How do you write the linear equation $- 12 x + 8 y = - 32$ $-12x+8y=-32$ in slope-intercept form?

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Answer 1 · 2016-12-25T11:50:41

The equation of the line $- 12 x + 8 y = - 32$ $-12 x + 8 y = - 32$ is expressed as $y = \frac{3}{2} x - 4$ $y = 3/2 x - 4$ in slope-intercept form.

Just clear the $y$ $"y"$ :

$- 12 x + 8 y = - 32 \Rightarrow 8 y = 12 x - 32 \Rightarrow$ $- 12 x + 8 y = - 32 rArr 8 y = 12 x - 32 rArr$

$\Rightarrow y = \frac{12 x - 32}{8} \Rightarrow y = \frac{12}{8} x - \frac{32}{8} \Rightarrow$ $rArr y = {12 x - 32}/8 rArr y = 12/8 x - 32/8 rArr$

$\Rightarrow y = \frac{3}{2} x - 4$ $rArr y = 3/2 x - 4$ .

How do you write the linear equation −12x+8y=−32-12x+8y=-32 in slope-intercept form?