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(D) Given equations are:$2x + 3y + 2z = 9 \quad (1)$$3x + 2y + 2z = 9 \quad (2)$$x - y + 4z = 8 \quad (3)$Subtracting equation $(1)$ from equation $(2

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has a unique solution

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(D) Given equations are:$2x + 3y + 2z = 9 \quad (1)$$3x + 2y + 2z = 9 \quad (2)$$x - y + 4z = 8 \quad (3)$Subtracting equation $(1)$ from equation $(2)$:$(3x + 2y + 2z) - (2x + 3y + 2z) = 9 - 9$$x - y = 0 \Rightarrow x = y$Substitute $x = y$ into equation $(3)$:$x - x + 4z = 8 \Rightarrow 4z = 8 \Rightarrow z = 2$Substitute $z = 2$ into equation $(1)$:$2x + 3y + 2(2) = 9$$2x + 3y = 5$Since $x = y$,we have $2x + 3x = 5 \Rightarrow 5x = 5 \Rightarrow x = 1$Thus,$x = 1, y = 1, z = 2$.The system has a unique solution $(1, 1, 2)$.

The following system of linear equations is given: $2x + 3y + 2z = 9$,$3x + 2y + 2z = 9$,and $x - y + 4z = 8$. Which of the following statements is true?

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