y=12x−1 in standard form?

y = (1/2)x - 1
2y = 2[(1/2)x - 1]
2y = x - 2
x - 2y - 2 = 0

Standard form : x - 2y - 2 = 0

Y=12x−1 in standard form,
So,
12x -y = 1. <----------------------/

　
Standard form: Ax + By = C
where A, B, C are mutually prime integers,
and A ≥ 0 (if A = 0, then B > 0)

y = 1/2 x − 1
2y = x − 2
−x + 2y = −2
x − 2y = 2

12x - y = 1
After your update
x - 2y = 2

y=12x−1

Explanation:

To put a polynomial into standard form, multiply out to get rid of the brackets, then group like items and put in descending order of powers.

But where having the y=mx+c

y=12X-1

The slope and y-intercept of the relation represented by the equation Y=12X-1 are

Y=12X-1

slope =m=12

c=-1

Therefore the standard form of the equation is "Y-12x+1=0". Hope this helps

Now

"Y-(1/2x)+1=0". (since 12x=1/2x )

The older 'standard form' is: y = mx + b

Which makes your equation already IN standard form.

'm' is the slope and 'b' is the y-intercept. Very easy and straightforward because the sign on 'm' & 'b' could be read directly from the equation.




The NEW 'standard form' is ax + by = c, which makes your equation become:

=> y = (1/2)x - 1
=> 2y = x - 2

=> -x + 2y = -2
=> x - 2y = 2 <--- final form

This is much more complicated and convoluted way to express linear equations, and the slope and y-intercept are now more tricky to read off this equation. This is a terrible way to write these equations. Don't bother using this form if you have a choice.

This is in standard form.

Standard form is : ax + by = c where a and b are prime integers and c is a constant , so y = 1/2x - 1 is 2y = x -2, or 2y -x =-2 which is x -2y = 2

In order to get the standard form of "Y=12x-1" you need to bring 12x and -1 to the other side. So it would equal "Y-12x+1=0, when you bring 12x to the other side it becomes -12x and when you bring -1 to the other side it becomes 1. Therefore the standard form of the equation is "Y-12x+1=0". Hope this helps!

Now that you updated your question the standard form is "Y-1/2x+1=0"

Yes