(a) Solve the inequality : \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\) (b) The diagram shows a rectangle PQRS from which a square of side x cm has been...

Answer Details

(a) Solving the inequality

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

Multiply every term by 8 (the LCM of the denominators) to clear fractions:

\[8(4) + 8\cdot\frac{3}{4}(x+2) \le 8\cdot\frac{3}{8}x + 8(1)\]

\[32 + 6(x+2) \le 3x + 8\]

\[32 + 6x + 12 \le 3x + 8\]

\[6x + 44 \le 3x + 8\]

Collect like terms:

\[6x - 3x \le 8 - 44\]

\[3x \le -36\]

\[x \le -12\]

(b) Finding x from the rectangle

From the diagram, rectangle PQRS has height \(|PS| = 20\ \text{cm}\). Along the base SR a square of side x cm is cut out, leaving 10 cm on the left and 10 cm on the right. So the full length of the rectangle is

\[|SR| = 10 + x + 10 = (20 + x)\ \text{cm}\]

Area of the whole rectangle:

\[(20 + x)\times 20 = 400 + 20x\]

Area of the square removed:

\[x \times x = x^2\]

The shaded (remaining) area is 484 cm\(^2\):

\[(400 + 20x) - x^2 = 484\]

\[-x^2 + 20x + 400 - 484 = 0\]

\[-x^2 + 20x - 84 = 0\]

\[x^2 - 20x + 84 = 0\]

Solve using factorisation (\(-6\) and \(-14\) multiply to 84 and add to \(-20\)):

\[(x - 6)(x - 14) = 0\]

\[x = 6 \quad\text{or}\quad x = 14\]

Both values are geometrically valid (the cut height stays within the 20 cm rectangle), so

\[x = 6\ \text{cm} \quad\text{or}\quad x = 14\ \text{cm}\]