Question 1171400


The sum of three numbers is {{{40}}}. 

{{{x+y+z=40}}}......eq.1

The second number is twice the first number. 

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

The sum of the first two numbers is {{{2}}} more than the third number. 

{{{x+y=z+2}}}......eq.3


{{{x+y+z=40}}}......eq.1, substitute {{{y}}} from eq.2

{{{x+2x+z=40}}}

{{{3x+z=40}}}....solve for {{{z}}}

{{{z=40-3x}}}............eq.1a

go to

{{{x+y=z+2}}}......eq.3,substitute {{{y}}} from eq.2 and {{{z}}} from eq.1a

{{{x+2x=40-3x+2}}}

{{{3x=42-3x}}}

{{{3x+3x=42}}}

{{{6x=42}}}

{{{x=7}}}

go to eq.2

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

{{{y=2*7}}}

{{{y=14}}}

go to eq,1a

{{{z=40-3x}}}............eq.1a

{{{z=40-3*7}}}

{{{z=40-21}}}

{{{z=19}}}

 the three numbers are:{{{7}}}, {{{14}}}, and {{{19}}}