QUESTION 2 2.1 Simplify the following. 2.1.1 \( \frac{x^{2}-4}{x+3} \div\left(\frac{x-2}{4 x} \times \frac{8 x^{4}}{x^{2}+3 x}\right) \) 2.1.2 \( \frac{3+\frac{2}{x-1}}{3 x-1} \) 2.2 Solve for \( x \) and \( y \) : \( y=x^{2}-3 x-5 \) \( y-2 x=1 \)

**2.1 Simplify the following** **2.1.1** Simplify \[ \frac{x^2-4}{x+3} \div \left(\frac{x-2}{4x} \times \frac{8x^4}{x^2+3x}\right) \] 1. Factor the numerator: \[ x^2-4 = (x-2)(x+2). \] So the expression becomes: \[ \frac{(x-2)(x+2)}{x+3} \div \left(\frac{x-2}{4x} \times \frac{8x^4}{x^2+3x}\right). \] 2. Compute the product inside the parentheses: \[ \frac{x-2}{4x} \times \frac{8x^4}{x^2+3x} = \frac{(x-2) \cdot 8x^4}{4x \cdot (x^2+3x)}. \] 3. Simplify constants and factor common terms: - \( \dfrac{8}{4} = 2 \). - Express \( x^2+3x \) as \( x(x+3) \). - Cancel one factor of \( x \) from \( x^4 \) (numerator) and \( x \) (denominator): \[ \frac{2(x-2)x^4}{x \cdot x(x+3)} = \frac{2(x-2)x^3}{x(x+3)} = \frac{2(x-2)x^2}{x+3}. \] 4. Replace the product in the original expression: \[ \frac{(x-2)(x+2)}{x+3} \div \frac{2(x-2)x^2}{x+3} = \frac{(x-2)(x+2)}{x+3} \times \frac{x+3}{2(x-2)x^2}. \] 5. Cancel common factors: - Cancel \( x+3 \). - Cancel \( x-2 \) (with the note \( x \neq 2 \)): \[ \frac{x+2}{2x^2}. \] Thus, the simplified form is: \[ \frac{x+2}{2x^2}. \] --- **2.1.2** Simplify \[ \frac{3+\frac{2}{x-1}}{3x-1} \] 1. Combine the terms in the numerator by finding a common denominator: \[ 3+\frac{2}{x-1} = \frac{3(x-1)}{x-1}+\frac{2}{x-1}=\frac{3x-3+2}{x-1}=\frac{3x-1}{x-1}. \] 2. The overall expression becomes: \[ \frac{\frac{3x-1}{x-1}}{3x-1} = \frac{3x-1}{(x-1)(3x-1)}. \] 3. Cancel the common factor \( 3x-1 \) (provided \( 3x-1 \neq 0 \)): \[ \frac{1}{x-1}. \] Thus, the simplified form is: \[ \frac{1}{x-1}. \] --- **2.2 Solve for \( x \) and \( y \):** \[ y = x^2-3x-5 \quad \text{and} \quad y-2x=1 \] 1. Express \( y \) from the second equation: \[ y-2x=1 \implies y=2x+1. \] 2. Substitute \( y=2x+1 \) into the first equation: \[ 2x+1 = x^2-3x-5. \] 3. Rearrange the equation to form a quadratic: \[ x^2-3x-5-2x-1=0 \quad \Longrightarrow \quad x^2-5x-6=0. \] 4. Factor the quadratic: \[ x^2-5x-6 = (x-6)(x+1)=0. \] 5. Solve for \( x \): \[ x-6=0 \implies x=6, \quad \text{or} \quad x+1=0 \implies x=-1. \] 6. Find corresponding \( y \) values using \( y=2x+1 \): - For \( x=6 \): \[ y=2(6)+1=13. \] - For \( x=-1 \): \[ y=2(-1)+1=-2+1=-1. \] Thus, the solutions are: \[ (x,y)=(6, 13) \quad \text{or} \quad (x,y)=(-1, -1). \]