Question 804940
{{{x + 3y = 15}}}....eq.1
{{{2x -3y = - 16}}}...eq.2
_______________________________add eq.1 and eq.2

{{{x+2x + 3y-3y = 15-16}}}

{{{3x  =  -1}}}

{{{x  =  -1/3}}}......plug in eq.1 or eq.2 and find {{{y}}}

{{{-1/3 + 3y = 15}}}....eq.1

{{{  3y = 15+1/3}}}

{{{  3y = 45/3+1/3}}}

{{{  3y = 46/3}}}

{{{  y = (46/3)/3}}}

{{{  y = 46/9}}}

so, intersection point is at ({{{-1/3}}},{{{46/9}}})


{{{ graph( 600, 600, -10, 10, -10, 10, -x/3+15/3,2x/3+16/3) }}}