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