Determine real values of x and y for which each statement is true
`(x+y)/(i)+ x- y+ 4=0`

Answer

To solve the equation \((x+y)/i + x - y + 4 = 0\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{x+y}{i} + x - y + 4 = 0 \]

### Step 2: Multiply through by \(i\)
To eliminate the fraction, we can multiply the entire equation by \(i\):
\[ (x+y) + i(x - y + 4i) = 0 \]

### Step 3: Separate real and imaginary parts
Now, we can rewrite the equation as:
\[ (x + y) + i(x - y + 4) = 0 \]
This means that the real part must equal 0 and the imaginary part must also equal 0. Therefore, we have two equations:
1. \(x + y = 0\) (Real part)
2. \(x - y + 4 = 0\) (Imaginary part)

### Step 4: Solve the first equation
From the first equation \(x + y = 0\), we can express \(x\) in terms of \(y\):
\[ x = -y \]

### Step 5: Substitute into the second equation
Now, substitute \(x = -y\) into the second equation:
\[ -x + 4 = 0 \implies -(-y) + 4 = 0 \implies y + 4 = 0 \]

### Step 6: Solve for \(y\)
Solving for \(y\):
\[ y = -4 \]

### Step 7: Find \(x\)
Now substitute \(y = -4\) back into \(x = -y\):
\[ x = -(-4) = 4 \]

### Step 8: Conclusion
Thus, the real values of \(x\) and \(y\) are:
\[ x = 4, \quad y = -4 \]

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