Answer :
we have
[tex]2x+3y=-12[/tex]
Isolate variable y
[tex]2x+3y=-12[/tex]
Subtract [tex]2x[/tex] both sides
[tex]3y=-2x-12[/tex]
Divide by [tex]3[/tex] both sides
[tex]y=-\frac{2}{3}x-4[/tex]
the slope m is equal to
[tex]m=-\frac{2}{3}[/tex]
therefore
the answer is the option B
[tex]-\frac{2}{3}[/tex]
The slope of the line is represented by the equation 2x+3y= -12 and 2x+3y= -12 is [tex]\frac{-2}{3}[/tex].
Given that,
2x + 3y = -12
Now here we isolated the variable y
2x + 3y = -12
Now subtract 2x from both sides
3y = -2x - 12
Now divide by 3 in both the sides.
[tex]y = \frac{-2}{3}x - 4[/tex]
Now
The slope m should be [tex]\frac{-2}{3}[/tex].
Therefore we can conclude that the slope of the line is represented by the equation 2x+3y= -12 and 2x+3y= -12 is [tex]\frac{-2}{3}[/tex].
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