The problem asks to find the real and imaginary parts of several expressions involving complex numbers $w$ and $x$, where: $w = -3 + i \quad \text{and} \quad x = 2 + 3i.$

Let's analyze each expression and find the real and imaginary parts:

1. $w + x$: $w + x = (-3 + i) + (2 + 3i) = (-3 + 2) + (i + 3i) = -1 + 4i.$

   • Real part: $-1$
   • Imaginary part: $4$

2. $w - x$: $w - x = (-3 + i) - (2 + 3i) = (-3 - 2) + (i - 3i) = -5 - 2i.$

   • Real part: $-5$
   • Imaginary part: $-2$

3. $wx$ (multiplication): $wx = (-3 + i)(2 + 3i).$ Using the distributive property: $wx = (-3 \cdot 2) + (-3 \cdot 3i) + (i \cdot 2) + (i \cdot 3i) = -6 - 9i + 2i + 3i^2.$ Since $i^2 = -1$, this becomes: $wx = -6 - 9i + 2i + 3(-1) = -6 - 7i - 3 = -9 - 7i.$

   • Real part: $-9$
   • Imaginary part: $-7$

4. $\frac{w}{x}$ (division): $\frac{w}{x} = \frac{-3 + i}{2 + 3i}.$ To simplify, we multiply the numerator and denominator by the complex conjugate of the denominator: $\frac{(-3 + i)(2 - 3i)}{(2 + 3i)(2 - 3i)}.$ The denominator simplifies to: $(2 + 3i)(2 - 3i) = 2^2 - (3i)^2 = 4 - 9(-1) = 4 + 9 = 13.$ Now, let's simplify the numerator: $(-3 + i)(2 - 3i) = (-3 \cdot 2) + (-3 \cdot -3i) + (i \cdot 2) - (i \cdot 3i) = -6 + 9i + 2i - 3(-1).$ $= -6 + 11i + 3 = -3 + 11i.$ So: $\frac{w}{x} = \frac{-3 + 11i}{13} = -\frac{3}{13} + \frac{11}{13}i.$

   • Real part: $-\frac{3}{13}$
   • Imaginary part: $\frac{11}{13}$

Would you like any additional details or explanations for these calculations?

Here are 5 related questions to expand your understanding:

1. What is the general formula for multiplying two complex numbers?
2. How can you find the complex conjugate of a given complex number?
3. Why do we multiply by the conjugate when dividing complex numbers?
4. What happens when you add or subtract two purely imaginary numbers?
5. Can a real number be considered a complex number? If so, how?

Tip: When dividing complex numbers, always multiply by the conjugate of the denominator to simplify the expression.