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**e^(i*pi)** $\displaystyle 2x^2+5x-3 = (x+3)(2x-1)$

$\displaystyle

\frac {4x^2 - x + 1}{2x^2 + 5x-3} - \frac{2x-9}{x+3} = \frac{5x-4}{2x-1} + \frac {7}{4}

$

$\displaystyle \frac{4x^2-x+1}{(x+3)(2x-1)} - (\frac{2x-9}{x+3} \times \frac{2x-1}{2x-1}) = \frac{5x-4}{2x-1} \times \frac{x+3}{x+3} + \frac{7}{4} \times \frac{(x+3)(2x-1)}{(x+3)(2x-1)}$

Combining:

$\displaystyle \frac{4x^2-x+1 - (4x^2-2x-18x+9)}{(x+3)(2x-1)} = \frac{5x^2+15x-4x-12+1.75(2x^2+5x-3)}{(x+3)(2x-1)}$

The denominators can now be cancelled. Don't forget that subtracting a negative is the same as adding a positive: $\displaystyle a-(-b) = a+b$