# What is the common point of the line y=1+2x and the parabola y=x^2+x+1?

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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

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

y(0) = 1, as expected.

So there are two common points for these equations

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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

x=1

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).

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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.