Question 936319


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

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

{{{z = x + 1}}}.......eq.3
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{{{x + y + z = 6}}}....eq.1...substitute {{{x}}} and {{{z}}}

{{{2y+ y + x + 1= 6}}}.....substitute {{{x}}} again

{{{2y+ y + 2y + 1= 6}}}...solve for {{{y}}}

{{{5y = 6-1}}}

{{{5y = 5}}}

{{{y = 5/5}}}

{{{highlight(y =1)}}}

go to {{{x = 2y}}}...........eq.2..substitute {{{y}}} and solve for {{{x}}}

 {{{x = 2*1}}}

 {{{highlight(x = 2)}}}

go to {{{z = x + 1}}}.......eq.3..substitute {{{x}}} and solve for {{{z}}}

{{{z = 2 + 1}}}

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