Find all $x$ such that neither $2x-5$ nor $-20x+13x-7$ is in the interval $(8,12)$.
First, $2x-5 \in (8,12)$ means that $8 < 2x-5 < 12$, so $13 < 2x < 17$, and $6.5 < x < 8.5$. Since we cannot have $x$ in the interval $(6.5, 8.5)$, it follows that the numbers which satisfy $2x-5 \in (8,12)$ are those that are equal to or less than $6.5$ or equal to or greater than $8.5$.
Now, $-20x+13x-7 \in (8,12)$ means that $8 < -20x+13x-7 < 12$. Rearranging this gives $15 < -7x < 19$, and dividing each term by $-7$ gives $-15/7 > x > -19/7$. Hence, any $x$ satisfying $-15/7 > x > -19/7$ is acceptable.
Graphing both inequalities, we have that $2x-5$ has all points to the left of $6.5$ shaded, while $-20x+13x-7$ has all points in between $-15/7$ and $-19/7$ shaded. To find the union of these shaded regions, which correspond to the numbers that satisfy neither $2x-5$ nor $-20x+13x-7$ in the interval $(8,12)$, subtract $x$ instead of adding. The answer is any $x$ that is $\boxed{\leq -19/7 \text{ or }\geq 8.5}$.
To find all values of $x$ such that neither $2x-5$ nor $-20x+13x-7$ is in the interval $(8,12)$, we need to solve two inequalities.
First, let's consider the inequality $2x-5 \notin (8,12)$. To find the values of $x$ that satisfy this inequality, we can set up two separate inequalities and solve for $x$:
1. $2x-5 < 8$
2. $2x-5 > 12$
Solving the first inequality for $x$:
$2x-5 < 8$
$2x < 8 + 5$
$2x < 13$
$x < \frac{13}{2}$
Solving the second inequality for $x$:
$2x-5 > 12$
$2x > 12 + 5$
$2x > 17$
$x > \frac{17}{2}$
Therefore, for the inequality $2x-5 \notin (8,12)$, the values of $x$ are $x < 6.5$ and $x > 8.5$.
Now, let's consider the inequality $-20x+13x-7 \notin (8,12)$. Again, we can set up two separate inequalities:
1. $-20x+13x-7 < 8$
2. $-20x+13x-7 > 12$
Simplifying the first inequality:
$-7x-7 < 8$
$-7x < 8 + 7$
$-7x < 15$
$x > -\frac{15}{7}$
Simplifying the second inequality:
$-7x+5 > 12$
$-7x > 12 - 5$
$-7x > 7$
$x < -1$
Therefore, for the inequality $-20x+13x-7 \notin (8,12)$, the values of $x$ are $x > -\frac{15}{7}$ and $x < -1$.
In conclusion, the values of $x$ for which neither $2x-5$ nor $-20x+13x-7$ is in the interval $(8,12)$ are $x < -1$, $-1 < x < -\frac{15}{7}$, $x < 6.5$, and $x > 8.5$.
To find all $x$ such that neither $2x-5$ nor $-20x+13x-7$ is in the interval $(8,12)$, we can set up two inequalities and solve for $x$.
First, let's consider $2x-5$. We want to find the values of $x$ for which $2x-5$ is not in the interval $(8,12)$. This means that $2x-5$ must be either less than or equal to 8, or greater than or equal to 12. We can write this as two separate inequalities:
$2x-5 \leq 8$ and $2x-5 \geq 12$
Solving the first inequality:
$2x-5 \leq 8$
Adding 5 to both sides:
$2x \leq 13$
Dividing both sides by 2:
$x \leq \frac{13}{2}$
So the first inequality gives us $x \leq \frac{13}{2}$.
Solving the second inequality:
$2x-5 \geq 12$
Adding 5 to both sides:
$2x \geq 17$
Dividing both sides by 2:
$x \geq \frac{17}{2}$
So the second inequality gives us $x \geq \frac{17}{2}$.
Now let's consider $-20x+13x-7$. We want to find the values of $x$ for which $-20x+13x-7$ is not in the interval $(8,12)$. This means that $-20x+13x-7$ must be either less than or equal to 8, or greater than or equal to 12. Again, we can write this as two separate inequalities:
$-20x+13x-7 \leq 8$ and $-20x+13x-7 \geq 12$
Simplifying the first inequality:
$-7x - 7 \leq 8$
Adding 7 to both sides:
$-7x \leq 15$
Dividing both sides by -7 (and flipping the inequality sign):
$x \geq -\frac{15}{7}$
So the first inequality gives us $x \geq -\frac{15}{7}$.
Simplifying the second inequality:
$-7x - 7 \geq 12$
Adding 7 to both sides:
$-7x \geq 19$
Dividing both sides by -7 (and flipping the inequality sign):
$x \leq -\frac{19}{7}$
So the second inequality gives us $x \leq -\frac{19}{7}$.
Now we have four conditions for $x$: $x \leq \frac{13}{2}$, $x \geq \frac{17}{2}$, $x \geq -\frac{15}{7}$, and $x \leq -\frac{19}{7}$. We can find the values of $x$ that satisfy all of these conditions by finding the intersection of the intervals defined by these conditions. After finding the intersection, we can express the final solution as an interval.
To find the intersection, we can plot the conditions on a number line:
-15/7 -19/7 17/2 13/2
Now we can see that the values of $x$ that satisfy all the conditions are $x \leq -\frac{19}{7}$ and $x \geq \frac{17}{2}$. In other words, $x$ is less than or equal to $-\frac{19}{7}$ or greater than or equal to $\frac{17}{2}$.
Therefore, the set of all $x$ such that neither $2x-5$ nor $-20x+13x-7$ is in the interval $(8,12)$ is $x \leq -\frac{19}{7}$ or $x \geq \frac{17}{2}$.