# Question #53d1f

Jul 15, 2017

the element $- 12$ is in the solution of the given inequality

#### Explanation:

We will verify the inequality for each term by substituting it in x, then

$\textcolor{red}{x = 18} \to \frac{2}{\cancel{3}} \cdot {\cancel{18}}^{6} + 3 < - 2 \cdot 18 - 7 \to 15 < - 43$ is not true

$\textcolor{red}{x = 6} \to \frac{2}{\cancel{3}} \cdot {\cancel{6}}^{2} + 3 < - 2 \cdot 6 - 7 \to 7 < - 19$ is not true

$\textcolor{red}{x = - 3} \to \frac{2}{\cancel{3}} \cdot {\cancel{- 3}}^{- 1} + 3 < - 2 \cdot \left(- 3\right) - 7 \to 1 < - 1$ is not true

$\textcolor{red}{x = - 12} \to \frac{2}{\cancel{3}} \cdot {\cancel{- 12}}^{- 4} + 3 < - 2 \cdot \left(- 12\right) - 7 \to - 5 < 17$ is true,

Then the element $- 12$ is in the solution of the given inequality.

On the other hand, if we solve the inequality, we will get:

$x < - \frac{15}{4}$

All the elements in A except $- 12$ are not in the solution