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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 $\alpha^2+\gamma^2=$
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$\frac{1}{x}+\frac{2}{y}-\frac{3}{z}-1=0, \frac{2}{x}-\frac{4}{y}+\frac{3}{z}-1=0$ and $\frac{3}{x}+\frac{6}{y}-\frac{6}{z}-4=0$ is $x=\alpha, y=\beta, z=\gamma$ then $\alpha^2+\gamma^2=$
Solution:
1090 Upvotes
Verified Answer
The correct answer is:
$5 \beta$
$\frac{1}{x}+\frac{2}{y}-\frac{3}{z}=1$ ...(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)
$\begin{aligned} & 4 / x=4 \Rightarrow x=1=\alpha \\ & \text { Put } x=1 \text { in (iv), } \\ & y=2=\beta \\ & \text { Put } x=1, y=2 \text { in (i) }\end{aligned}$
$\begin{aligned} & z=3=r \\ & \alpha^2+r^2=1^2+3^2=10=5 \beta\end{aligned}$
$\frac{2}{x}-\frac{4}{y}+\frac{3}{z}=1$ ...(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)
$\begin{aligned} & 4 / x=4 \Rightarrow x=1=\alpha \\ & \text { Put } x=1 \text { in (iv), } \\ & y=2=\beta \\ & \text { Put } x=1, y=2 \text { in (i) }\end{aligned}$
$\begin{aligned} & z=3=r \\ & \alpha^2+r^2=1^2+3^2=10=5 \beta\end{aligned}$
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