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Answer :
- Add $\frac{1}{2}x$ to both sides of the equation $\frac{1}{2}x = -\frac{1}{2}x$.
- Simplify the equation to get $x = 0$.
- The solution to the equation is $\boxed{0}$.
### Explanation
1. Analyze the equation
We are given the equation $\frac{1}{2}x = -\frac{1}{2}x$. Our goal is to find the value of $x$ that satisfies this equation.
2. Solve for x
To solve for $x$, we can add $\frac{1}{2}x$ to both sides of the equation:$$\frac{1}{2}x + \frac{1}{2}x = -\frac{1}{2}x + \frac{1}{2}x$$This simplifies to:$$x = 0$$
3. State the solution
Therefore, the value of $x$ that satisfies the equation $\frac{1}{2}x = -\frac{1}{2}x$ is $x = 0$.
### Examples
Imagine you're balancing a scale. On one side, you have half the weight of an unknown object, and on the other side, you have the negative of half the weight of the same object. The only way the scale can balance is if the object has no weight at all, meaning its weight is zero. This is similar to how the equation $\frac{1}{2}x = -\frac{1}{2}x$ is only true when $x = 0$. This concept is useful in physics for understanding equilibrium and balance in various systems.
- Simplify the equation to get $x = 0$.
- The solution to the equation is $\boxed{0}$.
### Explanation
1. Analyze the equation
We are given the equation $\frac{1}{2}x = -\frac{1}{2}x$. Our goal is to find the value of $x$ that satisfies this equation.
2. Solve for x
To solve for $x$, we can add $\frac{1}{2}x$ to both sides of the equation:$$\frac{1}{2}x + \frac{1}{2}x = -\frac{1}{2}x + \frac{1}{2}x$$This simplifies to:$$x = 0$$
3. State the solution
Therefore, the value of $x$ that satisfies the equation $\frac{1}{2}x = -\frac{1}{2}x$ is $x = 0$.
### Examples
Imagine you're balancing a scale. On one side, you have half the weight of an unknown object, and on the other side, you have the negative of half the weight of the same object. The only way the scale can balance is if the object has no weight at all, meaning its weight is zero. This is similar to how the equation $\frac{1}{2}x = -\frac{1}{2}x$ is only true when $x = 0$. This concept is useful in physics for understanding equilibrium and balance in various systems.
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