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(C) Given equations are:$I. x + 3y + 4z = 96$$II. 2x + 8z = 80$$III. 2x + 6y = 120$From equation $(III)$,$2x = 120 - 6y$,so $x = 60 - 3y$.Substitute $

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if $x  z$

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(C) Given equations are:$I. x + 3y + 4z = 96$$II. 2x + 8z = 80$$III. 2x + 6y = 120$From equation $(III)$,$2x = 120 - 6y$,so $x = 60 - 3y$.Substitute $x$ into equation $(II)$:$2(60 - 3y) + 8z = 80$$120 - 6y + 8z = 80$$8z - 6y = -40 \Rightarrow 4z - 3y = -20 \Rightarrow 3y = 4z + 20 \Rightarrow y = \frac{4z + 20}{3}$.Substitute $x$ and $y$ into equation $(I)$:$(60 - 3y) + 3y + 4z = 96$$60 + 4z = 96$$4z = 36 \Rightarrow z = 9$.Now,find $y$:$y = \frac{4(9) + 20}{3} = \frac{36 + 20}{3} = \frac{56}{3} \approx 18.67$.Now,find $x$:$x = 60 - 3(18.67) = 60 - 56 = 4$.Comparing the values: $x = 4, y = 18.67, z = 9$.Thus,$x  z$. Therefore,$x  z$.

Solve the given system of three equations and select the correct relationship between $x, y,$ and $z$ from the given options.
$I. x + 3y + 4z = 96$
$II. 2x + 8z = 80$
$III. 2x + 6y = 120$

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Solve the given system of three equations and select the correct relationship between $x, y,$ and $z$ from the given options.$I. x + 3y + 4z = 96$$II. 2x + 8z = 80$$III. 2x + 6y = 120$