Let’s solve the equation step by step:
[tex]\[
\frac{2x}{1 + 2y} = \frac{3x + 4i}{x - 3y}
\][/tex]
First, we want to cross-multiply to eliminate the fractions:
[tex]\[
2x(x - 3y) = (3x + 4i)(1 + 2y)
\][/tex]
Now, distribute on both sides:
[tex]\[
2x \cdot x - 2x \cdot 3y = 3x + 3x \cdot 2y + 4i + 4i \cdot 2y
\][/tex]
Simplify the equation:
[tex]\[
2x^2 - 6xy = 3x + 6xy + 4i + 8iy
\][/tex]
Combine like terms and move them to one side:
[tex]\[
2x^2 - 6xy - 3x - 6xy - 4i - 8iy = 0
\][/tex]
Group the real and imaginary parts separately:
Real part:
[tex]\[
2x^2 - 6xy - 3x = 0
\][/tex]
Imaginary part:
[tex]\[
0 = 4i + 8iy
\][/tex]
From the imaginary part equation, we get:
[tex]\[
4i + 8iy = 0
\][/tex]
To find [tex]\(y\)[/tex], factor out the common term [tex]\(4i\)[/tex]:
[tex]\[
4i(1 + 2y) = 0
\][/tex]
Since [tex]\(i\)[/tex] is imaginary and non-zero:
[tex]\[
1 + 2y = 0 \implies y = -\frac{1}{2}
\][/tex]
Now, substituting [tex]\( y = -\frac{1}{2} \)[/tex] back into the real part equation:
[tex]\[
2x^2 - 6x\left(-\frac{1}{2}\right) - 3x = 0
\][/tex]
Simplify the equation:
[tex]\[
2x^2 + 3x - 3x = 0 \implies 2x^2 = 0
\][/tex]
So,
[tex]\[
x = 0
\][/tex]
Thus, the solution to the equation [tex]\( \frac{2 x}{1+2 y}=\frac{3 x+i 4}{x-3 y} \)[/tex] given that [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are real is:
[tex]\[
(x, y) = (0, -\frac{1}{2})
\][/tex]
