Answer :
Sure, let's solve the given equation step by step. We're given:
[tex]\[ \frac{(1+i)x - 2i}{3+i} + \frac{(2 - 3i)y + i}{3-i} = i \][/tex]
The first step is to simplify each fraction. Let's simplify the denominator by multiplying the numerator and the denominator by the conjugate of the denominator.
### Simplifying the first term [tex]\(\frac{(1+i)x - 2i}{3+i}\)[/tex]:
Multiply the numerator and the denominator by the conjugate of [tex]\(3+i\)[/tex], which is [tex]\(3-i\)[/tex]:
[tex]\[ \frac{[(1+i)x - 2i](3-i)}{(3+i)(3-i)} \][/tex]
Since [tex]\( (3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 10 \)[/tex], we get:
[tex]\[ \frac{[(1+i)x - 2i](3-i)}{10} \][/tex]
Now expand the numerator:
[tex]\[ (1+i)x(3-i) - 2i(3-i) \][/tex]
Distribute [tex]\( (1+i)x \)[/tex]:
[tex]\[ = (3x - ix + 3ix - i^2 x) - 2i(3-i) \][/tex]
[tex]\[ = (3x + 2ix + x) - (6i + 2) \][/tex]
[tex]\[ = 4x + 2ix - 6i - 2 \][/tex]
Thus, the first term simplifies to:
[tex]\[ \frac{4x + 2ix - 6i - 2}{10} \][/tex]
This can be split into real and imaginary parts:
[tex]\[ \frac{4x - 2}{10} + \frac{2ix - 6i}{10} = \frac{4x - 2}{10} + \frac{2x - 6}{10}i \][/tex]
### Simplifying the second term [tex]\(\frac{(2 - 3i)y + i}{3-i}\)[/tex]:
Multiply the numerator and the denominator by the conjugate of [tex]\(3-i\)[/tex], which is [tex]\(3+i\)[/tex]:
[tex]\[ \frac{[(2 - 3i)y + i](3+i)}{(3-i)(3+i)} \][/tex]
Since [tex]\( (3-i)(3+i) = 3^2 - (-1) = 9 + 1 = 10 \)[/tex], we get:
[tex]\[ \frac{[(2 - 3i)y + i](3+i)}{10} \][/tex]
Now expand the numerator:
[tex]\[ (2 - 3i)y(3+i) + i(3+i) \][/tex]
Distribute [tex]\( (2 - 3i)y \)[/tex]:
[tex]\[ = (6y + 2iy - 9iy - 3i^2 y) + (3i + i^2) \][/tex]
[tex]\[ = (6y - 7iy + 3y) + (3i - 1) \][/tex]
[tex]\[ = 9y - 7iy + 3i - 1 \][/tex]
Thus, the second term simplifies to:
[tex]\[ \frac{9y - 1}{10} + \frac{-7iy + 3i}{10} \][/tex]
This can be split into real and imaginary parts:
[tex]\[ \frac{9y - 1}{10} + \frac{-7iy + 3i}{10} = \frac{9y - 1}{10} + \frac{-7y + 3}{10}i \][/tex]
### Combine and equate to [tex]\(i\)[/tex]:
Combining the simplified fractions:
[tex]\[ \left(\frac{4x - 2}{10} + \frac{9y - 1}{10}\right) + \left(\frac{2x - 6}{10} + \frac{-7y + 3}{10}\right)i = i \][/tex]
Equating the real parts and the imaginary parts separately:
Real part:
[tex]\[ \frac{4x - 2}{10} + \frac{9y - 1}{10} = 0 \][/tex]
Simplifies to:
[tex]\[ 4x - 2 + 9y - 1 = 0 \][/tex]
[tex]\[ 4x + 9y - 3 = 0 \quad \text{(1)} \][/tex]
Imaginary part:
[tex]\[ \frac{2x - 6}{10} + \frac{-7y + 3}{10} = 1 \][/tex]
Simplifies to:
[tex]\[ 2x - 6 - 7y + 3 = 10 \][/tex]
[tex]\[ 2x - 7y - 3 = 10 \][/tex]
[tex]\[ 2x - 7y = 13 \quad \text{(2)} \][/tex]
### Solve the system of equations:
We have the system of linear equations:
1. [tex]\( 4x + 9y = 3 \)[/tex]
2. [tex]\( 2x - 7y = 13 \)[/tex]
Multiply the second equation by 2:
[tex]\[ 4x - 14y = 26 \][/tex]
Subtract from the first equation:
[tex]\[ 4x + 9y - (4x - 14y) = 3 - 26 \][/tex]
[tex]\[ 4x + 9y - 4x + 14y = -23 \][/tex]
[tex]\[ 23y = -23 \][/tex]
[tex]\[ y = -1 \][/tex]
Substitute [tex]\(y = -1\)[/tex] into the first equation:
[tex]\[ 4x + 9(-1) = 3 \][/tex]
[tex]\[ 4x - 9 = 3 \][/tex]
[tex]\[ 4x = 12 \][/tex]
[tex]\[ x = 3 \][/tex]
Thus, the real values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are:
[tex]\[ \boxed{x = 3, y = -1} \][/tex]
[tex]\[ \frac{(1+i)x - 2i}{3+i} + \frac{(2 - 3i)y + i}{3-i} = i \][/tex]
The first step is to simplify each fraction. Let's simplify the denominator by multiplying the numerator and the denominator by the conjugate of the denominator.
### Simplifying the first term [tex]\(\frac{(1+i)x - 2i}{3+i}\)[/tex]:
Multiply the numerator and the denominator by the conjugate of [tex]\(3+i\)[/tex], which is [tex]\(3-i\)[/tex]:
[tex]\[ \frac{[(1+i)x - 2i](3-i)}{(3+i)(3-i)} \][/tex]
Since [tex]\( (3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 10 \)[/tex], we get:
[tex]\[ \frac{[(1+i)x - 2i](3-i)}{10} \][/tex]
Now expand the numerator:
[tex]\[ (1+i)x(3-i) - 2i(3-i) \][/tex]
Distribute [tex]\( (1+i)x \)[/tex]:
[tex]\[ = (3x - ix + 3ix - i^2 x) - 2i(3-i) \][/tex]
[tex]\[ = (3x + 2ix + x) - (6i + 2) \][/tex]
[tex]\[ = 4x + 2ix - 6i - 2 \][/tex]
Thus, the first term simplifies to:
[tex]\[ \frac{4x + 2ix - 6i - 2}{10} \][/tex]
This can be split into real and imaginary parts:
[tex]\[ \frac{4x - 2}{10} + \frac{2ix - 6i}{10} = \frac{4x - 2}{10} + \frac{2x - 6}{10}i \][/tex]
### Simplifying the second term [tex]\(\frac{(2 - 3i)y + i}{3-i}\)[/tex]:
Multiply the numerator and the denominator by the conjugate of [tex]\(3-i\)[/tex], which is [tex]\(3+i\)[/tex]:
[tex]\[ \frac{[(2 - 3i)y + i](3+i)}{(3-i)(3+i)} \][/tex]
Since [tex]\( (3-i)(3+i) = 3^2 - (-1) = 9 + 1 = 10 \)[/tex], we get:
[tex]\[ \frac{[(2 - 3i)y + i](3+i)}{10} \][/tex]
Now expand the numerator:
[tex]\[ (2 - 3i)y(3+i) + i(3+i) \][/tex]
Distribute [tex]\( (2 - 3i)y \)[/tex]:
[tex]\[ = (6y + 2iy - 9iy - 3i^2 y) + (3i + i^2) \][/tex]
[tex]\[ = (6y - 7iy + 3y) + (3i - 1) \][/tex]
[tex]\[ = 9y - 7iy + 3i - 1 \][/tex]
Thus, the second term simplifies to:
[tex]\[ \frac{9y - 1}{10} + \frac{-7iy + 3i}{10} \][/tex]
This can be split into real and imaginary parts:
[tex]\[ \frac{9y - 1}{10} + \frac{-7iy + 3i}{10} = \frac{9y - 1}{10} + \frac{-7y + 3}{10}i \][/tex]
### Combine and equate to [tex]\(i\)[/tex]:
Combining the simplified fractions:
[tex]\[ \left(\frac{4x - 2}{10} + \frac{9y - 1}{10}\right) + \left(\frac{2x - 6}{10} + \frac{-7y + 3}{10}\right)i = i \][/tex]
Equating the real parts and the imaginary parts separately:
Real part:
[tex]\[ \frac{4x - 2}{10} + \frac{9y - 1}{10} = 0 \][/tex]
Simplifies to:
[tex]\[ 4x - 2 + 9y - 1 = 0 \][/tex]
[tex]\[ 4x + 9y - 3 = 0 \quad \text{(1)} \][/tex]
Imaginary part:
[tex]\[ \frac{2x - 6}{10} + \frac{-7y + 3}{10} = 1 \][/tex]
Simplifies to:
[tex]\[ 2x - 6 - 7y + 3 = 10 \][/tex]
[tex]\[ 2x - 7y - 3 = 10 \][/tex]
[tex]\[ 2x - 7y = 13 \quad \text{(2)} \][/tex]
### Solve the system of equations:
We have the system of linear equations:
1. [tex]\( 4x + 9y = 3 \)[/tex]
2. [tex]\( 2x - 7y = 13 \)[/tex]
Multiply the second equation by 2:
[tex]\[ 4x - 14y = 26 \][/tex]
Subtract from the first equation:
[tex]\[ 4x + 9y - (4x - 14y) = 3 - 26 \][/tex]
[tex]\[ 4x + 9y - 4x + 14y = -23 \][/tex]
[tex]\[ 23y = -23 \][/tex]
[tex]\[ y = -1 \][/tex]
Substitute [tex]\(y = -1\)[/tex] into the first equation:
[tex]\[ 4x + 9(-1) = 3 \][/tex]
[tex]\[ 4x - 9 = 3 \][/tex]
[tex]\[ 4x = 12 \][/tex]
[tex]\[ x = 3 \][/tex]
Thus, the real values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are:
[tex]\[ \boxed{x = 3, y = -1} \][/tex]
