TS EAMCET · Maths · Complex Number
If \(\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1\), then \((x, y)\) is equal to
- A \(\left(\frac{7}{3}, \frac{-7}{15}\right)\)
- B \(\left(\frac{7}{3}, \frac{7}{15}\right)\)
- C \(\left(\frac{7}{5}, \frac{-7}{15}\right)\)
- D \(\left(\frac{7}{5}, \frac{7}{15}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{7}{3}, \frac{-7}{15}\right)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1 \\ & \Rightarrow \frac{[(1+i) x-i](2-i)}{\left(4-i^2\right)}+\frac{[(1+2 i) y+i](2+i)}{\left(4-i^2\right)}=1 \\ & \Rightarrow \frac{2(1+i) x-2 i-i(1+i) x+i^2}{4+1} \\ & +\frac{2(1+2 i) y+2 i+i(1+2 i)…
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