1/3x + 3y = -6: find what y is (ans supposed to be -1/9x-2)

[victor251]

1/3x+3y= -6, find what y is. (the1/3 is in fraction form)
The answer in suposto be -1/9x-2 (1/9 is in fraction form)
This is what I got:
x/3+3y= -6
minus x/3 from left side then minus x/3 from right side
3y=-6-x/3
then I divided 3 from 3y and divided 3 from the right side too
y=-6-x/3 (----------------suposto look like a fraction
3 dividing line between -6-x/6 and 3

I dont know what I did wrong in the question pls help

 

[Deleted member 4993]

Re: help me please

1/3x+3y= -6, find what y is. (the1/3 is in fraction form)
The answer in suposto be -1/9x-2 (1/9 is in fraction form)
This is what I got:
x/3+3y= -6
minus x/3 from left side then minus x/3 from right side
3y=-6-x/6 <--- this should be 3y = -6 - x/3
then I divided 3 from 3y and divided 3 from the right side too
y=-6-x/6 (----------------suposto look like a fraction
3 dividing line between -6-x/6 and 3

I dont know what I did wrong in the question pls help

 

[victor251]

Re: help me please

oh yeah my bad ill change that

 

[Denis]

1/3x+3y= -6, find what y is. (the1/3 is in fraction form)

Next time, show fractions like: (1/3)x or x/3

Always try to eliminate the fraction(s) as 1st step; multiply each term by 3:
x + 9y = -18

OK?