The inclination of the line $x - y + 3 = 0$ with the positive direction of $x - axis$ is 

To find the inclination of the line $x - y + 3 = 0$ with the positive direction of the $x$-axis, we first need to express the given line in slope-intercept form ($y = mx + c$), where $m$ represents the slope of the line.

The equation of the given line is $x - y + 3 = 0$, which can be rearranged as $y = x + 3$.

Comparing with $y = mx + c$, we have $m = 1$, indicating that the slope of the line is 1.

Now, the inclination of a line with slope $m$ with the positive direction of the $x$-axis is given by the angle $\theta = \arctan(m)$.

So, for our line with slope $m = 1$, the inclination angle $\theta$ is:

$</p><p style="line-height: 1.7" class="text-lg tracking-wide font-normal leading-relaxed">\theta = \arctan(1) = \frac{\pi}{4} \text{ radians} = 45^\circ</p><p style="line-height: 1.7" class="text-lg tracking-wide font-normal leading-relaxed">$

Therefore, the inclination of the line $x - y + 3 = 0$ with the positive direction of the $x$-axis is $45^\circ$.