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

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Tushar Chandra eNotes educator | Certified Educator

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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)


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giorgiana1976 | Student

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.


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


We'll factorize by x:


We'll put each factor as zero:



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=2*0+1, y=1

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



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).