3 years ago

How to solve number 13

1 answer:

8 0

$\bf \cfrac{3x-1}{x+2}-\cfrac{x-2}{x-1}\impliedby \stackrel{our~LCD~is}{(x+2)(x-1)} \\\\\\ \cfrac{(3x-1)(x-1)-\stackrel{\textit{difference of squares}}{(x-2)(x+2)}}{(x+2)(x-1)} \\\\\\ \cfrac{(3x^2-4x+1)-(x^2-2^2)}{(x+2)(x-1)}\implies \cfrac{(3x^2-4x+1)-(x^2-4)}{(x+2)(x-1)} \\\\\\ \cfrac{3x^2-4x+1-x^2+4}{x^2+x-2}\implies \cfrac{2x^2-4x+5}{x^2+x-2}$

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