Solving simultaneous equations.. help on a question please

ive been told to solve the simultaneous equations y=x^2 + x + 1 and x + 2y = 4

I first got the second equation to equal Y and then put them both equal to each other but then i found it difficult to rearrange and solve so i got X on its own in the second equation and then subbed it into the first equation, however, im again struggling to solve this so is anybody able to show me the method they would use and the steps to working it out? Thanks

As you said, rearrange the second eqn.
2y = 4 - x
Now, double the 1st eqn so it starts off 2y =
2y = 2x^2 + 2x + 2
Now, as 2y = 2y ....
4 - x = 2x^2 + 2x + 2
Collect all the terms on the RHS
0 = 2x^2 + 3x - 2
Now factorise
(2x ? ?)(x ? ?) = 0
There must be one neg. and one pos. number, and a 1 and a 2.
With a bit of trial and error, you can find it is
(2x - 1)(x + 2) = 0
so x = 1/2 and x = -2
Sub back into the eqns to find y

Let y = x^2 + x + 1 be Equation 1 and x + 2y = 4 be Equation 2.
You have got ready value of y in equation 1 which you can substitute in equation 2.
So you will get,
x + 2*(x^2 + x + 1) = 4
x + 2x^2 + 2x + 2 = 4
2x^2 + 3x + 2 = 4
2x^2 + 3x + 2 - 4 = 0
2x^2 + 3x - 2 = 0
2x^2 + 4x -x -2 = 0
2x*(x + 2) - 1*(x + 2) = 0
(x + 2)*(2x - 1) = 0
Either x + 2 = 0 or 2x -1 = 0
If x + 2 = 0 then x = -2 and
If 2x - 1 = 0 then x = 1/2
Putting the above values of x in equation 2 we get
Putting x = -2 we get -2 + 2y = 4
2y = 4 + 2 = 6
y = 6/2 = 3
One solution set is (-2, 3)
Putting x = 1/2 we get 1/2 + 2y = 4
2y = 4 -1/2 = (8-1)/2
2y = 7/2
y = 7/2*1/2
y = 7/4
Second solution set is (1/2, 7/4)

y=x^2 + x + 1 and x + 2y = 4
2-x/2=x^2+x+1
2x^2+3x-2=0
(2x -1 )(x +2 )=0
So x=1/2 or -2
So if x=1/2 then y=7/4
and x=-2 then y=3

x + 2y = 4
x = 4 - 2y

y = (4 - 2y)^2 + (4 - 2y) + 1
y = 16 - 16y + 4y^2 + 4 - 2y + 1
0 = 4y^2 - 18y + 21
y = 9/4 ± 3/2 i

x = 4 - 2(9/4 ± 3/2 i) = 4 - 9/2 ± 3i = -1/2 ± 3i