Question 1128391

{{{5x+20y=100}}} ...eq.1.....simplify, both sides divide by {{{5}}}
{{{x+3y=15}}}.......eq.2
--------------------
{{{x+4y=20}}} ...eq.1
{{{x+3y=15}}}.......eq.2
---------------------------subtract eq.2 from eq.1

{{{x+4y-(x+3y)=20-15}}}

{{{x+4y-x-3y=5}}}

{{{y=5}}}

go back to {{{x+3y=15}}}.......eq.2, substitute {{{y}}} value

{{{x+3*5=15}}}

{{{x+15=15}}}

{{{x=15-15}}}

{{{x=0}}}

=> intersection point of these two lines is at ({{{0}}},{{{5}}})


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(0,5,.12),locate(0,5.3,p(0,5)),
graph( 600, 600, -10, 10, -10, 10, -x/4+5, -x/3+5)) }}}