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 =...

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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)**