if the system of linear equations given by ( x + y + z = 3 ), ( 2x + 2y - z = 3 ), ( x + y - z = 1 ) is…

Explanation:

Step1: Subtract the third equation from the first equation

$$ \begin{align*} (x + y+z)-(x + y - z)&=3 - 1\ x + y+z-x - y + z&=2\ 2z&=2\ z&=1 \end{align*} $$

Step2: Substitute (z = 1) into the first two equations

The first equation becomes (x + y+1 = 3), so (x + y=2). The second equation becomes (2x + 2y-1 = 3), so (2x + 2y=4) (which is consistent with (x + y = 2) since (2(x + y)=4)).

Step3: Calculate (2x_0+2y_0 + z_0)

Since (x_0 + y_0=2) (from (x + y = 2)) and (z_0 = 1) $$2x_0+2y_0 + z_0=2(x_0 + y_0)+z_0$$ Substitute (x_0 + y_0 = 2) and (z_0=1) $$2\times2+1=5$$

Answer:

(5)