Solve the equation 1-i x-2 i-3-i-2-3 i y-i-3-i-i - x- y - R - i-1-

(1+i)x 2i/3+i+(2 3i)y+i/3 i=i ⇒(3 i)(1+i)x 2i(3 i)+(3+i)(2 3i)y+i(3+i)/9 i^2=i ⇒(3+2i i^2)x 6i+2i^2+(6 7i 3i^2)y+3i+i^2/10=i ⇒ (4+2i)x 3i 3+(9 7i)y=10i ⇒ (4+2i) ...

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Byju's Answer

Standard XII

Mathematics

Properties of Modulus

Solve the equ...

Question

Solve the equation $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i; x, y \in R; i = \sqrt{- 1} .$

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Solution

$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$
$\Rightarrow \frac{(3 - i) (1 + i) x - 2 i (3 - i) + (3 + i) (2 - 3 i) y + i (3 + i)}{9 - i^{2}} = i \Rightarrow \frac{(3 + 2 i - i^{2}) x - 6 i + 2 i^{2} + (6 - 7 i - 3 i^{2}) y + 3 i + i^{2}}{10} = i \Rightarrow (4 + 2 i) x - 3 i - 3 + (9 - 7 i) y = 10 i \Rightarrow (4 + 2 i) x + (9 - 7 i) y = 13 i + 3 \Rightarrow (2 x - 7 y) i + 4 x + 9 y = 13 i + 3 \Rightarrow 2 x - 7 y = 13 a n d 4 x + 9 y = 3$ [Equating the real parts and imaginary parts]
Solving them simultaneously, we get x =3, y =-1

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Solve the equation (1+i)x−2i3+i+(2−3i)y+i3−i=i;x,y∈R;i=√−1.

Solve the equation $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i; x, y \in R; i = \sqrt{- 1} .$