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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Aicharya · Accepted Answer

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} $

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,…

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