The problem consists of two parts. (a) Solve the inequality $4 + \frac{3}{4}(x+2) \leq \frac{3}{8}x + 1$. (b) A rectangle $PQRS$ has dimensions $20$ cm by $(10 + x + 10)$ cm, i.e., $20$ cm by $(20+x)$ cm. A square of side $x$ cm is cut out. The area of the shaded portion is $484$ cm$^2$. Find the value of $x$.

1. Problem Description

The problem consists of two parts.

(a) Solve the inequality $4 + \frac{3}{4}(x+2) \leq \frac{3}{8}x + 1$.

(b) A rectangle $PQRS$ has dimensions $20$ cm by $(10 + x + 10)$ cm, i.e., $20$ cm by $(20+x)$ cm. A square of side $x$ cm is cut out. The area of the shaded portion is $484$ cm$^2$. Find the value of $x$.

2. Solution Steps

(a) Solve the inequality:

$4 + \frac{3}{4}(x+2) \leq \frac{3}{8}x + 1$

$4 + \frac{3}{4}x + \frac{6}{4} \leq \frac{3}{8}x + 1$

$4 + \frac{3}{4}x + \frac{3}{2} \leq \frac{3}{8}x + 1$

$\frac{8}{2} + \frac{3}{2} + \frac{3}{4}x \leq \frac{3}{8}x + 1$

$\frac{11}{2} + \frac{3}{4}x \leq \frac{3}{8}x + 1$

Multiply by 8 to clear the fractions:

$8(\frac{11}{2} + \frac{3}{4}x) \leq 8(\frac{3}{8}x + 1)$

$44 + 6x \leq 3x + 8$

$6x - 3x \leq 8 - 44$

$3x \leq -36$

$x \leq -12$

(b) The area of the rectangle $PQRS$ is $20(20+x) = 400 + 20x$.

The area of the square cut out is $x^2$.

The area of the shaded portion is the area of the rectangle minus the area of the square:

$400 + 20x - x^2 = 484$

$-x^2 + 20x + 400 - 484 = 0$

$-x^2 + 20x - 84 = 0$

$x^2 - 20x + 84 = 0$

We can solve this quadratic equation using the quadratic formula or by factoring. Let's try factoring:

$(x - 6)(x - 14) = 0$

$x = 6$ or $x = 14$

Since the width of the rectangle is $20+x$, and we have two segments of length $10$, we must have $x < 20$. Both solutions are valid.

3. Final Answer

(a) $x \leq -12$

(b) $x = 6$ or $x = 14$