Question 934079
{{{y = x -8}}}....eq.1
{{{y = 3x}}}...eq.2...substitute in eq.1

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{{{3x= x -8}}}....eq.1

{{{3x-x =-8}}}

{{{2x =-8}}}

{{{x =-8/2}}}

{{{highlight(x =-4)}}}


{{{y = 3x}}}...eq.2...substitute {{{-4}}} for {{{x}}}

{{{y = 3(-4)}}}

{{{y = highlight(-12)}}}


{{{drawing( 600, 600, -10, 10, -15, 10,circle(-4,-12,.13),locate(-4,-12,p(-4,-12)), graph( 600, 600, -10, 10, -15, 10, 3x,x -8)) }}}