How do you find the slope and intercept of $12 x + 4 y - 2 = 0$ $12x+4y-2=0$ ?

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Answer 1 · 2018-04-19T03:13:43

Slope $= m = - 3$ $=m=-3$

$y -$ $y-$ intercept $= b = \frac{1}{2}$ $=b=1/2$

Solve the equation for $y :$ $y:$

$12 x + 4 y - 2 = 0$ $12x+4y-2=0$

Add $2$ $2$ to both sides:

$12 x + 4 y = 2$ $12x+4y=2$

Subtract $12 x$ $12x$ from both sides:

$4 y = 2 - 12 x$ $4y=2-12x$

Divide $4$ $4$ from both sides:

$y = \frac{2 - 12 x}{4}$ $y=(2-12x)/4$

Simplify:

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

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

This last equation is now in $y = m x + b$ $y=mx+b$ form where

$m$ $m$ is the slope (the coefficient of the variable $x$ $x$ )

$b$ $b$ is the $y -$ $y-$ intercept or the constant

From the equation:

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

Slope $= m = - 3$ $=m=-3$

$y -$ $y-$ intercept $= b = \frac{1}{2}$ $=b=1/2$

How do you find the slope and intercept of 12x+4y-2=012x+4y−2=012x+4y-2=0?