The problem contains three parts, where you need to solve inequalities for $x$. Let us solve each part step by step.

Part (a): Solve $3 + x < 2x - 1$

1. Rearrange terms to isolate $x$: $3 + x < 2x - 1 \implies 3 + 1 < 2x - x \implies 4 < x$ So, the solution is: $x > 4$

Part (b): Solve $x(x - 1) > 6$

1. Expand and rearrange the inequality: $x^2 - x - 6 > 0$
2. Factorize the quadratic: $x^2 - x - 6 = (x - 3)(x + 2)$ So, the inequality becomes: $(x - 3)(x + 2) > 0$
3. Solve for $x$ using the sign chart:
   • The critical points are $x = 3$ and $x = -2$.
   • Test intervals: $(-\infty, -2)$, $(-2, 3)$, $(3, \infty)$:
     • For $x \in (-\infty, -2)$, $(x - 3)(x + 2) > 0$: False
     • For $x \in (-2, 3)$, $(x - 3)(x + 2) > 0$: False
     • For $x \in (3, \infty)$, $(x - 3)(x + 2) > 0$: True So, the solution is: $x \in (-\infty, -2) \cup (3, \infty)$

Part (c): Solve $3 + x < 2x - 1$ and $x(x - 1) > 6$

From part (a), $x > 4$.
From part (b), $x \in (3, \infty)$.

The intersection of these solutions is: $x > 4$ So, the solution for part (c) is: $x > 4$

Final Solutions:

(a) $x > 4$
(b) $x \in (-\infty, -2) \cup (3, \infty)$
(c) $x > 4$

Would you like further details or have any questions?

Here are 5 related questions for practice:

1. Solve $5x - 7 > 3x + 1$.
2. Solve $x^2 - 4x + 3 < 0$.
3. Find the solution set for $x(x + 1) \geq 12$.
4. Solve $2x - 1 > 3$ and $x^2 - 5x + 6 < 0$ simultaneously.
5. Determine the critical points of $x^2 - 5x - 14$.

Tip: Always graph the quadratic inequalities to visualize the solution intervals!