How do you write $y + 3 = - \frac{1}{3} (2 x + 6)$ $y+3=-1/3(2x+6)$ in slope intercept form?

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Answer 1 · 2017-04-29T20:18:48

See the solution process below:

The slope-intercept form of a linear equation is: $y = m x + b$ $y = color(red)(m)x + color(blue)(b)$

Where $m$ $color(red)(m)$ is the slope and $b$ $color(blue)(b)$ is the y-intercept value.

Solving this for $y$ $y$ gives:

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

$y + 3 = (- \frac{1}{3} \cdot 2 x) + (- \frac{1}{3} \cdot 6)$ $y + 3 = (color(red)(-1/3) * 2x) + (color(red)(-1/3) * 6)$

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

$y + 3 = - \frac{2}{3} x - \frac{6}{3}$ $y + 3 = -2/3x - 6/3$

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

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

$y + 0 = - \frac{2}{3} x - 5$ $y + 0 = -2/3x - 5$

$y = - \frac{2}{3} x - 5$ $y = color(red)(-2/3)x - color(blue)(5)$

How do you write y+3=−13(2x+6)y+3=-1/3(2x+6) in slope intercept form?