Question 1004197


let the first number be {{{x}}}, the second number {{{y}}}, and third number {{{z}}}
the sum of three numbers is {{{129}}}; so, we have 

{{{x+y+z=129}}}.....eq.1


if the third number is {{{4}}} times the first, we have {{{z=4x}}}

if the first number is {{{9}}} more than the second, we have {{{y=x-9}}}

substitute it in eq.1

{{{x+(x-9)+4x=129}}}.....eq.1

{{{x+x-9+4x=129}}}

{{{6x=129+9}}}

{{{6x=138}}}

{{{x=138/6}}}

{{{highlight(x=23)}}}

now find other two numbers

{{{y=23-9}}}=>{{{highlight(y=14)}}}

{{{z=4*23}}}=>{{{highlight(z=92)}}}