QUESTION 2 2. Determine, without the use of a calculator, the value of \( x \) in each of the following: 2.1.1 \( \quad 2 x^{2}-5 x+3=0 \) \( 2.1 .2 \frac{x+1}{3}-\frac{x-2}{5}-2=0 \) 2.2 Given that: \( 2 x-3 y=3 \) and \( 2 x+y=7 \) Determine the values of \( x \) and \( y \) simultaneously. 2.3 The formula \( F=32+\frac{9 C}{5} \) is used for converting temperatures for degrees Celsius ( \( { }^{\circ} \mathrm{C} \) ) to degrees Fahrenheit ( \( { }^{\circ} \mathrm{F} \) ). Make \( C \) the subject of this formula.

**2.1.1** We start with the quadratic equation \[ 2x^2 - 5x + 3 = 0. \] The discriminant is \[ \Delta = (-5)^2 - 4\cdot2\cdot3 = 25 - 24 = 1. \] Using the quadratic formula \[ x = \frac{-b \pm \sqrt{\Delta}}{2a}, \] we have \[ x = \frac{5 \pm 1}{4}. \] Thus, the solutions are \[ x = \frac{5+1}{4} = \frac{6}{4} = \frac{3}{2} \quad \text{and} \quad x = \frac{5-1}{4} = \frac{4}{4} = 1. \] **2.1.2** Given the equation \[ \frac{x+1}{3} - \frac{x-2}{5} - 2 = 0, \] we eliminate the fractions by multiplying every term by the least common multiple of 3 and 5, which is 15: \[ 15 \left(\frac{x+1}{3}\right) - 15\left(\frac{x-2}{5}\right) - 15(2) = 0. \] This simplifies to \[ 5(x+1) - 3(x-2) - 30 = 0. \] Expanding the terms gives: \[ 5x + 5 - 3x + 6 - 30 = 0. \] Combine like terms: \[ 2x - 19 = 0. \] Solve for \(x\): \[ 2x = 19 \quad \Rightarrow \quad x = \frac{19}{2}. \] **2.2** We are given the system: \[ \begin{cases} 2x - 3y = 3, \\ 2x + y = 7. \end{cases} \] First, solve the second equation for \(y\): \[ y = 7 - 2x. \] Substitute this into the first equation: \[ 2x - 3(7 - 2x) = 3. \] Expanding and simplifying: \[ 2x - 21 + 6x = 3, \] \[ 8x - 21 = 3, \] \[ 8x = 24, \] \[ x = 3. \] Now, substitute \(x = 3\) back into \(y = 7 - 2x\): \[ y = 7 - 2(3) = 7 - 6 = 1. \] Thus, the solution is \[ x = 3, \quad y = 1. \] **2.3** We start with the temperature conversion formula: \[ F = 32 + \frac{9C}{5}. \] To solve for \(C\), first subtract 32 from both sides: \[ F - 32 = \frac{9C}{5}. \] Then, multiply both sides by \(\frac{5}{9}\) to isolate \(C\): \[ C = \frac{5}{9}(F - 32). \] Thus, expressed in terms of \(F\), the formula becomes \[ C = \frac{5(F - 32)}{9}. \]