It there is a common point, then the x and y in both equations must be equal. So, solve for x in the linear equation, and substitute into the second to solve for y:

y = 1 + 2x

x = (y - 1 ) / 2

y = x^2 + x + 1

y = [ (y-1)/2]^2 + (y - 1)/2 + 1

y = 1/4[y^2 -2y + 1] + (y - 1)/2 + 1

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

0 = 1/4y^2 - y +3/4

0 = y^2 - 4y + 3

0 = (y - 3)(y - 1)

y = 3, 1

Plug into the linear equation to find x:

x = 1, 0

Verify with the quadratic equation

y(1) = 1 + 1 + 1 = 3 , as expected

y(0) = 1, as expected.

So there are two common points for these equations

The common point that lies on the line and parabola in the same time is the intercepting point of the line and parabola.

So, the y coordinate of the point verify the equation of the line and the equatin of the parabola, in the same time.

2x+1=x^2+x+1

We'll move all term to one side and we'll combine like terms:

x^2-x=0

We'll factorize by x:

x*(x-1)=0

We'll put each factor as zero:

x=0

x-1=0

We'll add 1 both sides:

Now, we'll substitute the value of x in the equation of the line, because it is much more easier to compute y.

y=2x+1

x=0

y=2*0+1, y=1

So the first pair of coordinates of crossing point: A(0,1)

x=1

y=2*1+1=3

So the second pair of coordinates of crossing point: B(1,3).

**So, the common points are: A(0,1) and B(1,3). **

To find the common point of y = 1+x and y = x^2+x+1.

The two curves intersect at a point or points which are common points as they lie on both curves.

y = 1+2x..............(1)

y = x^2+x+1......(2).

Subract (1) from (2). This eliminates y resulting in equation of one single varible x.

0 = x^2+x+1- 2x-1.

0 = x^2-x .

So x(x-1) = 0.

x= 0 , or x= 1.

When x=0, y = x+1 = 0+1 = 1.

When x= 1, y = x+1 = 1+1 = 2.

Therefore (0,1) and (1,2) are the common points of both equation y = x+1 and the parabola y = x^2+1.