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TS EAMCET 9 May 2024 Shift 2 Solved Paper

Section: Mathematics

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Question : 8 of 160

Marks: +1, -0

If

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

2x+4y=

[TG EAPCET 9 May 2024 Shift 2]

$5$
$-2$
$1$
$-1$

Solution:

We have,

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

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

\;\Rightarrow \;\frac{[(2x+1)+(1-x)i](2-i)}{4-i^2} + \;\frac{[(y+1)+(-2y-1)i](1-2i)}{1-4i^2} = 1-2i

\;\Rightarrow [2(2x+1)+(1-x)+(y+1)+2(-2y-1)]

\;+ i[-2x-1+2(1-x)-2(y+1)+(-2y-1)]

\;= 5(1-2i)

\;\Rightarrow (3x-3y+2) + i(-4x-4y-2) = 5-10i

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