Find all values of $x$ which do not satisfy $$1 \le 4-3x+5x < 10-8x.$$ Express your answer in interval notation.
We have that $1 \le 4-3x+5x < 10-8x$ if and only if $1 \le 4+2x < 10-8x$. Next, we consider each of the two inequalities separately.
First, we solve $1 \le 4+2x$. Subtracting 4 from both sides gives $-3 \le 2x$, so dividing by 2 gives $-\frac{3}{2} \le x$.
Next, we solve $4+2x < 10-8x$. Adding $8x$ to both sides gives $8x+2x < 10-4$, or $10x<6$. Dividing by 10 gives $x<\frac{3}{5}$.
Therefore, $x$ satisfies both inequalities if and only if $-\frac35<x\le -\frac32$. Answer: $\boxed{\left(-\frac{3}{2}, -\frac{3}{5}\right)}$.
To find the values of $x$ that do not satisfy the inequality, we need to determine the values of $x$ that make the inequality false.
Given inequality is: $1 \le 4-3x+5x < 10-8x$.
To solve this inequality, we will break it down into two separate inequalities connected by the logical "and" operator, $\wedge$.
First, let's consider the left inequality:
$1 \le 4-3x+5x$
Step 1: Simplify the inequality
$1 \le 4+2x$
Step 2: Subtract 4 from both sides to isolate $x$
$1-4 \le 2x$
$-3 \le 2x$
Step 3: Divide both sides by 2, but be careful about the direction of the inequality since we're dealing with a negative coefficient.
$\displaystyle \frac{-3}{2} \geq x$
$x \leq -\frac{3}{2}$
The solution to the left inequality is $x \leq -\frac{3}{2}$.
Now let's consider the right inequality:
$4-3x+5x < 10-8x$
Step 1: Simplify the inequality
$4+2x < 10-8x$
Step 2: Combine like terms
$2x+8x < 10-4$
$10x < 6$
Step 3: Divide both sides by 10
$\displaystyle \frac{10x}{10} < \frac{6}{10}$
$x < \displaystyle \frac{3}{5}$
The solution to the right inequality is $x < \displaystyle \frac{3}{5}$.
Therefore, the combined solutions to both inequalities are $x \leq -\frac{3}{2}$ and $x < \displaystyle \frac{3}{5}$.
Expressing these solutions in interval notation, the values of $x$ that do not satisfy the inequality are:
$(-\infty,-\frac{3}{2}] \cup (-\infty,\frac{3}{5})$