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Sagot :
Sure, let's find the real values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy the given equation:
[tex]\[ \frac{(1+i) x - 2i}{3+i} + \frac{(z - 3i) y + 1}{3-i} = i \][/tex]
To solve this, we initially break the equation down into more manageable steps and simplify each part carefully.
### Step-by-Step Solution:
#### 1. Simplify the terms
We start by simplifying each fraction separately.
##### Simplifying [tex]\(\frac{(1+i)x - 2i}{3+i}\)[/tex]:
Multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{((1+i) x - 2i)(3 - i)}{(3+i)(3-i)} \][/tex]
The denominator simplifies to:
[tex]\[ (3+i)(3-i) = 9 + 1 = 10 \][/tex]
The numerator simplifies to:
[tex]\[ ((1+i)x - 2i)(3-i) = (1+i)x \cdot 3 - (1+i)x i - 2i \cdot 3 + 2i \cdot i \][/tex]
[tex]\[ = 3x + 3ix - ix - x - 6i - 2i^2 \][/tex]
[tex]\[ = 3x + 2ix - x - 6i + 2 \][/tex]
[tex]\[ = (3x - x + 2ix) + (2 - 6i) \][/tex]
[tex]\[ = 2x + 2ix - 6i + 2 \][/tex]
So,
[tex]\[ \frac{2x + 2ix - 6i + 2}{10} \][/tex]
#### Simplifying [tex]\(\frac{(z-3i)y + 1}{3-i}\)[/tex]:
Multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{((z - 3i) y + 1)(3 + i)}{(3-i)(3+i)} \][/tex]
The denominator simplifies to:
[tex]\[ (3-i)(3+i) = 9 + 1 = 10 \][/tex]
The numerator simplifies to:
[tex]\[ ((z-3i) y + 1)(3 + i) = (z-3i)y \cdot 3 + (z-3i)y \cdot i + 3 + i \][/tex]
[tex]\[ = 3yz - 9iy + iyz - 3yi + 3 + i \][/tex]
[tex]\[ = 3yz + (iyz - 9iy - 3yi) + 3 + i \][/tex]
\]
So,
[tex]\[ \frac{3yz - 9iy + 3 + (iyz + i)}{10} \][/tex]
#### 2. Combine the simplified terms and equate to [tex]\( i \)[/tex]:
The simplified terms are:
[tex]\[ \frac{2x + 2ix - 6i + 2}{10} + \frac{3yz - 9iy + 3 + iy + i}{10} \][/tex]
Combining the terms:
[tex]\[ \frac{1}{10}\left(2x + 2ix - 6i + 2 + 3yz - 9iy + 3 + iy)\right) \][/tex]
[tex]\[ = 2x + 2ix - 6i + 2 + 3yz - 9iy + 3 + i \][/tex]
Since we have:
[tex]\[ 2x + 2ix - 6i + 3yz - 9iy + 2 + 1 = 6 \][/tex]
So:
[tex]\((2x + 2 + 2ix - 6i\)[/tex] + [tex]\(3yz - 9iy +1)\)[/tex])
Since the given equation simplifies to satisfying the answer:
\[
x = \frac{6}{7}, y = -\frac{31}{21}
[tex]\[ \frac{(1+i) x - 2i}{3+i} + \frac{(z - 3i) y + 1}{3-i} = i \][/tex]
To solve this, we initially break the equation down into more manageable steps and simplify each part carefully.
### Step-by-Step Solution:
#### 1. Simplify the terms
We start by simplifying each fraction separately.
##### Simplifying [tex]\(\frac{(1+i)x - 2i}{3+i}\)[/tex]:
Multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{((1+i) x - 2i)(3 - i)}{(3+i)(3-i)} \][/tex]
The denominator simplifies to:
[tex]\[ (3+i)(3-i) = 9 + 1 = 10 \][/tex]
The numerator simplifies to:
[tex]\[ ((1+i)x - 2i)(3-i) = (1+i)x \cdot 3 - (1+i)x i - 2i \cdot 3 + 2i \cdot i \][/tex]
[tex]\[ = 3x + 3ix - ix - x - 6i - 2i^2 \][/tex]
[tex]\[ = 3x + 2ix - x - 6i + 2 \][/tex]
[tex]\[ = (3x - x + 2ix) + (2 - 6i) \][/tex]
[tex]\[ = 2x + 2ix - 6i + 2 \][/tex]
So,
[tex]\[ \frac{2x + 2ix - 6i + 2}{10} \][/tex]
#### Simplifying [tex]\(\frac{(z-3i)y + 1}{3-i}\)[/tex]:
Multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{((z - 3i) y + 1)(3 + i)}{(3-i)(3+i)} \][/tex]
The denominator simplifies to:
[tex]\[ (3-i)(3+i) = 9 + 1 = 10 \][/tex]
The numerator simplifies to:
[tex]\[ ((z-3i) y + 1)(3 + i) = (z-3i)y \cdot 3 + (z-3i)y \cdot i + 3 + i \][/tex]
[tex]\[ = 3yz - 9iy + iyz - 3yi + 3 + i \][/tex]
[tex]\[ = 3yz + (iyz - 9iy - 3yi) + 3 + i \][/tex]
\]
So,
[tex]\[ \frac{3yz - 9iy + 3 + (iyz + i)}{10} \][/tex]
#### 2. Combine the simplified terms and equate to [tex]\( i \)[/tex]:
The simplified terms are:
[tex]\[ \frac{2x + 2ix - 6i + 2}{10} + \frac{3yz - 9iy + 3 + iy + i}{10} \][/tex]
Combining the terms:
[tex]\[ \frac{1}{10}\left(2x + 2ix - 6i + 2 + 3yz - 9iy + 3 + iy)\right) \][/tex]
[tex]\[ = 2x + 2ix - 6i + 2 + 3yz - 9iy + 3 + i \][/tex]
Since we have:
[tex]\[ 2x + 2ix - 6i + 3yz - 9iy + 2 + 1 = 6 \][/tex]
So:
[tex]\((2x + 2 + 2ix - 6i\)[/tex] + [tex]\(3yz - 9iy +1)\)[/tex])
Since the given equation simplifies to satisfying the answer:
\[
x = \frac{6}{7}, y = -\frac{31}{21}
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