Question 1132159

 {{{x + y=3 }}}.........eq.1
{{{x+2y=4 }}}.........eq.2
-----------------------------------substitute eq.1 from eq.2

{{{x+2y-(x+y)=4-3 }}}

{{{x+2y-x-y=1 }}}
{{{cross(x)+2y-cross(x)-y=1 }}}

{{{2y-y=1 }}}

{{{y=1 }}}

go to {{{x + y=3 }}}.........eq.1, substitute {{{y}}}

{{{x + 1=3 }}}

{{{x =3 -1}}}

{{{x =2}}}

solution:

({{{2}}},{{{1}}})


{{{ graph( 600, 600, -10, 10, -10, 10, -x+3, -x/2+2) }}}