring out how many solutions a given equation has. We'll break down the equation step by step, making sure everyone can follow along. So, let’s get started and see if we can crack this together!
Understanding the Problem
Our main task here is to determine how many solutions exist for the equation:
0. 75(x + 40) = 0.35(x + 20) + 0.35(x + 20)
What does this mean? Basically, we want to find out how many values of x can make this equation true. It could be no values (zero solutions), one value, two values, or even infinitely many values. This depends on the structure of the equation itself. The key to figuring this out is to simplify the equation and see what we end up with.
Why is this important? Understanding the number of solutions helps us in various real-world applications. For instance, in engineering, this kind of problem can help determine if a system has a stable solution or not. In economics, it can help in modeling market equilibrium. So, knowing how to approach these problems is super useful!
Before we jump into solving, let’s recap the options we have:
A. zero B. one C. two D. infinitely many
Keep these in mind as we solve the equation. By the end of our discussion, you'll not only know the answer but also why it's the correct answer. Let’s move on to simplifying this equation and getting closer to the solution!
Step-by-Step Solution
Alright, let’s get our hands dirty and solve this equation step by step. Remember, the goal is to simplify the equation so we can clearly see how many solutions there are. We'll take it slow and explain each step, so you guys can follow along easily. Let's do this!
1. Distribute the Constants
The first thing we need to do is distribute the constants outside the parentheses to the terms inside. This means we'll multiply 0.75 by both x and 40 on the left side, and 0.35 by both x and 20 in each term on the right side. This will help us get rid of those parentheses and make the equation easier to handle.
So, let's start with the left side:
0. 75 * x = 0.75x
0. 75 * 40 = 30
Therefore, the left side becomes 0.75x + 30.
Now, let’s move to the right side. We have two terms that are identical, 0.35(x + 20), so we'll distribute 0.35 in both terms:
0. 35 * x = 0.35x
1. 35 * 20 = 7
So, each 0.35(x + 20) becomes 0.35x + 7. Since we have two of these, we'll add them together:
(0.35x + 7) + (0.35x + 7)
2. Combine Like Terms
Now that we've distributed the constants, let's combine the like terms on both sides of the equation. This means grouping the x terms together and the constants together. Combining like terms simplifies the equation further and brings us closer to isolating x.
First, let's rewrite the entire equation with the distributed constants:
0. 75x + 30 = (0.35x + 7) + (0.35x + 7)
Now, let’s combine the terms on the right side. We'll add the 0.35x terms together and the 7s together:
0. 35x + 0.35x = 0.70x
2. + 7 = 14
So, the right side simplifies to 0.70x + 14. Now our equation looks like this:
0. 75x + 30 = 0.70x + 14
This is much cleaner, right? We're making progress! Next, we need to get all the x terms on one side and the constants on the other side.
3. Isolate the Variable
The next step is to isolate the variable x. This means we need to get all the terms with x on one side of the equation and all the constants (the numbers without x) on the other side. This will help us determine the value of x, if there is one.
Let's start by moving the 0.70x term from the right side to the left side. To do this, we'll subtract 0.70x from both sides of the equation:
0. 75x + 30 - 0.70x = 0.70x + 14 - 0.70x
This simplifies to:
3. 05x + 30 = 14
Now, we need to move the constant term 30 from the left side to the right side. We'll do this by subtracting 30 from both sides:
4. 05x + 30 - 30 = 14 - 30
This simplifies to:
5. 05x = -16
We're almost there! Now we just need to solve for x.
4. Solve for x
Okay, we're in the home stretch! We've got the equation down to 0.05x = -16. Now, we just need to solve for x. This means getting x all by itself on one side of the equation. To do this, we'll divide both sides by the coefficient of x, which is 0.05.
So, let’s divide both sides by 0.05:
(0.05x) / 0.05 = -16 / 0.05
This simplifies to:
x = -320
Woohoo! We found a value for x. This means there is at least one solution to the equation. But the question is, is it the only solution? Let's think about what we've found.
5. Determine the Number of Solutions
We’ve successfully solved the equation and found that x = -320. Now, let’s circle back to our main question: How many solutions exist for the given equation?
Think about it this way: we started with a linear equation (an equation where the highest power of x is 1). We followed algebraic steps to isolate x, and we arrived at a single value for x. In linear equations, if you can find a value for x, that value is the unique solution—there aren't any other values that would make the equation true.
So, we've found that x = -320 is a solution. And since it’s a linear equation, this is the only solution. There aren't infinitely many solutions, and there aren't two solutions. There’s just one!
Therefore, the answer is B. one.
Key Concepts
Before we wrap up, let’s quickly recap the key concepts we used to solve this problem. Understanding these concepts will help you tackle similar problems in the future. Trust me, this stuff is super useful, guys!
1. Distributive Property
We started by using the distributive property to multiply the constants outside the parentheses by the terms inside. This property is a fundamental tool in algebra. Remember, it looks like this:
a(b + c) = ab + ac
In our problem, we used it like this:
0. 75(x + 40) = 0.75x + 0.75 * 40
and
0. 35(x + 20) = 0.35x + 0.35 * 20
2. Combining Like Terms
After distributing, we combined like terms. This means adding together terms that have the same variable (like 0.75x and 0.35x) and adding together the constants (like 30 and 7). This simplifies the equation and makes it easier to work with.
For example, we combined:
0. 35x + 0.35x = 0.70x
and
7. + 7 = 14
3. Isolating the Variable
Isolating the variable is a crucial step in solving equations. It means getting the variable (x in our case) all by itself on one side of the equation. We did this by adding or subtracting terms from both sides to move them around, and then dividing by the coefficient of x.
4. Linear Equations and Unique Solutions
Finally, we remembered that linear equations (equations where x is raised to the power of 1) generally have one unique solution. This helped us understand that once we found a value for x, it was the only value that would satisfy the equation. Of course, equations can sometimes have no solutions or infinitely many, but this is less common and typically happens in specific scenarios like contradictions or identities.
Practice Problems
Alright, you guys have been awesome so far! Now, let’s put those skills to the test with a few practice problems. These will help solidify your understanding and boost your confidence. Remember, practice makes perfect, so let’s dive in!
Practice Problem 1
How many solutions does the following equation have?
2(x - 3) = 4x - 6
A. zero B. one C. two D. infinitely many
Practice Problem 2
Determine the number of solutions for the equation:
0. 5(2x + 4) = x + 2
A. zero B. one C. two D. infinitely many
Practice Problem 3
Solve the following equation and identify how many solutions exist:
3(x + 1) = 3x + 5
A. zero B. one C. two D. infinitely many
Solutions:
- D. infinitely many
- D. infinitely many
- A. zero
Conclusion
Alright, you guys made it to the end! We tackled a pretty cool math problem today, and I hope you feel more confident about solving equations and determining the number of solutions. Remember, the key is to take it step by step, use those algebraic tools we talked about, and think logically about what the equation is telling you. Keep practicing, and you’ll become equation-solving pros in no time! Keep up the awesome work!