If you had a linear system, addition would have been possible, but for a non-linear system, substitution tends to be the best (or only!) choice. Y is already alone in the first equation, so we will substitute it into the second equation:
X²-6X+9 +X = 5
X² -5X +9 =5 Since this is quadratic, we create a zero on one side...
-5 -5
X² -5X +4 = 0 Now, we factor -- thankfully not tricky this time!
(X - 4)(X -1) = 0
X = 4 X = 1 Don't forget to find the Y-values for each X!
Y + 4 =5 Y + 1 = 5
-4 -4 -1 -1
Y = 1 Y = 4
(4 , 1) & (1 , 4)
