Question 1154405
{{{3x+6y=12}}}.....eq.1
{{{3x+y=8}}}......eq.2
------------------------------subtract eq.2 from eq.1

{{{3x+6y-(3x+y)=12-8}}}

{{{cross(3x)+6y-cross(3x)-y=4}}}

{{{6y-y=4}}}

{{{5y=4}}}

{{{highlight(y=4/5)}}} or {{{y=0.8}}}


go to

{{{3x+y=8}}}......eq.2, substitute {{{y}}}

{{{3x+4/5=8}}}......solve for {{{x}}}


{{{3x=8-4/5}}}

{{{3x=40/5-4/5}}}

{{{3x=36/5}}}

{{{x=(36/5)/3}}}

{{{x=36/(5*3)}}}

{{{x=cross(36)12/(5*cross(3))}}}

{{{highlight(x=12/5)}}} or {{{y=2.4}}}