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

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### 2 Answers

We have to solve for x and y given that 2x-y+1=0 and x^2+x-y+1=0.

2x - y + 1 = 0

=> y = 2x + 1

substitute this in x^2+x-y+1=0.

x^2 + x - y + 1 = 0

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

=> x^2 - x = 0

=> x(x - 1) =0

=> x = 0 and x = 1

y = 2x + 1

=> y = 1 and y = 3

**Therefore the curves meet at ( 0,1) and (1, 3)**

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:

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).**