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:
Found 3 solutions by Aswathy, lwsshak3, ccs2011:
Answer by Aswathy(23) (Show Source): You can put this solution on YOUR website!
(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) (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) (Show Source): You can put this solution on YOUR website!
factor denominators
Combine fractions, get common denominator by multiplying by
Distribute and add like terms in numerator
Factor numerator
Cancel
Multiply by (y+3) on both sides
Subtract 3 on both sides
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