Q46 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics

If the solution of the system of simultaneous equations $\frac{1}{x}+\frac{2}{y}-\frac{3}{z}-1=0, \frac{2}{x}-\frac{4}{y}+\frac{3}{z}-1=0$ and $\frac{3}{x}+\frac{6}{y}-\frac{6}{z}-4=0$ is $x=\alpha, y=\beta, z=\gamma$ then $\alpha^2+\gamma^2=$

  • Correct answer
    A
    $5 \beta$
  • B
    $\beta^2$
  • C
    $3 \beta$
  • D
    $2 \beta^2$

Solution

$\frac{1}{x}+\frac{2}{y}-\frac{3}{z}=1$ ...(i) $\frac{2}{x}-\frac{4}{y}+\frac{3}{z}=1$ ...(ii) $\frac{3}{x}+\frac{6}{y}-\frac{6}{z}=4$ ...(iii) Equation (i) + Equation (ii) $\frac{3}{x}-\frac{2}{y}=2$ ...(iv) Equation (ii) $\times 2+$ Equation (iii) $\frac{7}{x}-\frac{2}{y}=6$ ...(v) Equation (v) - Equation (iv) $\begin{aligned} & 4 / x=4 \Rightarrow x=1=\alpha \\ & \text { Put } x=1 \text { in (iv), } \\ & y=2=\beta \\ & \text { Put } x=1, y=2 \text { in (i) }\end{aligned}$ $\begin{aligned} & z=3=r \\ & \alpha^2+r^2=1^2+3^2=10=5 \beta\end{aligned}$

Hence, the correct answer is option: 'A'.

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