Question 732660
The system of equations {{{system(y= 6-x,y= x^2-6x+6)}}} has 2 solutions.
Substituting {{{6-x}}} for {{{y}}} in {{{ y= x^2-6x+6}}} you get
{{{6-x=x^2-6x+6}}} --> {{{-x=x^2-6x}}} --> {{{x^2-5x=0}}} --> {{{x(x-5)=0}}}
So the solutions are {{{x=0}}} and {{{x=5}}} , 2 solutions.
For {{{x=0}}} , {{{y=6-0}}} --> {{{y=6}}} gives you point (0,6).
For {{{x=5}}} , {{{y=6-5}}} --> {{{y=1}}} gives you point (5,1).
Those 2 points are the intersections of the straight line represented by the linear function {{{y=6-x}}} and the parabola represented by the quadratic function {{{y= x^2-6x+6}}}
{{{graph(300,300,-5,10,-5,10,6-x,x^2-6x+6)}}}