Find real number 'x' and 'y' such that
`3x+2iy-ix+5y=7+5i`.

Answer

To find the real numbers \( x \) and \( y \) such that \( 3x + 2iy - ix + 5y = 7 + 5i \), we will equate the real and imaginary parts of the complex numbers.

### Step 1: Rearrange the equation
We start with the equation:
\[ 3x + 2iy - ix + 5y = 7 + 5i \]
We can rearrange the left side to group the real and imaginary parts:
\[ (3x + 5y) + (2y - x)i = 7 + 5i \]

### Step 2: Equate real parts
From the equation, we can see that the real part on the left side is \( 3x + 5y \) and the real part on the right side is \( 7 \). Thus, we can set up the first equation:
\[ 3x + 5y = 7 \quad \text{(1)} \]

### Step 3: Equate imaginary parts
Next, we look at the imaginary parts. The imaginary part on the left side is \( 2y - x \) and the imaginary part on the right side is \( 5 \). Therefore, we set up the second equation:
\[ 2y - x = 5 \quad \text{(2)} \]

### Step 4: Solve the system of equations
Now we have a system of two equations:
1. \( 3x + 5y = 7 \)
2. \( 2y - x = 5 \)

We can solve these equations simultaneously. Let's solve equation (2) for \( x \):
\[ x = 2y - 5 \quad \text{(3)} \]

### Step 5: Substitute equation (3) into equation (1)
Now, substitute equation (3) into equation (1):
\[ 3(2y - 5) + 5y = 7 \]
Expanding this gives:
\[ 6y - 15 + 5y = 7 \]
Combining like terms:
\[ 11y - 15 = 7 \]
Adding 15 to both sides:
\[ 11y = 22 \]
Dividing by 11:
\[ y = 2 \]

### Step 6: Substitute \( y \) back to find \( x \)
Now that we have \( y \), we can substitute it back into equation (3) to find \( x \):
\[ x = 2(2) - 5 = 4 - 5 = -1 \]

### Final Result
Thus, the values of \( x \) and \( y \) are:
\[ x = -1, \quad y = 2 \]