# How do you solve the inequality  1/(x+1)>3/(x-2)?

Jan 19, 2016

$x < - \frac{5}{2} \textcolor{w h i t e}{\times}$ or $\textcolor{w h i t e}{\times} - 1 < x < 2$

#### Explanation:

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

$x + 1 \ne 0 \iff x \ne - 1$

$x - 2 \ne 0 \iff x \ne 2$

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

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: $\textcolor{w h i t e}{\times x} x > 2$:

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

$x - 2 > 3 \left(x + 1\right)$

$x - 2 > 3 x + 3$

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

$- 2 x > 5$

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

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

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

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case 2: $\textcolor{w h i t e}{\times x} - 1 < x < 2$:

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

$\textcolor{w h i t e}{i} x - 2 < 3 \left(x + 1\right)$

$\textcolor{w h i t e}{x} - 2 x < 5$

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

$\textcolor{w h i t e}{\times x} x > - \frac{5}{2}$

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

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case 3: $\textcolor{w h i t e}{\times x} x < - 1$:

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:

$x - 2 > 3 x + 3$

$\textcolor{w h i t e}{i} - 2 x > 5$

$\textcolor{w h i t e}{\times i} x < - \frac{5}{2}$

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

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In total, the solution is

$x < - \frac{5}{2} \textcolor{w h i t e}{\times}$ or $\textcolor{w h i t e}{\times} - 1 < x < 2$

or, if you prefer a different notation,

$x \in \left(- \infty , - \frac{5}{2}\right) \cup \left(- 1 , 2\right)$.

Jan 26, 2016

]-oo, -5/2[ uu ]-1, 2[

#### Explanation:

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

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

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

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

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

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

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

$\left(x + 1\right) \left(x - 2\right)$ corresponds to a parabola which gives positive values in the ineterval  ]-oo, -1 [ uu ] 2, +oo[ and negative values in the interval ]-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:

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

which gives:

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

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

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

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

which gives

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

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

Joining the solutions of the two cases we obtain:

]-oo, -5/2[ uu ]-1, 2[