The problem is to express the complex number $\frac{4-i}{3+i}$ in the form $a+bi$, where $a$ and $b$ are real numbers.

1. Problem Description

The problem is to express the complex number $\frac{4-i}{3+i}$ in the form $a+bi$, where $a$ and $b$ are real numbers.

2. Solution Steps

To express the given complex number in the form $a+bi$, we need to eliminate the complex number from the denominator. We can do this by multiplying both the numerator and denominator by the conjugate of the denominator.

The conjugate of $3+i$ is $3-i$.

Multiply both the numerator and denominator by $3-i$:

$$\frac{4-i}{3+i} \times \frac{3-i}{3-i}$$

$$= \frac{(4-i)(3-i)}{(3+i)(3-i)}$$

Expand the numerator:

$$(4-i)(3-i) = 4(3) + 4(-i) - i(3) - i(-i) = 12 - 4i - 3i + i^2$$

Since $i^2 = -1$, we have:

$$12 - 7i - 1 = 11 - 7i$$

Expand the denominator:

$$(3+i)(3-i) = 3(3) + 3(-i) + i(3) + i(-i) = 9 - 3i + 3i - i^2$$

Since $i^2 = -1$, we have:

$$9 - (-1) = 9 + 1 = 10$$

Therefore,

$$\frac{(4-i)(3-i)}{(3+i)(3-i)} = \frac{11-7i}{10} = \frac{11}{10} - \frac{7}{10}i$$

3. Final Answer

The expression of $\frac{4-i}{3+i}$ in the form $a+bi$ is $\frac{11}{10} - \frac{7}{10}i$.