A curve C has parametric equations x = 2t - 1, y = 4t - 7 + rac{3}{t}, t eq 0 Show that the Cartesian equation of the curve C can be written in the form y = rac{2x^2 + ax + b}{x + 1}, x eq -1 where a and b are integers to be found. - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 1

Now, simplifying each term we have: $y = 2(x + 1) - 7 + \frac{6}{x + 1}$ This simplifies to: $y = 2x + 2 - 7 + \frac{6}{x + 1}$ So: $y = 2x - 5 + \frac{6}{x + 1}$

To combine into a single fraction, we rewrite y as: $y = \frac{(2x - 5)(x + 1) + 6}{x + 1}$ Distributing gives: $y = \frac{2x^2 - 5x + 2x - 5 + 6}{x + 1}$ Thus: $y = \frac{2x^2 - 3x + 1}{x + 1}$ Here we see that a = -3 and b = 1.