Question 935661

given:
({{{3}}},{{{2}}},{{{-5}}})=> {{{x=3}}}, {{{y=2}}}, and {{{z=-5}}}


{{{ax+by+cz=r}}} ...we can set {{{a=1}}}, {{{b=1}}}, and {{{c=1}}}, then plug in {{{x=3}}}, {{{y=2}}}, and {{{z=-5}}}, we will have


{{{1*3+1*2-1*5=r }}}

{{{3+2-5=r }}}
{{{5-5=r }}}
{{{0=r}}}

so, first equation is:

{{{x+y+z=0}}}.......1

now set {{{a=3}}}, {{{b=2}}}, and {{{c=4}}}, then plug in {{{x=3}}}, {{{y=2}}}, and {{{z=-5}}}, we will have

{{{3*3+2*2+4(-5)=r}}}
{{{9+4-20=r}}}
{{{-7=r}}}

second equation is:

{{{3x+2y+4z=-7}}}.........2

now set {{{a=1}}}, {{{b=5}}}, and {{{c=-5}}}, then plug in {{{x=3}}}, {{{y=2}}}, and {{{z=-5}}}, we will have

{{{3+5*2-5(-5)=r}}}

{{{3+10+25=r}}}

{{{38=r}}}

third equation is:

{{{x+5y-z=38}}}........3



so, your system is:

{{{x+y+z=0}}}.......1
{{{3x+2y+4z=-7}}}.........2
{{{x+5y-z=38}}}........3



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