solve the system

Question

  y=-x+7 and y=0.5(x-3)^2

Answer 1

y=-x+7

y=(x-3)²/2=(x²-6x+9)/2

So -x+7=(x²-6x+9)/2 or

-2x+14=x²-6x+9 or

0=x²-4x-5 competing the square we have

(x-2)²-9=0 or x-2=±3 or x=5, or x=-1

In which cases y=2 or y= 8

(5,2) and (-1,8) are the solutions