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2 Answers

Hi Darian,
 
1/3(8y-2)+1/9(12y+10)
 
By using distributive property,we get
 
8y/3-2/3+12y/9+10/9
 
8y/3+4y/3-2/3+10/9
 
12y/3-6/9+10/9
4y+4/9
 
4(y+1/9)
Hi Darian;
[(1/3)(8y-2)]+[(1/9)(12y+10)]
We would like the denominator in both bracketed equations to be the same.
Let's multiply the first bracketed equation by 3/3
[(3/3)(1/3)(8y-2)]+[(1/9)(12y+10)]
[(3/9)(8y-2)]+[(1/9)(12y+10)]
[(24y/9)-(6/9)]+[(12y/9)+(10/9)]
(36y/9)+(4/9)
4y+(4/9)
or
4[y+(1/9)]