Question 978378

let one number be {{{x}}}, second number {{{y}}}, and third number {{{z}}}

given:
the sum of three numbers is {{{65}}}: {{{x+y+z=65}}}
One number is ten more than a second number:   {{{x=y+10}}}=> {{{y=x-10}}}

It is also twice the third:   {{{x=2z}}}=>{{{z=x/2}}}

so, we have:

{{{x+(x-10)+x/2=65}}}.........solve for {{{x}}}

{{{x+x-10+x/2=65}}}

{{{2x+x/2=65+10}}}

{{{4x/2+x/2=75}}}

{{{5x/2=75}}}

{{{5x=75*2}}}

{{{x=150/5}}}

{{{highlight(x=30)}}}

find other two numbers:
{{{y=x-10}}}=>{{{y=30-10}}}=>{{{highlight(y=20)}}}
{{{z=x/2}}}=>{{{z=30/2}}}=>{{{highlight(z=15)}}}

so, your three numbers are: {{{highlight(15)}}},{{{highlight(20)}}}, and {{{highlight(30)}}}