To solve the equation
[tex]\[ \frac{i x}{1 + 2y} = \frac{3x + 4i}{x - 3y} \][/tex]
where [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are real, we will follow these steps:
1. Cross Multiply to Eliminate the Denominators:
First, let's cross multiply to get rid of the denominators, which gives us:
[tex]\[ (i x) (x - 3y) = (3 x + 4i) (1 + 2y) \][/tex]
2. Distribute Each Term:
Next, distribute the terms on both sides of the equation. This yields:
[tex]\[ i x^2 - 3i xy = 3x + 6xy + 4i + 8iy \][/tex]
3. Separate Real and Imaginary Parts:
Since [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are real, and the equation must hold true for both real and imaginary parts separately, we now separate them:
For the real parts, we get:
[tex]\[ 0 = 3x + 6xy \][/tex]
For the imaginary parts, we get:
[tex]\[ i x^2 - 3i xy = 4i + 8i y \][/tex]
Dividing out the [tex]\( i \)[/tex] from the imaginary part equation simplifies to:
[tex]\[ x^2 - 3xy = 4 + 8y \][/tex]
4. Solve the Real Part Equation:
From the real part equation:
[tex]\[ 0 = 3x + 6xy \][/tex]
Factor out [tex]\( 3x \)[/tex]:
[tex]\[ 3x(1 + 2y) = 0 \][/tex]
Hence, [tex]\( x = 0 \)[/tex] or [tex]\( 1 + 2y = 0 \)[/tex].
If [tex]\( 1 + 2y = 0 \)[/tex], then:
[tex]\[ y = -\frac{1}{2} \][/tex]
5. Substitute into the Imaginary Part Equation:
Substituting [tex]\( y = -\frac{1}{2} \)[/tex] into the equation [tex]\( x^2 - 3xy = 4 + 8y \)[/tex]:
[tex]\[ x^2 - 3x(-\frac{1}{2}) = 4 + 8(-\frac{1}{2}) \][/tex]
Simplifying this gives us:
[tex]\[ x^2 + \frac{3x}{2} = 4 - 4 \][/tex]
[tex]\[ x^2 + \frac{3x}{2} = 0 \][/tex]
Factor out the [tex]\( x \)[/tex]:
[tex]\[ x(x + \frac{3}{2}) = 0 \][/tex]
Therefore, [tex]\( x = 0 \)[/tex] or [tex]\( x = -\frac{3}{2} \)[/tex].
6. Check the Solutions:
- If [tex]\( x = 0 \)[/tex] and [tex]\( y \)[/tex] can be any real number.
- If [tex]\( y = -\frac{1}{2} \)[/tex], then [tex]\( x = 0 \)[/tex] or [tex]\( x = -\frac{3}{2} \)[/tex].
Finally, since [tex]\( x \)[/tex] and [tex]\( y \)[/tex] must be real, we conclude that the equation has no real solutions that satisfy both conditions based on the earlier derived constraints.
Thus, the answer is that there are no real solutions for the given equation.
