How do you write the slope intercept form of the line $3x-2y=-16$ ?

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Answer 1 · 2017-05-21T00:03:34

$y = 3/2x + 8$

Simply make y the subject of the equation:

$-2y = -16 - 3x$

$y = 8 + 3/2x$

$y = 3/2x + 8$

Answer 2 · 2017-05-21T00:04:04

$y=3/2x+8$

Slope intercept form is defined as $y=mx+b$ where $m$ is the slope an $b$ is the $y$ -intercept.

In this case we have $3x-2y=-16$ . To write this in slope intercept form we must solve for $y$ . Thus,

Subtract $3x$ to both sides:

$cancel(3x-3x)-2y=-3x-16$

$-2y=-3x-16$

Divide $-2$ to both sides:

$cancel(-2/-2)y=(-3x-16)/-2$

Simplify:

$y=3/2x+8$

This is now written in slope intercept form where $m$ (the slope) is $3/2$ and our $y$ -intercept is $8$

How do you write the slope intercept form of the line 3x-2y=-163x-2y=-16?