How do solve $\frac{2}{x + 1} \geq \frac{1}{x - 2}$ $2/(x+1)>=1/(x-2)$ and write the answer as a inequality and interval notation?

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Answer 1 · 2017-01-07T12:25:00

$- 1 < x < 2 or x \geq 5$ $-1 < x<2 or x>=5$

or, in interval notation:

$] - 1; 2 [\cup [5; \infty [$ $]-1;2[uu[5;oo[$

The inequality can be rewritten as:

$\frac{2}{x + 1} - \frac{1}{x - 2} \geq 0$ $2/(x+1)-1/(x-2)>=0$

and then

$\frac{2 (x - 2) - (x + 1)}{(x + 1) (x - 2)} \geq 0$ $(2(x-2)-(x+1))/((x+1)(x-2))>=0$

$\frac{2 x - 4 - x - 1}{(x + 1) (x - 2)} \geq 0$ $(2x-4-x-1)/((x+1)(x-2))>=0$

$\frac{x - 5}{(x + 1) (x - 2)} \geq 0$ $(x-5)/((x+1)(x-2))>=0$

Let's study the sign of each binomial:

1) $x - 5 \geq 0 \to x \geq 5$ $x-5>=0->x>=5$
2) $x + 1 > 0 \to x > - 1$ $x+1>0->x> -1$
3) $x - 2 > 0 \to x > 2$ $x-2>0->x>2$

  __-1__2__5__

1) $-$ $-$ $-$ $-$ $-$ $-$ o $+$ $+$
2) $-$ $-$ $+$ $+$ $+$ $+$ $+$ $+$
3) $-$ $-$ $-$ $-$ $+$ $+$ $+$ $+$

Then the inequality is verified when the fraction is greater than 0, that's:

$- 1 < x < 2 or x \geq 5$ $-1 < x<2 or x>=5$

or, in interval notation:

$] - 1; 2 [\cup [5; \infty [$ $]-1;2[uu[5;oo[$

How do solve 2x+1≥1x−22/(x+1)>=1/(x-2) and write the answer as a inequality and interval notation?