Answer :

Alright, let's find the real part of the expression [tex]\( \frac{2}{3+i} + \frac{3}{2+i} \)[/tex].

### Step-by-Step Solution:

1. Rationalizing the Denominators:

First, we rationalize each of the fractions by multiplying the numerator and the denominator by the conjugate of the denominator.

For the first term [tex]\( \frac{2}{3+i} \)[/tex]:

- The conjugate of [tex]\( 3+i \)[/tex] is [tex]\( 3-i \)[/tex].
- Multiply both numerator and denominator by [tex]\( 3-i \)[/tex]:

[tex]\[ \frac{2}{3+i} \cdot \frac{3-i}{3-i} = \frac{2(3-i)}{(3+i)(3-i)} \][/tex]

- Simplify the numerator:

[tex]\[ 2(3-i) = 6 - 2i \][/tex]

- Simplify the denominator using the difference of squares formula [tex]\((a+b)(a-b) = a^2 - b^2\)[/tex]:

[tex]\[ (3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 10 \][/tex]

- So, we have:

[tex]\[ \frac{2(3-i)}{(3+i)(3-i)} = \frac{6-2i}{10} = 0.6 - 0.2i \][/tex]

2. For the second term [tex]\( \frac{3}{2+i} \)[/tex]:

- The conjugate of [tex]\( 2+i \)[/tex] is [tex]\( 2-i \)[/tex].
- Multiply both numerator and denominator by [tex]\( 2-i \)[/tex]:

[tex]\[ \frac{3}{2+i} \cdot \frac{2-i}{2-i} = \frac{3(2-i)}{(2+i)(2-i)} \][/tex]

- Simplify the numerator:

[tex]\[ 3(2-i) = 6 - 3i \][/tex]

- Simplify the denominator using the difference of squares formula:

[tex]\[ (2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 5 \][/tex]

- So, we have:

[tex]\[ \frac{3(2-i)}{(2+i)(2-i)} = \frac{6-3i}{5} = 1.2 - 0.6i \][/tex]

3. Adding the Rationalized Fractions:

Now we sum the two fractions [tex]\( 0.6 - 0.2i \)[/tex] and [tex]\( 1.2 - 0.6i \)[/tex]:

[tex]\[ (0.6 - 0.2i) + (1.2 - 0.6i) = (0.6 + 1.2) + (-0.2i - 0.6i) \][/tex]

- Combine the real parts:

[tex]\[ 0.6 + 1.2 = 1.8 \][/tex]

- Combine the imaginary parts:

[tex]\[ -0.2i - 0.6i = -0.8i \][/tex]

So, the combined result is:

[tex]\[ (0.6 - 0.2i) + (1.2 - 0.6i) = 1.8 - 0.8i \][/tex]

4. Extracting the Real Part:

Finally, the real part of [tex]\( 1.8 - 0.8i \)[/tex] is simply:

[tex]\[ \boxed{1.8} \][/tex]