How do you determine the value(s) of k such that the system of linear equations has the indicated number of solutions: no solutions for x + 2y + kz = 6 and 3x + 6y + 8z = 4?

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Answer 1 · 2017-08-20T15:40:54

Find the values that have a solution and exclude them.

$k!=(14-8z)/(3z)$ for the equations to NOT to have solutions

Note: $z!=0$ as it becomes undefined. (not allowed to divide by 0)

Given:

$x+2y+kz=6" "....................Equation(1)$
$3x+6y+8z=4" "...................Equation(2)$

$color(blue)("Consider "Equation(1))$

Manipulation gives:
$y= -1/2 x-1/2 kz+3" "...............Equation(1_a)$

$color(blue)("Consider "Equation(1))$
Manipulation gives:

$y=-3/6 x-8/6 z+4/6$

$y=-1/2 x-4/3 z+2/3" ".........................Equation(2_a)$

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Set $Equation(1_a)=Equation(2_a)$ through $y$

$cancel( -1/2 x)-1/2 kz+3" " =" "cancel(-1/2 x)-4/3 z+2/3$

$1/2 kz" "=" "3-2/3+4/3 z$

$kz=2(7/3+4/3 z)$

$k=(14-8z)/(3z)$ for the equation to have solutions. So we have:

$k!=(14-8z)/(3z)$ for the equation NOT to have solutions