Question 934077

{{{y = x + 4}}}....eq.1
{{{y = -x}}}.......eq.2
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start with

{{{y = x + 4}}}....eq.1 ...substitute {{{-x}}} for {{{y}}} from eq.2

{{{-x = x + 4}}}...solve for {{{x}}}

{{{-4 = x + x}}}

{{{-4 = 2x}}}

{{{-4/2 = x}}}

{{{highlight(x=-2)}}}

go to eq.2 substitute {{{-2}}} for {{{x}}}

{{{y = -x}}}.......eq.2

{{{y = -(-2)}}}

{{{highlight(y = 2)}}}

so, intersection point is ({{{-2}}},{{{2}}})


{{{drawing( 600, 600, -10, 10, -10, 10,circle(-2,2,.13),locate(-2,2,p(-2,2)), graph( 600, 600, -10, 10, -10, 10, -x, x + 4)) }}}