How do you write #x = 1/3 y – 2/3 # in standard form?

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Answer 1 · 2018-04-15T09:11:17

See a solution process below..

The standard form formula goes like this;

#Ax + Bx = C#

Where;

#A, B and C# are coefficients..

#x = 1/3y - 2/3#

First, multiply through by their LCM, in this case is #3#

#3(x) = 3(1/3y) - 3(2/3)#

#3x = 3y -2#

#3x - 3y = -2, "is in standard form"#