Find all $x$ such that neither $2+x+5x$ nor $2-x-8-3x$ is in the interval $(-2,11]$. Express your answer as an interval or as a union of intervals.
First, we set $2+x+5x<-2$. Simplifying the left-hand side gives $6x<-4$, so $x<-\frac{2}{3}$.
Next, we set $2-x-8-3x>11$. Simplifying the left-hand side gives $-4x<-21$, so $x>\frac{21}{4}$.
Therefore, neither inequality is satisfied if $-\frac{2}{3}<x<\frac{21}{4}$, so the values of $x$ that satisfy neither inequality are $x\leq-\frac{2}{3}$ or $x\geq\frac{21}{4}$. In interval notation, this is $\boxed{(-\infty, -\frac23] \cup [\frac{21}{4}, \infty)}$.
To find all $x$ such that neither expression $2+x+5x$ nor $2-x-8-3x$ is in the interval $(-2,11]$, we can proceed as follows:
First, let's simplify the expressions:
$2+x+5x = 2+6x$
$2-x-8-3x = -6-4x$
Now, we need to find the values of $x$ for which these two expressions are not in the interval $(-2, 11]$.
For the first expression, $2+6x$, it will be outside the interval $(-2,11]$ if it is either less than or equal to $-2$ or greater than $11$. In other words, we want to find:
$2+6x \leq -2$ or $2+6x > 11$
Solving the first inequality, $2+6x \leq -2$:
$2+6x \leq -2 \implies 6x \leq -4 \implies x \leq -\frac{4}{6} \implies x \leq -\frac{2}{3}$
Solving the second inequality, $2+6x > 11$:
$2+6x > 11 \implies 6x > 9 \implies x > \frac{9}{6} \implies x > \frac{3}{2}$
Now, let's consider the second expression, $-6-4x$. It will be outside the interval $(-2,11]$ if it is either less than or equal to $-2$ or greater than $11$. In other words, we want to find:
$-6-4x \leq -2$ or $-6-4x > 11$
Solving the first inequality, $-6-4x \leq -2$:
$-6-4x \leq -2 \implies -4x \leq 4 \implies x \geq -1$
Solving the second inequality, $-6-4x > 11$:
$-6-4x > 11 \implies -4x > 17 \implies x < -\frac{17}{4}$
To find the values of $x$ where neither expression is in the interval $(-2,11]$, we can take the union of the solutions to the inequalities we found:
$x \leq -\frac{2}{3}$ or $x > \frac{3}{2}$ or $x \geq -1$ or $x < -\frac{17}{4}$
We can express this as an interval or union of intervals:
$x \in (-\infty,-\frac{2}{3}] \cup (\frac{3}{2}, \infty) \cup [-1, -\frac{17}{4})$
Therefore, the set of all $x$ that satisfy the given conditions is the interval $(-\infty,-\frac{2}{3}] \cup (\frac{3}{2}, \infty) \cup [-1, -\frac{17}{4})$.
To find all $x$ such that neither $2+x+5x$ nor $2-x-8-3x$ is in the interval $(-2,11]$, we can solve each inequality separately and then find their intersection.
First, let's solve $2+x+5x \notin (-2,11]$. We can rewrite this inequality as $6x + 2 \notin (-2,11]$. We want to find the values of $x$ for which $6x+2$ is not within the interval $(-2,11]$.
If $6x+2$ is less than or equal to $-2$, we have $6x+2 \leq -2$. Solving this inequality, we get $x \leq -\frac{4}{6} = -\frac{2}{3}$.
If $6x+2$ is greater than 11, we have $6x+2 > 11$. Solving this inequality, we get $x > \frac{9}{6} = \frac{3}{2}$.
Therefore, the values of $x$ for which $6x+2$ is not in the interval $(-2,11]$ are $x \leq -\frac{2}{3}$ or $x > \frac{3}{2}$.
Next, let's solve $2-x-8-3x \notin (-2,11]$. Simplifying, we get $-4x-6 \notin (-2,11]$. We want to find the values of $x$ for which $-4x-6$ is not within the interval $(-2,11]$.
If $-4x-6$ is less than or equal to $-2$, we have $-4x-6 \leq -2$. Solving this inequality, we get $x \geq -\frac{1}{2}$.
If $-4x-6$ is greater than 11, we have $-4x-6 > 11$. Solving this inequality, we get $x < -\frac{17}{4} = -4.25$.
Therefore, the values of $x$ for which $-4x-6$ is not in the interval $(-2,11]$ are $x \geq -\frac{1}{2}$ or $x < -4.25$.
To find the intersection of the solutions, we take the values of $x$ that satisfy both inequalities. Therefore, the values of $x$ for which neither $2+x+5x$ nor $2-x-8-3x$ is in the interval $(-2,11]$ are $x \leq -\frac{2}{3}$ or $x > \frac{3}{2}$ and $x \geq -\frac{1}{2}$ or $x < -4.25$.
Expressing this as a union of intervals, the solution is $(-\infty, -4.25) \cup \left[-\frac{2}{3}, -\frac{1}{2}\right) \cup \left(\frac{3}{2}, \infty\right)$.