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If the system of linear equations given by $x+y+z=3,2 x+2 y-z=3, x+y-z=1$ is consistent and if $\left(x_0, y_0, z_0\right)$ is a solution, then $2 x_0+2 y_0+z_0=$
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The correct answer is:
5
In consistent adding Eqs. (i) and (iii), we get
$$
2 x+2 y=4 \Rightarrow x+y=2
$$
put in Eq. (i)
$$
2+z=3 \Rightarrow z=1
$$
Then,
$$
\begin{aligned}
& 2 x_0+2 y_0+z_0 \\
& =2\left(x_0+y_0\right)+z_0 \\
& =2 \times 2+1 \quad \left[\because x+y=2 \Rightarrow x_0+y_0=2\right] \\
& =5
\end{aligned}
$$
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