How do you write #y=4/3x+2/3# in standard form?

1 Answer
Jun 28, 2015

#4x-3y=-2#

Explanation:

The standard form of a linear equation is:
#Ax + By = C#
#A# can not be negative. #A#, #B# and #C# should all be integers.

The first thing we should do is move the #x# over to the left part of the equation. You can do this by substracting #4/3x# from both parts:
#y - 4/3x = 4/3x + 2/3 - 4/3x#
#y - 4/3x = 2/3#

By reordering, you get:
#-4/3x + y = 2/3#

Now we need to make sure that the number that's before the #x# (#A#) is positive. You can do this by multiplying both parts by #-1#:

#-1*(-4/3x + y) = -1*2/3#
#4/3x - y=-2/3#

Now, all we need to do is make A, B and C integers. You can always do this by multiplying by the LCM of all the denominators (#3#, #1# and #3#). This LCM is #3#:

#3*(4/3x - y)=3*(-2/3)#
#4x-3y=-2#