If frac{(1+i) x-i}{2+i}+frac{(1+2 i) y+i}{2-i | vedclass

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(A) Given the equation: $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$ Multiply by the common denominator $(2+i)(2-i) = 4 - i^2 = 5$: $[(1+i)x - i]

Vedclass · Accepted Answer

$\left(\frac{7}{3}, \frac{-7}{15}\right)$

Vedclass · Answer

(A) Given the equation: $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$ Multiply by the common denominator $(2+i)(2-i) = 4 - i^2 = 5$: $[(1+i)x - i](2-i) + [(1+2i)y + i](2+i) = 5$ Expanding the terms: $[(1+i)(2-i)x - i(2-i)] + [(1+2i)(2+i)y + i(2+i)] = 5$ $[(2 - i + 2i - i^2)x - (2i - i^2)] + [(2 + i + 4i + 2i^2)y + (2i + i^2)] = 5$ Since $i^2 = -1$: $[(3+i)x - (2i - 1)] + [(5i)y + (2i - 1)] = 5$ $(3+i)x - 2i + 1 + 5iy + 2i - 1 = 5$ $(3x + ix) + 5iy = 5$ $3x + i(x + 5y) = 5 + 0i$ Comparing real and imaginary parts: $3x = 5 \Rightarrow x = \frac{5}{3}$ (Wait,re-evaluating the expansion) Let's re-expand carefully: $(1+i)(2-i) = 2 - i + 2i - i^2 = 3 + i$ $i(2-i) = 2i - i^2 = 2i + 1$ $(1+2i)(2+i) = 2 + i + 4i + 2i^2 = 5i$ $i(2+i) = 2i + i^2 = 2i - 1$ Equation: $(3+i)x - (2i+1) + (5i)y + (2i-1) = 5$ $(3+i)x + 5iy - 2 = 5$ $(3+i)x + 5iy = 7$ $3x + i(x + 5y) = 7 + 0i$ $3x = 7 \Rightarrow x = \frac{7}{3}$ $x + 5y = 0$ $\Rightarrow \frac{7}{3} + 5y = 0$ $\Rightarrow 5y = -\frac{7}{3}$ $\Rightarrow y = -\frac{7}{15}$ Thus,$(x, y) = \left(\frac{7}{3}, -\frac{7}{15}\right)$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

If $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$,then $(x, y)$ is equal to

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