How do you write $4y = 3/16x - 0.8$ in standard form?

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Answer 1 · 2015-05-23T13:31:02

Standard form of a linear equation is (normally taken to be):
$Ax+By = C$ (sometimes with the restriction that $A>0 " & "AepsilonZZ$ )

Given $4y = 3/16x-0.8$

Subtracting $3/16x$ from both sides
$-3/16x + 4y = -0.8$

Multiplying both sides by $(-16)$ (see "sometimes" restriction above)

$3x - 64y = 12.8$
(which is in the requested "standard form")

How do you write 4y = 3/16x - 0.84y = 3/16x - 0.8 in standard form?