Solving Rational Inequality

To solve

 $\frac{x+12}{𝑥+2}≤2$  , first, we transform the given inequality into its general form which is...

•  $\frac{x+12}{𝑥+2}+2≤0$  

•  $\frac{x+12}{𝑥+2}-2≤0$  

•  $2\left(\frac{x+12}{𝑥+2}\right)≤0$  

•  $\frac{1}{2}\left(\frac{x+12}{𝑥+2}\right)≤0$  

Next, we combine the left hand side of

 $\frac{x+12}{𝑥+2}-2≤0$  into a single rational expression which is...

•  $\frac{x+12+2\left(x+2\right)}{𝑥+2}≤0$  

•  $\frac{x-12+2\left(x-2\right)}{𝑥+2}≤0$  

•  $\frac{x+12-2\left(x+2\right)}{𝑥+2}≤0$  

•  $\frac{x+12-2\left(x+2\right)}{𝑥-2}≤0$  

Then, we use distributive property of multiplication to make

 $\frac{x+12-2\left(x+2\right)}{𝑥+2}≤0$  into...

•  $\frac{x+12-2x+2}{𝑥+2}≤0$  

•  $\frac{x-12-2x-2}{𝑥+2}≤0$  

•  $\frac{x+12-2x+4}{𝑥+2}≤0$  

•  $\frac{x+12-2x-4}{𝑥+2}≤0$  

Combining like terms will simplify

 $\frac{x+12-2x-4}{𝑥+2}≤0$  into...

•  $\frac{x+8}{𝑥+2}≤0$  

•  $\frac{x-8}{𝑥+2}≤0$  

•  $\frac{-x+8}{𝑥+2}≤0$  

•  $\frac{-x-8}{𝑥+2}≤0$  

Based from

 $\frac{-x+8}{𝑥+2}≤0$  , what are the critical values of the given rational inequality?

•  $8\ \&\ 2$  

•  $-8\ \&\ -2$  

•  $-8\ \&\ 2$  

•  $8\ \&\ -2$  

Finally, using test points, what is the solution to the rational inequality

 $\frac{-x+8}{𝑥+2}≤0$  ?

•  $\left(-\infty,\ -2\right)\cup\left[8,\ \infty\right)$  

•  $\left(-\infty,\ -2\right)$  

•  $\left[8,\ \infty\right)$  

•  $\left(-2,\ 8\right]$