Question 1133558
{{{x + y = -5 }}}.......eq.1
{{{3x -4y = 6}}}.......eq.2
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start with 
{{{x + y = -5 }}}.......eq.1...solve fo {{{x}}}

{{{x = -y-5 }}}.....eq.1a..substitute in eq.2


{{{3(-y-5) -4y = 6}}}.......eq.2 ..solve fo {{{y}}}

{{{-3y-15 - 4y = 6}}}

{{{-7y-15  = 6}}}

{{{-6-15  = 7y}}}

{{{-21  = 7y}}}

{{{y=-21/7}}}

{{{y=-3}}}


go back to eq.1a, substitute {{{y}}}

{{{x = -(-3)-5 }}}.....eq.1a

{{{x = 3-5 }}}

{{{x = -2 }}}



{{{ graph( 600, 600, -10, 10, -10, 10, -x-5, 3x/4-6/4) }}}