How do you solve this simultaneous equation?

Question

Find a point of intersection of
y=2-x and (x-3)^2 + y^2=5

Please show me your working!!

Answer 1

(x - 3)² + (2 - x)² = 5
x² - 6x + 9 + 4 - 4x + x² = 5
2x² - 10x + 8 = 0
x² - 5x + 4 = 0
(x - 1).(x - 4) = 0
x = 1 , x = 4
Points of intersection are (1 , 1) and (4 , - 2)

Answer 2

y = 2-x lets call this eq. 1

(x-3)^2 + y^2=5 call this eq 2

putting the value of y from eq1 into eq 2

we have

(x-3)^2 + (2-x)^2=5
expanding we get

x^2+ 9 -6x + 4+x^2 -4x = 5
reducing we'll have

2x^2 -10x +13-5 = 0
=>2x^2 -10x + 8 = 0
factorizing we'll get
2x^2 -8x -2x +8 = 0
=> 2x(x-4) - 2(x-4) = 0
=> (x-4)(2x - 2) = 0
=> x =4,1

Thus the two equations intersect at two points
which can be obtained by putting the values of x above into eq1 or eq 2 (I'll chose eq 1 as its simple )
Thus for x=4
eq 1 => y = 2-x = 2-4 =-2
This Gives first point of intersection as P1(4,-2)
Similarly for x =1
eq 1 => y = 2-x = 2-1 = 1
This Gives second point of intersection as P2(1,1)

Thus the points of intersection are P1(4,-2) & P2(1,1)

That's it!!