SOLUTION: Please help me solve this equation: {{{5/(y-3) - 30/(y^2-9)=1}}}

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Question 476228: Please help me solve this equation: 5%2F%28y-3%29+-+30%2F%28y%5E2-9%29=1
Found 3 solutions by Aswathy, lwsshak3, ccs2011:
Answer by Aswathy(23) About Me  (Show Source):
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5%2F%28y-3%29+-+30%2F%28y%5E2-9%29=1
(5/y-3)-{30/(y+3)(y-3)}=1 {By Algebraic identity (a^2-b^2)=(a+a)(a-b)}
(5y+15-30)/{(y+3)(y-3)}=1
(5y-15)/{(y+3)(y-3)}=1
{{5(y-3)}/{(y+3)(y-3)}}=1
(5)/(y+3)=1 (By cancelling out y-3 in numerator and denominator)
5=y+3
5-3=y
2=y
y=2

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
solve this equation: 5/(y-3) - 30/(y^2-9)=1
**
5/(y-3)-30/(y^2-9)=1
5/(y-3)-30/(y+3)(y-3)=1
LCD:(y+3)(y-3)
5(y+3)-30=(y+3)(y-3)
5y+15-30=y^2-9
y^2-5y+6=0
(y-2)(y-3)=0
y=2
or
y=3 (reject, (y-3)≠0

Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
5%2F%28y-3%29+-+30%2F%28y%5E2-9%29=1
factor denominators
5%2F%28y-3%29+-+30%2F%28%28y%2B3%29%28y-3%29%29=1
Combine fractions, get common denominator by multiplying by %28y%2B3%29%2F%28y%2B3%29
5%28y%2B3%29%2F%28%28y%2B3%29%28y-3%29%29+-+30%2F%28%28y%2B3%29%28y-3%29%29=1
%285%28y%2B3%29+-+30%29%2F%28%28y%2B3%29%28y-3%29%29+=+1
Distribute and add like terms in numerator
%285y+-+15%29%2F%28%28y%2B3%29%28y-3%29%29=+1
Factor numerator
5%28y-3%29%2F%28%28y%2B3%29%28y-3%29%29=+1
Cancel
5%2F%28y%2B3%29=+1
Multiply by (y+3) on both sides
5+=+y%2B3
Subtract 3 on both sides
2+=+y