How do you solve the inequality $\frac{1}{x + 1} > \frac{3}{x - 2}$ $1/(x+1)>3/(x-2)$ ?

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How do you solve the inequality $\frac{1}{x + 1} > \frac{3}{x - 2}$ $1/(x+1)>3/(x-2)$ ?

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Answer 1 · 2016-01-19T15:51:34

$x < - \frac{5}{2} \times$ $x < - 5/2 color(white)(xx)$ or $\times - 1 < x < 2$ $color(white)(xx)-1 < x < 2$

Explanation:

First of all, note that your inequality is only defined if your denominators are not equal to zero:

$x + 1 \neq 0 \Leftrightarrow x \neq - 1$ $x + 1 != 0 <=> x != -1$

$x - 2 \neq 0 \Leftrightarrow x \neq 2$ $x - 2 != 0 <=> x != 2$

Now, your next step would be to "get rid" of the fractions. This can be done if multiplying both sides of the inequality with $x + 1$ $x+1$ and $x - 2$ $x-2$ .

However, you need to be careful since if you multiply an inequality with a negative number, you must flip the inequality sign.

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Let's consider the different cases:

case 1: $\times x x > 2$ $color(white)(xxx) x > 2$ :

Both $x + 1 > 0$ $x + 1 > 0$ and $x - 2 > 0$ $x - 2 > 0$ hold. Thus, you get:

$x - 2 > 3 (x + 1)$ $x - 2 > 3 (x + 1)$

$x - 2 > 3 x + 3$ $x - 2 > 3x + 3$

... compute $- 3 x$ $-3x$ and $+ 2$ $+2$ on both sides...

$- 2 x > 5$ $-2x > 5$

... divide by $- 2$ $-2$ on both sides. As $- 2$ $-2$ is a negative number, you must flip the inequality sign...

$x < - \frac{5}{2}$ $x < - 5/2$

However, there is no $x$ $x$ that satisfies both the condition $x > 2$ $x > 2$ and $x < - \frac{5}{2}$ $x < - 5/2$ . Thus, there is no solution in this case.

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case 2: $\times x - 1 < x < 2$ $color(white)(xxx) -1 < x < 2$ :

Here, $x + 1 > 0$ $x + 1 > 0$ but $x - 2 < 0$ $x - 2 < 0$ . Thus, you need to flip the inequality sign once and you get:

$i x - 2 < 3 (x + 1)$ $color(white)(i) x - 2 < 3 (x + 1)$

$x - 2 x < 5$ $color(white)(x) -2x < 5$

... divide by $- 2$ $-2$ and flip the inequality sign again...

$\times x x > - \frac{5}{2}$ $color(white)(xxx) x > -5/2$

The inequality $x > - \frac{5}{2}$ $x > -5/2$ is true for all $x$ $x$ in the interval $- 1 < x < 2$ $-1 < x < 2$ . Thus, in this case, we have the solution $- 1 < x < 2$ $-1 < x < 2$ .

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case 3: $\times x x < - 1$ $color(white)(xxx) x < -1$ :

Here, both denominators are negative. Thus, if you multiply the inequality with both of them, you need to flip the inequality sign twice and you will get:

$x - 2 > 3 x + 3$ $x - 2 > 3x + 3$

$i - 2 x > 5$ $color(white)(i) -2x > 5$

$\times i x < - \frac{5}{2}$ $color(white)(xxi) x < - 5/2$

As the condition $x < - \frac{5}{2}$ $x < -5/2$ is more restrictive than the condition $x < - 1$ $x < -1$ , the solution for this case is $x < - \frac{5}{2}$ $x < - 5/2$ .

=========================================

In total, the solution is

$x < - \frac{5}{2} \times$ $x < - 5/2 color(white)(xx)$ or $\times - 1 < x < 2$ $color(white)(xx)-1 < x < 2$

or, if you prefer a different notation,

$x \in (- \infty, - \frac{5}{2}) \cup (- 1, 2)$ $x in (- oo, -5/2) uu (-1, 2)$ .

Answer 2 · 2016-01-26T22:28:28

$] - \infty, - \frac{5}{2} [\cup] - 1, 2 [$ $]-oo, -5/2[ uu ]-1, 2[$

Explanation:

$\frac{1}{x + 1} > \frac{3}{x - 2}$ $1/(x+1)>3/(x-2)$

let pass everithing to the left side of the inequality by subtracting $\frac{3}{x - 2}$ $3/(x-2)$ :

$\frac{1}{x + 1} - \frac{3}{x - 2} > 0$ $1/(x+1)-3/(x-2)>0$

Now we must, put all the inequation we the same denominator. The part with (x+1) we multiply by $\frac{x - 2}{x - 2}$ $(x-2)/(x-2)$ (which is 1!) and vice-versa:

$\frac{x - 2}{(x + 1) (x - 2)} - \frac{3 (x + 1)}{(x + 1) (x - 2)} > 0$ $(x-2)/((x+1)(x-2))-(3(x+1))/((x+1)(x-2))>0$

We did the trick before, to have all the inequation with the same denominator:

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

$(x + 1) (x - 2)$ $(x+1)(x-2)$ corresponds to a parabola which gives positive values in the ineterval $] - \infty, - 1 [\cup] 2, + \infty [$ $]-oo, -1 [ uu ] 2, +oo[$ and negative values in the interval $] - 1, 2 [$ $]-1, 2[$ . Remeber that x cannot be -1 or 2 due to giving denominator zero.

In the first case (denominator positive) we can simplify the inequation into:

$- 2 x - 5 > 0$ $-2x-5>0$ and $x \in] - \infty, - 1 [\cup] 2, + \infty [$ $x in ]-oo, -1 [ uu ] 2, +oo[$

which gives:

$x < - \frac{5}{2}$ $x<-5/2$ and $x \in] - \infty, - 1 [\cup] 2, + \infty [$ $x in ]-oo, -1 [ uu ] 2, +oo[$ .

The interception of intervals above gives $x < - \frac{5}{2}$ $x<-5/2$ .

In the second case, the denominator is negative, so for the result giving a positive number, the numerator must be negative:

$- 2 x - 5 < 0$ $-2x-5<0$ and $x \in] - 1, 2 [$ $x in ]-1, 2[$

which gives

$x ≻ \frac{5}{2}$ $x>-5/2$ . and $x \in] - 1, 2 [$ $x in ]-1, 2[$

The interception of intervals gives $x \in] - 1, 2 [$ $x in ]-1, 2[$

Joining the solutions of the two cases we obtain:

$] - \infty, - \frac{5}{2} [\cup] - 1, 2 [$ $]-oo, -5/2[ uu ]-1, 2[$

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