To solve the inequality $|7 - 2x| > (x + 1)|1 - x|$, we need to consider different cases based on the behavior of absolute value expressions.

Step 1: Analyze the absolute values

We need to look at when each expression inside the absolute value is positive or negative.

1. $|7 - 2x|$:

   • $7 - 2x \geq 0 \Rightarrow x \leq \frac{7}{2}$
   • $7 - 2x < 0 \Rightarrow x > \frac{7}{2}$

2. $|1 - x|$:

   • $1 - x \geq 0 \Rightarrow x \leq 1$
   • $1 - x < 0 \Rightarrow x > 1$

This gives us different intervals to consider for the absolute value functions.

Step 2: Solve on different intervals

We'll now solve the inequality piece by piece on different intervals, considering the signs of the expressions.

Case 1: $x \leq 1$

In this region, both $7 - 2x$ and $1 - x$ are positive, so we can rewrite the inequality without absolute values: $7 - 2x > (x + 1)(1 - x)$ Expanding the right-hand side: $7 - 2x > (x + 1)(1 - x) = x - x^2 + 1 - x = 1 - x^2$ So, the inequality becomes: $7 - 2x > 1 - x^2$ Move all terms to one side: $x^2 - 2x - 6 > 0$ Factor the quadratic: $(x - 3)(x + 2) > 0$ This inequality holds when $x < -2$ or $x > 3$. However, since we are considering $x \leq 1$ in this case, the only solution is $x < -2$.

Case 2: $1 < x \leq \frac{7}{2}$

In this region, $7 - 2x$ is positive and $1 - x$ is negative, so the inequality becomes: $7 - 2x > -(x + 1)(x - 1)$ Expanding the right-hand side: $7 - 2x > -(x^2 - 1) = -x^2 + 1$ So the inequality becomes: $7 - 2x > 1 - x^2$ Move all terms to one side: $x^2 - 2x - 6 > 0$ We already factored this as $(x - 3)(x + 2) > 0$, and this inequality holds when $x < -2$ or $x > 3$. In the interval $1 < x \leq \frac{7}{2}$, the solution is $x > 3$, so $3 < x \leq \frac{7}{2}$.

Case 3: $x > \frac{7}{2}$

In this region, both $7 - 2x$ and $1 - x$ are negative, so the inequality becomes: $-(7 - 2x) > -(x + 1)(x - 1)$ Simplifying the inequality: $-7 + 2x > -x^2 + 1$ Move all terms to one side: $x^2 + 2x - 8 > 0$ Factor the quadratic: $(x - 2)(x + 4) > 0$ This inequality holds when $x > 2$ or $x < -4$. Since we are considering $x > \frac{7}{2}$, the solution is $x > 2$, but there are no new solutions from this region.

Step 3: Combine the solutions

From the cases, we have two intervals where the inequality holds:

1. $x < -2$
2. $3 < x \leq \frac{7}{2}$

Therefore, the solution to the inequality is: $x \in (-\infty, -2) \cup (3, \frac{7}{2}]$

Would you like further details or have any questions?

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Here are five related questions:

1. How do you solve an inequality involving more than two absolute values?
2. What is the significance of breaking an absolute value inequality into cases?
3. How would the solution change if the inequality was $|7 - 2x| < (x + 1)|1 - x|$?
4. How do you solve inequalities that contain polynomials of degree higher than two?
5. What is the role of factoring when solving polynomial inequalities?

Tip: When solving absolute value inequalities, always split the equation into cases based on when the expression inside the absolute value is positive or negative.