<s>[INST] To find the x and y-intercepts of the line $ax+by+c=0$ Please check if I've solved the problem in the correct way:
The problem is as follows:

Find the points at which the line $ax+by+c=0$ crosses the x and y-axes. (Assume that $a \neq 0$ and $b \neq 0$.

My solution:
We have to find the x and y-intercepts of the line. At the 'x-intercept' the ordinate must be equal to $0$ and at the 'y-intercept' the abscissa must be equal to $0$.
Now we solve the equation $ax+by+c=0$ for $y$:
$ax + by + c=0$
$ax + by = -c$
$by = -ax -c$
$y = \frac {-ax-c}{b}$
$\because x = 0$ at y-intercept,
$\therefore y = \frac {-a(0)}{b} -\frac{c}{b}$
$y = -\frac cb$.
The point at which the line crosses the y-axis is $(0,-\frac cb)$
Now we solve the equation $ax+by+c=0$ for $x$:
$ax+by+c=0$
$ax+by=-c$
$ax = -by-c$
$x = \frac {-by-c}{a}$
$\because y = 0$ at x-intercept
$\therefore x = \frac {-b(0)}{a} -\frac{c}{a}$
$x = -\frac ca$
The point at which the line crosses the x-axis is $(-\frac ca,0)$
 [/INST] easier way is this equation of line intercept form
$$ \frac xa+\frac yb=1$$
so write the given eqn in this form
$$ax+by+c=0$$
$$ax+by=-c$$
$$\frac {ax}{-c}+\frac {by}{-c}=1$$
$$\frac {x}{\frac{-c}{a}}+\frac {y}{\frac {-c}{b}}=1$$
so intercept at X axis is (-c/a,0) and Y axis is (0,-c/b)
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