Question 1151190

The sum of three numbers is {{{110}}}. 

{{{x+y+z=110}}}

The second number is {{{10}}} less than the first.
{{{y=x-10}}}

 The third number is {{{4}}} times the first. 

{{{z=4x}}}

{{{x+y+z=110}}}......substitute  {{{y}}} and {{{z}}}

{{{x+x-10+4x=110}}}.......solve for {{{x}}}

{{{6x=110+10}}}

{{{6x=120}}}

{{{x=20}}} 

find next one

{{{y=20-10}}}

{{{y=10}}} 

{{{z=4*20}}}

{{{z=80}}} 

so, the numbers are:   {{{10}}},{{{20}}}, and  {{{80}}}