The problem consists of two parts. (a) Solve the inequality: $4 + \frac{3}{4}(x + 2) \le \frac{3}{8}x + 1$. (b) A rectangle PQRS has dimensions 20 cm by (10+10) cm = 20 cm. Two squares of side x cm have been cut from the rectangle. The area of the shaded portion is 484 cm$^2$. Find the values of x.

1. Problem Description

The problem consists of two parts.

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

(b) A rectangle PQRS has dimensions 20 cm by (10+10) cm = 20 cm. Two squares of side x cm have been cut from the rectangle. The area of the shaded portion is 484 cm$^2$. Find the values of x.

2. Solution Steps

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

Step 1: Expand the terms in the inequality.

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

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

Step 2: Combine the constant terms on the left side.

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

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

Step 3: Subtract $\frac{3}{8}x$ from both sides.

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

$\frac{11}{2} + \frac{6}{8}x - \frac{3}{8}x \le 1$

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

Step 4: Subtract $\frac{11}{2}$ from both sides.

$\frac{3}{8}x \le 1 - \frac{11}{2}$

$\frac{3}{8}x \le \frac{2}{2} - \frac{11}{2}$

$\frac{3}{8}x \le -\frac{9}{2}$

Step 5: Multiply both sides by $\frac{8}{3}$.

$x \le -\frac{9}{2} \cdot \frac{8}{3}$

$x \le -\frac{72}{6}$

$x \le -12$

(b) Find the values of x.

Step 1: Find the area of the rectangle PQRS.

The length is 20 cm and the width is 10 cm + 10 cm = 20 cm.

Area of rectangle PQRS = length * width = $20 \cdot 20 = 400$ cm$^2$.

Step 2: Find the area of the two squares.

Each square has side x cm, so the area of each square is $x^2$ cm$^2$.

The area of the two squares is $2x^2$ cm$^2$.

Step 3: The area of the shaded portion is the area of the rectangle minus the area of the two squares.

Area of shaded portion = Area of rectangle - Area of two squares

$484 = 400 - 2x^2$ (There must be a typo, the shaded area should be less than the rectangle)

Assuming the shaded area is actually

3

1

6. $316=400-2x^2$

$2x^2 = 400-316$

$2x^2 = 84$

$x^2=42$

$x = \sqrt{42} \approx 6.48$

Correcting the given equation:

Let $A_{shaded}$ be the area of the shaded region. We are given that $A_{shaded} = 484 \text{ cm}^2$. The total area of the rectangle $PQRS$ is $20 \cdot (10+10) = 20 \cdot 20 = 400 \text{ cm}^2$.

Since two squares of side $x$ are cut out, the area of the two squares is $2x^2$.

Then, the area of the shaded region should be $400 - 2x^2$. So we must have

$400 - 2x^2 = A_{shaded}$

If $A_{shaded}=484$, then

$400 - 2x^2 = 484$

$-2x^2 = 84$

$x^2 = -42$, which is impossible since $x$ is a real number.

Since the squares are cut from the rectangle, it must be that $A_{shaded} < 400$. Therefore there must be an error in the problem statement.

3. Final Answer

(a) $x \le -12$

(b) Assuming the shaded area is actually 316, $x = \sqrt{42}$ cm. There is an error in the problem statement. It is impossible to have a shaded area of

4

8

4.