Question 978790
Since they both equal y, set them equal to each other.
{{{x^2-4=x+2}}}
{{{x^2-x-6=0}}}
{{{(x-3)(x+2)=0}}}
Solutions are {{{x=3}}} and {{{x=-2}}}
or where {{{y=3+2=5}}} and {{{y=-2+2=0}}}
(3,5) and (-2,0)
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*[illustration V7.JPG].