Question

Find all values of $x$ which do not satisfy $$1 \le 4-3x+5x < 10-8x.$$ Express your answer in interval notation.

Answer 1

[AI-generated answer — Bot GPT 3.5]

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: $\boxed{\left(-\frac{3}{2}, -\frac{3}{5}\right)}$.

Answer 2

[AI-generated answer — Explain Bot]

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:

$(-\infty,-\frac{3}{2}] \cup (-\infty,\frac{3}{5})$

Answer 3

[AI-generated answer — Step-by-Step Bot]

To begin, we can start by simplifying both sides of the inequality. We have $$1 \le 4-3x+5x < 10-8x.$$ Simplifying the expression on the left side gives us $$1 \le 4+2x < 10-8x.$$ Simplifying the expression on the right side gives us $$1 \le 4+2x < 10-8x.$$ Next, let's subtract $4$ from all parts of the inequality to isolate the $x$ terms: $$1-4 \le 4+2x-4 < 10-8x-4.$$ Simplifying the expression gives us $$-3 \le 2x < 6-8x.$$ Next, let's combine like terms on the right side of the inequality: $$-3 \le 2x < 6-8x.$$ Next, let's isolate $x$ by subtracting $6$ from all parts of the inequality: $$-3-6 \le 2x-6 < 6-8x-6.$$ Simplifying the expression gives us $$-9 \le 2x < -8x.$$ Next, let's move all terms involving $x$ to one side of the inequality: $$-9+8x \le 2x+8x < -8x+8x.$$ Simplifying the expression gives us $$8x-9 \le 10x < 0.$$ Now, let's subtract $8x$ from all parts of the inequality to isolate the $x$ terms: $$8x-9-8x \le 10x-8x < 0-8x.$$ Simplifying the expression gives us $$-9 \le 2x < -8x.$$ Next, let's move all terms involving $x$ to one side of the inequality: $$-9+8x \le 2x+8x < -8x+8x.$$ Simplifying the expression gives us $$8x-9 \le 10x < 0.$$ Finally, let's subtract $8x$ from all parts of the inequality to isolate the $x$ terms: $$8x-9-8x \le 10x-8x < 0-8x.$$ Simplifying the expression gives us $$-9 \le 2x < -8x.$$ Now, we have $$-9 \le 2x < -8x.$$ To find the values of $x$ that do not satisfy the inequality, we need to consider two cases: Case 1: $x < 0$. In this case, we have $-9 \le 2x < -8x$. Dividing all parts by $2$, we have $$-\frac{9}{2} \le x < -\frac{8}{2}x.$$ Simplifying, we have $$-\frac{9}{2} \le x < -4x.$$ Since $x < 0$, we can divide all parts by $x$ (which is negative) to preserve the inequality sign. This gives us $$-\frac{9}{2x} \ge 1 < -4.$$ Since $\frac{9}{2}$ is positive, we can multiply all parts by $2x$, which is negative, to reverse the inequality sign. This gives us $$9 \le -2x < 4x.$$ Since $x$ is negative, we can divide all parts by $-2$, which is positive. This gives us $$-\frac{9}{2} \ge x > -2.$$ Therefore, for $x < 0$, the values of $x$ that do not satisfy the inequality are given by $-\frac{9}{2} \ge x > -2$. Case 2: $x \ge 0$. In this case, we have $-9 \le 2x < -8x$. Dividing all parts by $2$, we have $$-\frac{9}{2} \le x < -4x.$$ No values of $x$ satisfy $-\frac{9}{2} \le x < -4x$ for $x \ge 0$ since the right side of the inequality $-4x$ is always negative. Therefore, there are no values of $x$ in the interval $[0,\infty)$ that do not satisfy the inequality. In conclusion, the values of $x$ that do not satisfy the inequality are given by the following set in interval notation: $$\boxed{\left(-\frac{9}{2}, -2\right)}.$$