Question

How do you solve $\dfrac{x+7}{x-3}>0$ using a sign chart?

Answer 1

Hint: We first try to understand the conditions for the polynomials $\left( x+7 \right)$ and $\left( x-3 \right)$ so that the total outcome for $\dfrac{x+7}{x-3}$ is positive. We assign signs for individual terms and find the domain for the polynomial.

Complete step by step solution:
We have been given an inequality of $\dfrac{x+7}{x-3}>0$.
We have to find the possibilities for the numerator and the denominator for their respective signs so that the total outcome for $\dfrac{x+7}{x-3}$ is positive.
In a division the outcome will be positive when the numerator and the denominator have the same signs. We have two types of possibilities.
In the first case $x+7>0$ and $x-3>0$.
We solve these two equations to get $x>-7$ and $x>3$.
The combined solution for the outcomes will be $x>3$.
In the second case $x+7<0$ and $x-3<0$.
We solve these two equations to get $x<-7$ and $x<3$.
The combined solution of these two conditions is $x<-7$.
Therefore, the final solution domain for $x$ will be $x<-7$ or $x>3$.
The sign chart for the inequality will be

	$x<-7$	$-7< x< 3$	$x>3$
$\left( x+7 \right)$	-	+	+
$\left( x-3 \right)$	-	-	+

Note: We need to be careful about the sign. In case of opposite signs, the total outcome would have been negative. We can also express the solution domain for $x$ in the form of a complement set where $x\in \mathbb{R}\backslash \left[ -7,3 \right]$.