How many common points have the curve entities?2x-y+1=0 and x^2+x-y+1=0

We notice that the curve entities are a line and a
parabola. We'll re-write their equations:

y = 2x + 1, line
equation

y =x^2 + x + 1, parabola
equation

The common points, which are located on the line
and parabola in the same time, are the intercepting points of the line and
parabola.

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

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

We'll move
all term to the left 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:

 x=1

Now,
we'll substitute the value of x in the equation of the line, because it is much more
easier to determine
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, there are 2 common points and
they are: (0,1) and (1,3).