SOLUTION: Solve for Y: (2)/(3)+3y=5y-(2)/(15)
Your supposed to eliminate the fractions first. I keep coming up with 5/2 even though I know the answer is 2/5. I just cant see what I'm doi
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-> SOLUTION: Solve for Y: (2)/(3)+3y=5y-(2)/(15)
Your supposed to eliminate the fractions first. I keep coming up with 5/2 even though I know the answer is 2/5. I just cant see what I'm doi
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Question 518996: Solve for Y: (2)/(3)+3y=5y-(2)/(15)
Your supposed to eliminate the fractions first. I keep coming up with 5/2 even though I know the answer is 2/5. I just cant see what I'm doing wrong
I first eliminate 2/3 by multiplying both sides by 3/2. This gives me:
3y=5y-1/5
Move the 5y over and get -2y
multiply by the recip -1/2
that gives me 5/2! Grrr Found 2 solutions by scott8148, jessica43:Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website! So you were correct in working with the fractions first, however, you cannot just multiply both sides by 3/2. Multiply by 3/2 would get rid of the 2/3, but it would also multiply both the 3y and the 5y by 3/2, which would make the problem even more complicated.
Instead, start by rewriting the fractions with the same denominator so you can easily add or subtract them from each other. In this case:
2/3 = 10/15
So now you have:
10/15 + 3y = 5y - 2/15
12/15 + 3y = 5y (added 2/15 to each side)
12/15 = 2y (subtracted 3y from each side)
now simplify the fraction
12/15 = 4/5
so you have:
4/5 = 2y
4/10 = y
And since 4/10 = 2/5, you have the answer you thought you would!
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