How do you write $y + \frac{1}{3} = \frac{1}{4} (x - 3)$ $y+1/3 = 1/4(x-3)$ in standard form?

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Answer 1 · 2015-05-31T10:08:46

$y + \frac{1}{3} = \frac{1}{4} (x - 3)$ $y+1/3=1/4(x-3)$

To get rid of the fractions, multiply both sides $12$ $12$ :

$12 (y + \frac{1}{3}) = 12 \times \frac{1}{4} (x - 3)$ $12(y+1/3)=12xx1/4(x-3)$

$12 y + 4 = 3 (x - 3)$ $12y+4=3(x-3)$

$12 y + 4 = 3 x - 9$ $12y+4=3x-9$

Standard form :
$A x + B y = C$ $Ax+By=C$

Rearrange to get:

$3 x - 12 y = 13$ $color(green)[3x-12y=13$

How do you write y+13=14(x−3)y+1/3 = 1/4(x-3) in standard form?