Question 1177247

{{{2x+y=0}}} ..............1)
{{{x+ 2y + kz = -1}}}..............2)
 {{{x +k^2*z = 1}}}..............3)
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{{{2x+y=0}}} ..............1), solve for {{{y}}}

{{{y=-2x}}}..............1a)

go to

{{{x+ 2y + kz = -1}}}..............2), substitute{{{ y}}} from 1a)

{{{x+ 2(-2x)+ kz = -1}}}.............., solve for {{{x}}}

{{{x-4x+ kz = -1}}}

{{{-3x+ kz = -1}}}

{{{kz+1 = 3x}}}

{{{x=(kz+1)/3}}}...........2a)


go to

 {{{x +k^2*z = 1}}}..............3), substitute {{{x}}}

 {{{(kz+1)/3 +k^2*z = 1}}}........., solve for {{{z}}}

 {{{kz+1 +3k^2*z = 3}}}

 {{{kz+1 +3k^2*z =3-1}}}

 {{{z(k+3k^2) =2}}}

{{{z =2/(k+3k^2)}}}

 {{{z =2/k(1+3k)}}}.........3a)


express {{{x}}} in terms of {{{k}}}


{{{x=(kz+1)/3}}}...........2a), substitute {{{z}}}

{{{x=(k(2/k(1+3k))+1)/3}}}

{{{x=((2/(1+3k))+1)/3}}}

{{{x=((2/(1+3k))+(1+3k)/(1+3k))/3}}}

{{{x=((2+1+3k)/(1+3k))/3}}}

{{{x=((3+3k)/(1+3k))/3}}}

{{{x=(3(1+k)/(1+3k))/3}}}

{{{x=(1+k)/(1+3k)}}}.....................2b)


go to

{{{y=-2x}}}..............1a), substitute {{{x}}}

{{{y=-2(1+k)/(1+3k)}}}

{{{y=-(2+2k)/(1+3k)}}}...............1b)




The solutions to the system of equations are: 

{{{x=(1+k)/(1+3k)}}}

{{{y=-2(1+k)/(1+3k)}}}

{{{z =2/k(1+3k) }}}

-> for {{{k<>0}}} and {{{k<>-1/3}}}