How do you simplify $\frac{x}{x - 1} - \frac{4}{x + 2}$ $(x)/(x-1) - (4)/(x+2)$ ?

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Answer 1 · 2016-08-10T21:22:16

$\frac{x^{2} - 2 x + 4}{(x - 1) (x + 2)}$ $(x^2 -2x+4)/((x-1)(x+2))$

To add fractions, find the LCD and then make equivalent fractions.

$\frac{x}{x - 1} - \frac{4}{x + 2} LCD = (x - 1) (x + 2)$ $(x)/(x-1) - (4)/(x+2) " LCD" = (x-1)(x+2)$

= $\frac{x (x + 2) - 4 (x - 1)}{(x - 1) (x + 2)}$ $(x(x+2)- 4(x-1))/((x-1)(x+2))$

= $\frac{x^{2} + 2 x - 4 x + 4}{(x - 1) (x + 2)}$ $(x^2 +2x- 4x+4)/((x-1)(x+2))$

= $\frac{x^{2} - 2 x + 4}{(x - 1) (x + 2)}$ $(x^2 -2x+4)/((x-1)(x+2))$

How do you simplify xx−1−4x+2(x)/(x-1) - (4)/(x+2)?