If the values x=alpha, y=beta, z=gamma satisf | vedclass

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(C) Given the system of equations:$1) x+2y+3z=4$$2) 3x+y+z=3$$3) x+3y+3z=2$Subtract equation $(1)$ from equation $(3)$:$(x+3y+3z) - (x+2y+3z) = 2 - 4$

Vedclass · Accepted Answer

$1-2\beta$

Vedclass · Answer

(C) Given the system of equations:$1) x+2y+3z=4$$2) 3x+y+z=3$$3) x+3y+3z=2$Subtract equation $(1)$ from equation $(3)$:$(x+3y+3z) - (x+2y+3z) = 2 - 4$$y = -2$Substitute $y = -2$ into equations $(1)$ and $(2)$:$x + 2(-2) + 3z = 4 \implies x + 3z = 8$$3x + (-2) + z = 3 \implies 3x + z = 5$From $3x + z = 5$,we get $z = 5 - 3x$.Substitute $z$ into $x + 3z = 8$:$x + 3(5 - 3x) = 8$$x + 15 - 9x = 8$$-8x = -7 \implies x = \frac{7}{8}$Now find $z$:$z = 5 - 3(\frac{7}{8}) = 5 - \frac{21}{8} = \frac{40-21}{8} = \frac{19}{8}$We have $\alpha = \frac{7}{8}, \beta = -2, \gamma = \frac{19}{8}$.Calculate $3\alpha + \gamma$:$3(\frac{7}{8}) + \frac{19}{8} = \frac{21}{8} + \frac{19}{8} = \frac{40}{8} = 5$.Check the options for $\beta = -2$:$A) \beta = -2$$B) 2\beta = -4$$C) 1 - 2\beta = 1 - 2(-2) = 5$$D) 2\beta + 1 = 2(-2) + 1 = -3$Since $3\alpha + \gamma = 5$ and $1 - 2\beta = 5$,the correct option is $C$.

If the values $x=\alpha, y=\beta, z=\gamma$ satisfy all the $3$ equations $x+2y+3z=4$,$3x+y+z=3$ and $x+3y+3z=2$,then $3\alpha+\gamma=$

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