Question 930698
{{{4x+3y=15}}}....eq.1 

{{{2x-5y=1}}}.....eq.2. ..both sides multiply by {{{-2}}}
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{{{4x+3y=15}}}....eq.1 

{{{-4x+10y=-2}}}.....eq.2
_____________________________add both eq.1 and eq.2

{{{4x+3y+(-4x+10y)=15+(-2)}}}

{{{cross(4x)+3y-cross(4x)+10y=15-2}}}

{{{3y+10y=13}}}

{{{13y=13}}}

{{{y=13/13}}}

{{{highlight(y=1)}}}

go to eq.2, substitute {{{1}}} for {{{y}}}


{{{2x-5*1=1}}}.....eq.2

{{{2x-5=1}}}

{{{2x=1+5}}}

{{{2x=6}}}

{{{x=6/2}}}

{{{highlight(x=3)}}}

solution- or intersection point is ({{{3}}},{{{1}}})


{{{ graph( 600, 600, -10, 10, -10, 10, -4x/3+5,2x/5-1/5) }}}