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]
