How do you convert $8x-4y=16$ into slope-intercept form?

Hint:We have been given the equation of a straight-line which is in the standard form. In order to convert it into the slope-intercept form, we must have proper knowledge of the various forms of the equation of straight-line including the standard form and the slope-intercept form. Accordingly, we shall proceed further and make changes to our equation.

Complete step-by-step answer:
The standard form of a line is given as:
$ax+by=c=0$
Where,
$a=$ coefficient of x-variable
$b=$ coefficient of y-variable
$c=$ constant term
We can put various values of x or y-variable to find that particular point on line. If we input the value of both the x and y-component of the point, we can also verify whether that point belongs to that particular line or not.
The slope-intercept form of a line is expressed as:
$y=mx+c$
Where,
$m=$ slope of line
$c=$ intercept of the line
We shall make changes to our given equation, $8x-4y=16$accordingly.
Taking the term with y-variable to right hand side of equation and the constant term to the left hand side of the equation, we get
$\begin{align}
  & \Rightarrow 8x-16=4y \\
 & \Rightarrow 4y=8x-16 \\
\end{align}$
We will now divide the whole equation by 4 to make the coefficient of y equal to 1:
$\Rightarrow \dfrac{4y}{4}=\dfrac{8x}{4}-\dfrac{16}{4}$
$\Rightarrow y=2x-4$
Therefore, the equation, $8x-4y=16$is converted into its slope-intercept form as $y=2x-4$.

Note:
One thing to be taken care of is that the coefficient of y-variable is always 1 in the slope-intercept form of a straight line. Therefore, we must divide the entire equation with the coefficient of y to make it equal to one. Also, the coefficient of x-variable is the slope of the line.