Question 768648
{{{y=2x }}}.......eq.1...this one is already solved for {{{y}}}, just substitute it in eq.2
{{{x=2y-2}}}.....eq.2
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{{{x=2*2x-2}}}.....eq.2..solve for {{{x}}}

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

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

{{{2=3x}}}

{{{2/3=x}}}......go back to eq.1 and solve for {{{y}}}

{{{y=2*(2/3) }}}.......eq.1

{{{y=4/3 }}}

{{{drawing( 600, 600, -10, 10, -10, 10,circle(2/3,4/3,0.1),locate(2/3,4/3, p(2/3,4/3)), graph( 600, 600, -10, 10, -10, 10,2x, x/2+1)) }}}