#### Answer

$\dfrac{2}{3+i}=\dfrac{3}{5}-\dfrac{1}{5}i$

#### Work Step by Step

$\dfrac{2}{3+i}$
Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator:
$\dfrac{2}{3+i}=\dfrac{2}{3+i}\cdot\dfrac{3-i}{3-i}=\dfrac{2(3-i)}{3^{2}-i^{2}}=\dfrac{6-2i}{9-i^{2}}=...$
Substitute $i^{2}$ with $-1$ and simplify:
$...=\dfrac{6-2i}{9-(-1)}=\dfrac{6-2i}{9+1}=\dfrac{6-2i}{10}=\dfrac{6}{10}-\dfrac{2}{10}i=\dfrac{3}{5}-\dfrac{1}{5}i$