Question 1204629

{{{y = x - 8}}}.....eq.1
{{{y = -2x + 1}}}.....eq.2
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since left sides equal, right sides must be equal too

so, we have

{{{ x - 8=-2x+1}}}

{{{ x +2x=8+1}}}

{{{ 3x=9}}}

{{{ x=3}}}


go to eq.1, substitute {{{x}}}

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

{{{y = - 5}}}


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(3,-5,.12), locate(3,-5,p(3,-5)),
graph( 600, 600, -10, 10, -10, 10, x - 8, -2x + 1)) }}}