How do you write the equation in slope intercept form given x-3y=6?

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Answer 1 · 2017-04-08T17:13:48

$y = \frac{1}{3} x - 2$ $y = 1/3x - 2$

Given: $x - 3 y = 6$ $x - 3y = 6$

Put in slope-intercept form $y = m x + b$ $y = mx + b$ :

Add $3 y$ $3y$ to both sides of the equation to make it positive:

$x - 3 y + 3 y = 6 + 3 y$ $x - 3y + 3y = 6 + 3y$
$x = 6 + 3 y$ $x = 6 + 3y$

Subtract $6$ $6$ from both sides of the equation:

$x - 6 = 6 - 6 + 3 y$ $x - 6 = 6 - 6 + 3y$
$x - 6 = 3 y$ $x - 6 = 3y$

Divide both sides of the equation by $3$ $3$ :

$\frac{x}{3} - \frac{6}{3} = \frac{3 y}{3}$ $x/3 - 6/3 = (3y)/3$
$\frac{1}{3} x - 2 = y$ $1/3x - 2 = y$

So $y = \frac{1}{3} x - 2$ $y = 1/3x - 2$