How do you find the slope and intercept of $3 y + 5 = 0$ $3y+5=0$ ?

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Answer 1 · 2018-03-27T17:46:39

slope $= 0; y - intercept = (0, - \frac{5}{3})$ $= 0; " " y-"intercept" = (0, -5/3)$

Given: $3 y + 5 = 0$ $3y + 5 = 0$

Solve for $y$ $y$ :
$3 y = - 5$ $" "3y = -5$

$y = - \frac{5}{3}$ $" " y = -5/3$

A line has the form: $y = m x + b$ $y = mx + b$ ;

where $m$ $m$ = slope, and $b$ $b$ is the $y$ $y$ -intercept: $(0, b)$ $(0, b)$

In the line equation $y = - \frac{5}{3}$ $y = -5/3$ ; $m = 0$ $" "m = 0$

This means the line is a horizontal line at $y = - \frac{5}{3}$ $y = -5/3$

The $y$ $y$ -intercept $= (0, - \frac{5}{3})$ $= (0, -5/3)$

How do you find the slope and intercept of 3y+5=03y+5=03y+5=0?