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.

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**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  
$$
C = \frac{5(F - 32)}{9}.
$$
**2.1.1** The solutions are $$ x = \frac{3}{2} $$ and $$ x = 1 $$. **2.1.2** $$ x = \frac{19}{2} $$. **2.2** $$ x = 3 $$ and $$ y = 1 $$. **2.3** $$ C = \frac{5(F - 32)}{9} $$.

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