Question 908080

let's numbers be: first is {{{x}}}, second is {{{y}}}, and third is {{{z}}}

given:

The sum of the three numbers is {{{6}}}. 

{{{x+y+z=6}}}....eq.1

The third number ({{{z}}}) is the sum of the first ({{{x}}}) and second ({{{y}}})numbers. 

{{{z=x+y}}} ....eq.2

The first{{{x}}}  number is {{{one}}}{{{ more}}} than the third number {{{z}}}.

{{{x=z+1}}} ....eq.3

so, we have system:

{{{x+y+z=6}}}....eq.1
{{{z=x+y}}} ....eq.2
{{{x=z+1}}} ....eq.3
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start with
{{{x+y+z=6}}} ....eq.1 and substitute {{{z}}} from eq.2

{{{x+y+(x+y)=6}}}

{{{2(x+y)=6}}}

{{{x+y=3}}}  or {{{3=x+y}}} ......1a

go to {{{z=x+y}}} ....eq.2 and you will see that {{{3=x+y}}}...1a and eq.2 have same values on right side which means 

{{{highlight(z=3)}}}

now go to {{{x=z+1}}} ....eq.3, plug in {{{z}}} and find {{{x}}}

{{{x=3+1}}} ....eq.3

{{{highlight(x=4)}}}

and, finally go to {{{x+y+z=6}}}....eq.1,plug in {{{z}}} and {{{x}}},than find {{{y}}}

{{{4+y+3=6}}}
{{{y+7=6}}}
{{{y=6-7}}}
{{{highlight(y=-1)}}}