Question:
If \( (3i + 1)x + (4i + 4)y + 5 = 0 \) with \( x, y \) being real and \( i = \sqrt{-1} \), then x = _________.
If \( (3i + 1)x + (4i + 4)y + 5 = 0 \) with \( x, y \) being real and \( i = \sqrt{-1} \), then x = _________.
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For equations involving complex numbers, separate the real and imaginary parts and solve the system of equations.
Updated On: Dec 26, 2025
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Correct Answer: 2.4
Solution and Explanation
The given equation is:
\[
(3i + 1)x + (4i + 4)y + 5 = 0
\]
Separate the real and imaginary parts:
\[
(3i)x + x + (4i)y + 4y + 5 = 0
\]
Group real and imaginary terms:
\[
(x + 4y + 5) + i(3x + 4y) = 0
\]
For the equation to hold, both the real and imaginary parts must be equal to zero. So:
\[
x + 4y + 5 = 0 \quad \text{(1)}
\]
\[
3x + 4y = 0 \quad \text{(2)}
\]
From equation (2), solve for \( y \):
\[
y = -\frac{3}{4}x
\]
Substitute this into equation (1):
\[
x + 4\left(-\frac{3}{4}x\right) + 5 = 0
\]
\[
x - 3x + 5 = 0
\]
\[
-2x + 5 = 0
\]
\[
x = \frac{5}{2} = 2.5
\]
Thus, \( x = \boxed{2.5} \).
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