How do you write y-1/4 = -3( x + 1/4 ) in standard form?

Sep 22, 2015

$12 x + 4 y = - 2$
Note: I'm assuming that by "standard form", you mean $A x + B y = C$.

Explanation:

First we begin by distributing the -3 on the right side of the equation:
$y - \frac{1}{4} = - 3 x - \frac{3}{4}$
From here, there are a few ways you can proceed. You can work with the fractions, but I prefer to get rid of them, in this case by multiplying by $4$:
$4 y - 1 = - 12 x - 3$
Now it's just a little algebra:
$4 y = - 12 x - 2$ (adding $1$ to both sides)
$12 x + 4 y = - 2$ (adding $12 x$ to both sides)

This is the official standard form of a linear equation, but most of the time you see them in slope-intercept form, $y = m x + b$. If this is what you meant by "standard form", or if you just want to know just in case, with a little work you can get there:
$12 x + 4 y = - 2$
$4 y = - 2 - 12 x$ (subtracting $12 x$ from both sides)
$y = - 3 x - \frac{1}{2}$ (dividing by $4$)
And there you go.