Question 959420

The mean of three numbers is {{{31}}}. 
{{{(x+y+z)/3=31}}}
{{{x+y+z=31*3}}}
{{{x+y+z=93}}}

if the second is {{{1}}} more than twice the first, we have
{{{y=2x+1}}}
if the third is {{{4}}} less than {{{3}}} times the first, we have
{{{z=3x-4}}}

substitute it in {{{x+y+z=93}}}

{{{x+2x+1+3x-4=93}}}........solve for {{{x}}}

{{{6x-3=93}}}

{{{6x=93+3}}}

{{{6x=96}}}

{{{x=96/6}}}

{{{highlight(x=16)}}}

now find {{{y}}} and {{{z}}}

{{{y=2x+1}}}
{{{y=2*16+1}}}
{{{highlight(y=33)}}}

{{{z=3x-4}}}
{{{z=3*16-4}}}
{{{z=48-4}}}
{{{highlight(z=44)}}}