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)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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