Question
Question: How do you convert \(8x-4y=16\) into slope-intercept form?...
How do you convert 8x−4y=16 into slope-intercept form?
Solution
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=16accordingly.
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
⇒8x−16=4y⇒4y=8x−16
We will now divide the whole equation by 4 to make the coefficient of y equal to 1:
⇒44y=48x−416
⇒y=2x−4
Therefore, the equation, 8x−4y=16is 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.