Solve this simultaneous equation? (urgent please)

Normally i can do these but this one has completely thrown me? I know youre meant to substitute, i think...but ive tried it and it just doesnt work :/ if anyone can solve it and explain how id be much grateful

y = x² + x + 1
x + 2y =4

Thankyou!

y = x^2 + x + 1

x + 2(x^2 + x + 1 ) = 4
x + 2x^2 + 2x + 2 - 4 = 0
2x^2 +3x - 2 = 0

(2x - 1)(x + 2) = 0

2x-1 = 0; x+2= 0
x = 1/2 ; x = - 2

y= 1/2^2 + 1/2 +1 = 1/4 + 2/4 + 4/4= 7/4

y = (-2)^2 + (-2) + 1= 4 - 2 +1 = 3


answers are { 1/2 ; 7/4 } ; { - 2 ; 3 }

Well let's see
y = (4 - x)/2 so then
(4-x)/2 = x^2 + x +1 then
4-x = 2x^2+2x+2 so then 2x^2 + 3x - 2 = 0
factorise this to give (2x - 1)(x + 2) = 0 so this gives
x = - 2 or x = 1/2
so then you can work out what y is then - OK?

We know,
y = x^2 + x + 1 = 2-(x/2)

=> x^2 + (3/2)x - 1 = 0
=> 2x^2 + 3x - 2 = 0
=> (2x-1)(x+2) = 0
=> x = 1/2 OR x = -2

Can any body change my question mark on my username