# How do you solve 2/(x-5)>1/(x+4) using a sign chart?

Jan 30, 2017

$\left\{x | \left(\text{-13" < x < "-4}\right) \cup \left(x > 5\right)\right\} .$

Also written as $x \in \left(\text{-13", "-4}\right) \cup \left(5 , \infty\right)$.

#### Explanation:

In order to do a sign analysis, we need to be able to re-write this inequality as a product/quotient of factors on one side, and 0 on the other. That way, the sign of each factor will contribute to whether the whole product/quotient is above or below 0.

We start by moving all terms to the left side, making the right side 0:

$\frac{2}{x - 5} > \frac{1}{x + 4} \text{ "=>" } \frac{2}{x - 5} - \frac{1}{x + 4} > 0$

Next, we combine these two fractions into one through a common denominator, $\left(x - 5\right) \left(x + 4\right)$:

$\frac{2}{x - 5} - \frac{1}{x + 4} > 0$

$\implies \frac{2 \left(x + 4\right) - 1 \left(x - 5\right)}{\left(x - 5\right) \left(x + 4\right)} > 0$

Note: this is perfectly okay, since we're multiplying both terms on the left by something equal to 1, thus the sign of each term isn't changing.

Simplify:

$\implies \frac{2 x + 8 - x + 5}{\left(x - 5\right) \left(x + 4\right)} > 0$

$\implies \frac{x + 13}{\left(x - 5\right) \left(x + 4\right)} > 0$

The LHS is now a product/quotient of factors. The $x$-values that make these factors 0 will be our possible sign-changing points. Those values are $x \in \left\{\text{-13, -4, 5}\right\}$.

We can now create our sign chart:

$\left[\begin{matrix}\null & | & \null & \text{-13" & \null & "-4" & \null & 5 & \null \\ x+13 & | & \null & \null & \null & \null & \null & \null & \null \\ x-5 & | & \null & \null & \null & \null & \null & \null & \null \\ x+4 & | & \null & \null & \null & \null & \null & \null & \null \\ "=======" & | & "==" & "==" & "==" & "==" & "==" & "==" & "==} \\ \frac{x + 13}{\left(x - 5\right) \left(x + 4\right)} & | & \null & \null & \null & \null & \null & \null & \null\end{matrix}\right]$

Next, we fill in the table with $+ / -$ signs, depending on where each factor is positive/negative. For instance, $x + 13$ is negative for $x$-values below $\text{-13}$, and it's positive everywhere else:

$\left[\begin{matrix}\null & | & \null & \text{-13" & \null & "-4" & \null & 5 & \null \\ x+13 & | & - & \null & + & \null & + & \null & + \\ x-5 & | & \null & \null & \null & \null & \null & \null & \null \\ x+4 & | & \null & \null & \null & \null & \null & \null & \null \\ "=======" & | & "==" & "==" & "==" & "==" & "==" & "==" & "==} \\ \frac{x + 13}{\left(x - 5\right) \left(x + 4\right)} & | & \null & \null & \null & \null & \null & \null & \null\end{matrix}\right]$

If we do this for each factor, our table will look like this:

$\left[\begin{matrix}\null & | & \null & \text{-13" & \null & "-4" & \null & 5 & \null \\ x+13 & | & - & \null & + & \null & + & \null & + \\ x-5 & | & - & \null & - & \null & - & \null & + \\ x+4 & | & - & \null & - & \null & + & \null & + \\ "=======" & | & "==" & "==" & "==" & "==" & "==" & "==" & "==} \\ \frac{x + 13}{\left(x - 5\right) \left(x + 4\right)} & | & \null & \null & \null & \null & \null & \null & \null\end{matrix}\right]$

The last row in the chart is filled by multiplying the signs of each column:

$\left[\begin{matrix}\null & | & \null & \text{-13" & \null & "-4" & \null & 5 & \null \\ x+13 & | & - & \null & + & \null & + & \null & + \\ x-5 & | & - & \null & - & \null & - & \null & + \\ x+4 & | & - & \null & - & \null & + & \null & + \\ "=======" & | & "==" & "==" & "==" & "==" & "==" & "==" & "==} \\ \frac{x + 13}{\left(x - 5\right) \left(x + 4\right)} & | & - & \null & + & \null & - & \null & +\end{matrix}\right]$

This last row now tells us where the fraction (x+13)/((x-5)(x+4) is positive/negative. The fraction is greater than 0 when $x$ is between -13 and -4, and also when $x$ is above 5. Thus, our solution is

$\left\{x | \left(\text{-13" < x < "-4}\right) \cup \left(x > 5\right)\right\} .$