SOLUTION: 3y2+13y+14/y 2+3y+2*2y2-3y-2/3y2+22+35 cant figure out how to multiply to find the answer

Question 414534: 3y2+13y+14/y 2+3y+2*2y2-3y-2/3y2+22+35
cant figure out how to multiply to find the answer
Answer by woolybear(13) (Show Source): You can put this solution on YOUR website!
3y^(2)+13y+(14)/(y^(2))+3y+2*2y^(2)-3y-(2)/(3)*y^(2)+22+35
Multiply 2 by 2y^(2) to get 4y^(2).
3y^(2)+13y+(14)/(y^(2))+3y+4y^(2)-3y-(2)/(3)*y^(2)+22+35
Multiply -(2)/(3) by y^(2) to get -(2y^(2))/(3).
3y^(2)+13y+(14)/(y^(2))+3y+4y^(2)-3y-(2y^(2))/(3)+22+35
Since 3y^(2) and 4y^(2) are like terms, add 4y^(2) to 3y^(2) to get 7y^(2).
7y^(2)+13y+(14)/(y^(2))+3y-3y-(2y^(2))/(3)+22+35
Since 13y and 3y are like terms, add 3y to 13y to get 16y.
7y^(2)+16y+(14)/(y^(2))-3y-(2y^(2))/(3)+22+35
Since 16y and -3y are like terms, add -3y to 16y to get 13y.
7y^(2)+13y+(14)/(y^(2))-(2y^(2))/(3)+22+35
Add 35 to 22 to get 57.
7y^(2)+13y+(14)/(y^(2))-(2y^(2))/(3)+57
To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is 3y^(2). Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.
7y^(2)*(3y^(2))/(3y^(2))-(2y^(2))/(3)*(y^(2))/(y^(2))+13y*(3y^(2))/(3y^(2))+(14)/(y^(2))*(3)/(3)+57*(3y^(2))/(3y^(2))
Complete the multiplication to produce a denominator of 3y^(2) in each expression.
(21y^(4))/(3y^(2))-(2y^(4))/(3y^(2))+(39y^(3))/(3y^(2))+(42)/(3y^(2))+(171y^(2))/(3y^(2))
Combine the numerators of all expressions that have common denominators.
(21y^(4)-2y^(4)+39y^(3)+42+171y^(2))/(3y^(2))
Combine all like terms in the numerator.
(19y^(4)+39y^(3)+171y^(2)+42)/(3y^(2))
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