Question 955207

{{{x+y=13}}}....eq.1

{{{x-y=15}}}....eq.2
-------------------------- by  substitution

{{{x+y=13}}}....eq.1...solve for {{{x}}}

{{{x=13-y}}}....substitute in eq.2


{{{13-y-y=15}}}....eq.2...solve for {{{y}}}

{{{13-2y=15}}}

{{{13-15=2y}}}

{{{-2=2y}}}

{{{y=-1}}}

go to {{{x=13-y}}} substitute {{{-1}}} for {{{y}}}, and solve for {{{x}}}

{{{x=13-(-1)}}}

{{{x=13+1}}}

{{{x=14}}}