Determine real values of x and y for which each statement is true
`-(x+3y) i+ (2-y+1)= (8)/(i)`

Answer

To solve the equation \(- (x + 3y)i + (2 - y + 1) = \frac{8}{i}\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ - (x + 3y)i + (2 - y + 1) = \frac{8}{i} \]
We can rewrite the right-hand side:
\[ \frac{8}{i} = 8 \cdot \frac{-i}{-1} = -8i \]
So, the equation becomes:
\[ - (x + 3y)i + (3 - y) = -8i \]

### Step 2: Separate real and imaginary parts
Now, we separate the real and imaginary parts:
- Real part: \(3 - y\)
- Imaginary part: \(- (x + 3y)\)

This gives us the system of equations:
1. \(3 - y = 0\) (real part)
2. \(- (x + 3y) = -8\) (imaginary part)

### Step 3: Solve the first equation
From the first equation:
\[ 3 - y = 0 \implies y = 3 \]

### Step 4: Substitute \(y\) into the second equation
Now, substitute \(y = 3\) into the second equation:
\[ - (x + 3(3)) = -8 \]
This simplifies to:
\[ - (x + 9) = -8 \]
Multiplying both sides by -1 gives:
\[ x + 9 = 8 \]

### Step 5: Solve for \(x\)
Now, solve for \(x\):
\[ x = 8 - 9 \implies x = -1 \]

### Final Solution
The real values of \(x\) and \(y\) are:
\[ x = -1, \quad y = 3 \]