Rewriting Equations: Finding Equivalent Solutions

Hey guys! Let's dive into some math and figure out how to rewrite equations while keeping the same solutions. We'll be using properties of equations to manipulate the given equation and identify those that are equivalent. This is super important because it helps us solve problems more easily and understand the relationships between different equations. Understanding the properties of equations, like the commutative, associative, and distributive properties, is key to success in algebra and beyond. So, let's get started and make sure you're confident with these concepts!

Understanding the Original Equation and Properties

Alright, let's start with the original equation: $\frac{3}{5} x+\frac{2}{3}+x=\frac{1}{2}-\frac{1}{5} x$. Our mission is to transform this equation using mathematical properties without changing its solution. Remember, the solution to an equation is the value of 'x' that makes the equation true. To keep the solution the same, every move we make must be mathematically sound. This means sticking to the basic rules of algebra.

The core properties we'll be using include the commutative property (which allows us to change the order of addition and multiplication), the associative property (which lets us group terms differently in addition and multiplication), and the distributive property (which allows us to multiply a term across terms inside parentheses). Let's go through the original equation and identify each step in detail.

First, combine like terms on the left side of the equation. We can combine $\frac{3}{5} x$ and $x$. Since $x$ is the same as $\frac{5}{5} x$, adding the two gives us $\frac{8}{5} x$. So, we have $\frac{8}{5} x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5} x$. Then, we can add $\frac{1}{5} x$ to both sides, which gets us $\frac{9}{5} x+\frac{2}{3}=\frac{1}{2}$. We can subtract $\frac{2}{3}$ from both sides as well. Finally, we isolate x by multiplying both sides by the reciprocal of $\frac{9}{5}$, which is $\frac{5}{9}$.

It is important to understand why each step is valid. For example, adding or subtracting the same value from both sides maintains equality. Similarly, multiplying or dividing both sides by the same non-zero value also preserves the solution. These principles are fundamental in equation manipulation.

Applying Properties: A Step-by-Step Guide

Now, let's look at each option provided and determine if it has the same solution as our original equation. We'll methodically use algebraic manipulation to confirm.

Analyzing the Options

Let's analyze each option, comparing it back to the original equation $\frac{3}{5} x+\frac{2}{3}+x=\frac{1}{2}-\frac{1}{5} x$ to determine which equations share the same solution.

Option A: $\frac{8}{5} x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5} x$

Notice that in option A, the terms $\frac{3}{5} x$ and $x$ have been combined on the left side to get $\frac{8}{5} x$. This is a valid application of the commutative property and is perfectly acceptable. Then, no other operations have been performed. This is a crucial first step in simplifying the equation. Thus, by combining like terms, the solution to equation A remains identical to our original equation. So, option A has the same solution as the original equation. Let's make sure that we understand the steps carefully so that we can easily find solutions.

Option B: $18 x+20+30 x=15-6 x$

This option appears very different at first glance, but let's see if we can transform it into something familiar. Let's convert all terms to a common denominator to make it simple. First, let's simplify and combine like terms. If we multiply our original equation by 30 to get rid of the fractions, we'll get something similar to this.

• Original Equation multiplied by 30:
  • $30 * (\frac{3}{5} x)+30 * (\frac{2}{3})+30 * x=30 * (\frac{1}{2})-30 * (\frac{1}{5} x)$
  • $18 x+20+30 x=15-6 x$

By simplifying the original equation, we have successfully created an equivalent equation with the same solution. So, yes, it seems that option B is also correct!

Option C: $9 x+10+15 x=7.5-3 x$

Again, let's follow the same strategy and try to convert the option C equation to the original equation form or compare it with option B to make our life simple. Comparing option C to our original equation, can we get there? If we multiply by 15, we get:

• $15 * (\frac{3}{5} x)+15 * (\frac{2}{3})+15 * x=15 * (\frac{1}{2})-15 * (\frac{1}{5} x)$
  • $9 x+10+15 x=7.5-3 x$

Comparing this with the original equation, we can see that all the operations are correct, so option C is also correct.

Option D: $9 x+6+15 x=15-3 x$

To check if option D has the same solution, we can use the same strategy again. To see this, we can compare it with the original equation and its multiple forms. Multiplying our original equation by 15 again, we get:

• $15 * (\frac{3}{5} x)+15 * (\frac{2}{3})+15 * x=15 * (\frac{1}{2})-15 * (\frac{1}{5} x)$
  • $9 x+10+15 x=7.5-3 x$

Comparing option D with this simplified form, we can clearly see that it is not similar and does not have the same solution, which makes this option incorrect.

Conclusion: Selecting the Right Options

So, guys, based on our analysis, we can determine which equations have the same solution as the original equation, $\frac{3}{5} x+\frac{2}{3}+x=\frac{1}{2}-\frac{1}{5} x$. Remember, we're looking for equations that, when solved, would give us the same 'x' value.

We found that options A, B, and C are correct, and option D is incorrect. Therefore, the answer choices are:

• A: $\frac{8}{5} x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5} x$
• B: $18 x+20+30 x=15-6 x$
• C: $9 x+10+15 x=7.5-3 x$

These options all represent valid transformations of the original equation using algebraic properties, ensuring that the solution remains unchanged. I hope you guys enjoyed this. Let me know if you have any questions!