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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 ?

Option: 1

If α = β and α ≠ 7, then the system has a unique solution


Option: 2

If α = β = 7, then the system has no solution
 


Option: 3

 For every point (α, β) ≠ (7,7) on the line x − 2y + 7 = 0, the system has infinitely many solutions
 


Option: 4

There is a unique point (α, β) on the line x + 2y + 18 = 0 for which the system has infinitely many solutions


Answers (1)

best_answer

\begin{aligned} & x+y+z=6 \\ & \alpha x+\beta y+7 z=3 \\ & x+2 y+3 z=14 \end{aligned}

equation (3) – equation (1)

\begin{aligned} & y+2 z=8 \\ & y=8-2 z \end{aligned}

From (1) x = –2 + z
Value of x and y put in equation (2)

\begin{aligned} & \alpha(-2+z)+\beta(8-2 z)+7 z=3 \\ & -2 \alpha+\alpha z+8 \beta-2 \beta z+7 z=3 \\ & (\alpha-2 \beta+7) z=2 \alpha-8 \beta+3 \end{aligned}

if  \alpha -2\beta +7\neq 0  then system has unique solution
if  (\alpha -2\beta +7=0)  and  2\alpha -8\beta +3\neq 0  then system has no solution
if  (\alpha -2\beta +7=0)  and  2\alpha -8\beta +3=0  then system has infinite solution

Posted by

Sanket Gandhi

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