Find all $x$ so that neither $-2(6+2x)\le -16-x+8-5x$ nor $-3x\ge 18-7x+25$.
To deal with the inequality $-2(6+2x)\le -16-x+8-5x$, we simplify the left side and the right side separately. On the left side, we distribute the $-2$ to get $-12-4x$. On the right side, we combine like terms and get $-8-6x$. Thus our inequality is \begin{align*}
-12-4x\le-8-6x.
\end{align*}We add $4x$ to both sides and add $8$ to both sides to obtain \begin{align*}
12&\le 2x.
\end{align*}Dividing both sides by $2$ and reversing the inequality gives $x\ge 6$.
To deal with the inequality $-3x\ge 18-7x+25$, we simplify the left side and the right side separately. On the right side, we combine like terms and get $43-7x$. Thus our inequality is \begin{align*}
-3x\ge 43-7x.
\end{align*}Adding $3x$ to both sides and adding $7x$ to both sides to obtain \begin{align*}
43&\le 10x.
\end{align*}Dividing both sides by $10$ gives $x\ge \frac{43}{10}=\frac{86}{20}=\boxed{\frac{43}{10}}$.
Finally, we notice that the problem statement asks for values of $x$ that satisfy neither of the inequalities. To do this, we look for values of $x$ that satisfy both $x\ge 6$ and $x\ge \frac{43}{10}$. When $x\ge \frac{43}{10}$, it is guarantted that $x\ge 6$, so the only solutions to both inequalities are $x\ge \frac{43}{10}$. Therefore, the final solution is $\boxed{x>\frac{43}{10}}$.
To find all values of $x$ that satisfy both inequalities, we need to solve each inequality separately and then find the intersection of their solution sets.
Let's start with the first inequality:
$$-2(6+2x)\le -16-x+8-5x$$
We simplify this as follows:
$$-12 - 4x \le -16 - x + 8 - 5x$$
$$-12 - 4x \le -8 - 6x$$
$$-4x \le -6x - 4$$
$$-2x \le -4$$
$$2x \ge 4$$
$$x \ge 2$$
So, the first inequality is satisfied for all $x$ greater than or equal to 2.
Now, let's move on to the second inequality:
$$-3x \ge 18 - 7x + 25$$
We simplify this as follows:
$$-3x \ge 43 - 7x$$
$$4x \ge 43$$
$$x \ge \frac{43}{4}$$
So, the second inequality is satisfied for all $x$ greater than or equal to $\frac{43}{4}$.
To find the values of $x$ that satisfy both inequalities, we need to find the intersection of the solution sets. Since the first inequality is satisfied for $x \ge 2$ and the second inequality is satisfied for $x \ge \frac{43}{4}$, the intersection of their solution sets is the set of numbers greater than or equal to the larger value, which is $\frac{43}{4}$.
Therefore, the values of $x$ that satisfy both inequalities are $x \ge \frac{43}{4}$.
To solve the inequality $-2(6+2x) \leq -16 - x + 8 - 5x$, we follow these steps:
Step 1: Distribute the $-2$ on the left side of the inequality:
$-12 - 4x \leq -16 - x + 8 - 5x$
Step 2: Combine like terms on the right side of the inequality:
$-12 - 4x \leq -8 - 6x$
Step 3: Add $6x$ to both sides of the inequality:
$2x - 12 \leq -8$
Step 4: Add $12$ to both sides of the inequality:
$2x \leq 4$
Step 5: Divide both sides of the inequality by $2$:
$x \leq 2$
Therefore, the solutions for the inequality $-2(6+2x) \leq -16 - x + 8 - 5x$ is $x \leq 2$.
Now, let's solve the inequality $-3x \geq 18 - 7x + 25$:
Step 1: Combine like terms on the right side of the inequality:
$-3x \geq 43 - 7x$
Step 2: Add $7x$ to both sides of the inequality:
$4x \geq 43$
Step 3: Divide both sides of the inequality by $4$ (Remember to flip the inequality sign since we are dividing by a negative number):
$x \leq \frac{43}{4}$
Therefore, the solution for the inequality $-3x \geq 18 - 7x + 25$ is $x \leq \frac{43}{4}$.
Combining the solutions for both inequalities, we find that $x$ satisfies neither if it does not simultaneously satisfy $x \leq 2$ and $x \leq \frac{43}{4}$. Therefore, the solution set for the given inequalities is:
$\boxed{x \leq 2 \text{ and } x \leq \frac{43}{4}}$