Question 775466
1. {{{4x+3y+3z=-8}}}
2. {{{2x+y+z=-4}}}
3. {{{3x-2y+(m^2-6)z=m-4 }}}
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1. {{{4x+3y+3z=-8}}}
2. {{{2x+y+z=-4}}}...multiply by 2 
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1. {{{4x+3y+3z=-8}}}
2. {{{4x+2y+2z=-8}}}......and subtract from 1.
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{{{4x+3y+3z-(4x+2y+2z)=-8-(-8)}}}

{{{4x+3y+3z-4x-2y-2z=-8+8}}}
{{{y+z=0}}}.......solve for y

{{{y=-z}}}........go to 2. .substitute y

{{{2x-z+z=-4}}}
{{{2x=-4}}}
{{{x=-4/2}}}
{{{highlight(x=-2)}}}.......first solution

go to 3.,substitute x and y solve for z


{{{3(-2)-2(-z)+(m^2-6)z=m-4 }}}

{{{-6+2z+(m^2-6)z=m-4 }}}

{{{2z+(m^2-6)z=m-4 +6}}}

{{{(m^2-4)z=m+2}}}

{{{z=(m  +2)/(m^2-4)}}}

{{{z=(m  +2)/((m-2)(m+2))}}}

{{{z=cross((m  +2))1/((m-2)cross((m+2)))}}}

{{{z=1/(m-2)}}}    => {{{y=-1/(m-2)}}} since {{{y=-z}}} 

so, solution will exist for all real numbers m except {{{m=2}}} 

if {{{(m-2) =0}}} => {{{m=2}}}=>{{{z=1/(2-2)=1/0}}}........there will be no solutions
if {{{m=1}}}=>{{{z=1/(1-2)=-1}}} and {{{y=1}}}
or
if {{{m=3}}}=>{{{z=1/(3-2)=1}}} and {{{y=-1}}}

so, real integer solutions are:
{{{x=-2}}}, {{{y=1}}},{{{z=-1}}} for {{{m=1}}}
and
{{{x=-2}}}, {{{y=-1}}},{{{z=1}}} for {{{m=3}}}

and
{{{x=-2}}}, {{{y=-1/2}}},{{{z=1/2}}} for {{{m=4}}}

and so on......your answer is:

c. an infinite number of solution