Question 1157953
{{{100x+95y+10z= 54(x+y+z)}}}

{{{100x+95y+10z= 54x+54y+54z}}}

{{{100x+95y+10z-54x-54y-54z=0}}}

{{{46x+41y-44z=0}}}........solve for {{{z}}}

{{{46x+41y=44z}}}

{{{z=46x/44+41y/44}}}

{{{z=23x/22+41y/44}}}...........eq.1



{{{95x+10y+5z= 33(x+y+z)}}}

{{{95x+10y+5z= 33x+33y+33z}}}

{{{95x+10y+5z-33x-33y-33z=0}}}

{{{62x - 23y - 28z = 0}}}........solve for {{{z}}}

{{{62x - 23y =28z }}}

{{{z=62x/28 - 23y/28}}}

{{{z=31x/14 - 23y/28}}}...........eq.2



from eq.1 and eq.2 we have


{{{23x/22+41y/44=31x/14 - 23y/28}}}.......solve for {{{y}}}

{{{23y/28+41y/44=31x/14 - 23x/22}}}

{{{23y/28+41y/44=31x/14 - 23x/22}}}

{{{(135y)/77=(90 x)/77}}}

{{{135y=77(90 x)/77}}}

{{{135y=90x}}}

{{{y=90x/135}}}

{{{y= 2x/3}}}.............1a




go to {{{z=31x/14 - 23y/28}}}...........eq.2, substitute {{{y}}}


{{{z=31x/14 - (23(2x/3))/28}}}

{{{z=31x/14 - (46x/3)/28}}}

{{{z=31x/14 - 23x/42}}}

{{{z=5x/3}}}.............1b



Real solution:

{{{y= 2x/3}}}
{{{z=5x/3}}}


Integer solution:
{{{y= 2x}}}  
{{{z=5x }}}


=>since {{{x}}} could be any number {{{3n}}} (because of fraction in real solution), integer solution is


{{{x = 3n}}}, {{{y = 2n}}}, {{{z = 5n}}}, where {{{n}}} element {{{Z}}}


if we start with {{{n=1}}}, solution is {{{highlight(x = 3)}}}, {{{highlight(y = 2)}}}, {{{highlight(z = 5)}}}


check if statement is true if {{{x = 3}}}, {{{y = 2}}}, {{{z = 5}}} :


{{{100x+95y+10z= 54(x+y+z)}}}

{{{100*3+95*2+10*5= 54(3+2+5)}}}

{{{300+190+50= 54(10)}}}

{{{540= 540}}}=>true


let's check if  {{{n=2}}}, solution is {{{x = 6}}}, {{{y = 4}}}, {{{z = 10}}}

{{{100x+95y+10z= 54(x+y+z)}}}

{{{100*6+95*4+10*10= 54(6+4+10)}}}

{{{600+380+100= 54(20)}}}

{{{1080= 1080}}}=>true