<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] Your work is exemplary, Samama. You know your definitions well, and your answers are entirely correct.
A nice "shortcut" is to take advantage of what you already know: 


*

*the $x$ intercept is the value of $x$ when $y = 0$, and 

*the $y$ intercept is the value of $y$ when $x = 0$.

*$ax + by + c = 0 \iff ax + by = -c$


x intercept
We can use  $\;ax + by = -c\;$ to solve for $x$ when $y = 0$, by plugging in $0$ for $y,\;$ right at the start:
$$ax + \underbrace{by}_{y = 0} = -c \iff ax = - c \iff x = \frac {-c}{a}$$
y intercept
And we can do the same to solve for $y$ when $x = 0$, by plugging in $0$ for $x,\,$ right at the start:
$$\underbrace{ax}_{x = 0} + by = -c \;\;\iff\;\; by = -c \;\;\iff \;\; y = \frac{-c}{b}$$
</s>