For the system of linear equations x+y+z=6; a | vedclass

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(D) The given system of equations is:$x+y+z=6$ $(1)$$\alpha x+\beta y+7z=3$ $(2)$$x+2y+3z=14$ $(3)$Subtracting $(1)$ from $(3)$,we get $y+2z=8$,so $y=

Vedclass · Accepted Answer

For every point $(\alpha, \beta) 
eq (7,7)$ on the line $x-2y+7=0$,the system has infinitely many solutions.

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(D) The given system of equations is:$x+y+z=6$ $(1)$$\alpha x+\beta y+7z=3$ $(2)$$x+2y+3z=14$ $(3)$Subtracting $(1)$ from $(3)$,we get $y+2z=8$,so $y=8-2z$. Substituting this into $(1)$,$x+(8-2z)+z=6 \Rightarrow x=z-2$.Substituting $x=z-2$ and $y=8-2z$ into $(2)$:$\alpha(z-2)+\beta(8-2z)+7z=3$$(\alpha-2\beta+7)z = 2\alpha-8\beta+3$.For a unique solution,the coefficient of $z$ must be non-zero: $\alpha-2\beta+7 
eq 0$.For infinitely many solutions,both sides must be zero: $\alpha-2\beta+7=0$ and $2\alpha-8\beta+3=0$.Solving these: $2\alpha-4\beta+14=0$ and $2\alpha-8\beta+3=0$. Subtracting gives $4\beta+11=0 \Rightarrow \beta=-11/4$,then $\alpha=-25/2$. This is a unique point,not on the line $x-2y+7=0$.Thus,option $D$ is $NOT$ true.

For the system of linear equations $x+y+z=6$; $\alpha x+\beta y+7z=3$; $x+2y+3z=14$,which of the following is $NOT$ true?

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