If 1- i x -2 i -3- i -2-3 i y - i -3- i - i- then values of x and y areA- y -43-c- x -16-7A- y -43-63- x -16-7B- y -43-63- x -16-7C- x-1- y -0D- x -3- y -1

(1+i)x 2i3+i+(2 3i)y+i3 i=i ⇒[x+(x 2)i](3 i)(3^2 i^2)+[2y+(1 3y)i](3+i)3^2 i^2=i ⇒4x 2+(2x 6)i +9y 1+(3 7y)i10=i ⇒ 4x+9y 3+(2x 7y 3)i=10i Compare both the side ...

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Byju's Answer

Standard XII

Mathematics

Inequality of 2 Complex Numbers

If 1+ i x -2 ...

Question

If $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$ , then values of $x$ and $y$ are

y = \frac{43}{63}, x = - \frac{16}{7}

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y = - \frac{43}{63}, x = \frac{16}{7}

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x = 3, y = - 1

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x = 1, y = 0

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Solution

The correct option is C $x = 3, y = - 1$
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$
$\Rightarrow \frac{[x + (x - 2) i] (3 - i)}{(3^{2} - i^{2})} + \frac{[2 y + (1 - 3 y) i] (3 + i)}{3^{2} - i^{2}} = i$
$\Rightarrow \frac{4 x - 2 + (2 x - 6) i + 9 y - 1 + (3 - 7 y) i}{10} = i$
$\Rightarrow 4 x + 9 y - 3 + (2 x - 7 y - 3) i = 10 i$
Compare both the sides
$4 x + 9 y = 3$ and $2 x - 7 y = 13$
By solving above equations
$y = - 1$ and $x = 3$

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Similar questions

If $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$ , then the real values of x and y are given by :

Q. If

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i

, then

(x, y) =

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i

. Find

x, y

Q. The values of

x

and

y

satisfying the equation

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i

, are

Q. 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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Inequality of 2 Complex Numbers

Standard XII Mathematics

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If (1+i)x−2i3+i+(2−3i)y+i3−i=i, then values of x and y are

The correct option is C x=3,y=−1(1+i)x−2i3+i+(2−3i)y+i3−i=i ⇒[x+(x−2)i](3−i)(32−i2)+[2y+(1−3y)i](3+i)32−i2=i ⇒4x−2+(2x−6)i+9y−1+(3−7y)i10=i ⇒4x+9y−3+(2x−7y−3)i=10i Compare both the sides 4x+9y=3 and 2x−7y=13 By solving above equations y=−1 and x=3

If $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$ , then values of $x$ and $y$ are