How do you subtract #\frac { 9y } { y ^ { 2} - 25} - \frac { 8} { y + 5}#?

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Answer 1 · 2017-05-31T16:20:47

#(y+40)/(y^2-25)#

difference of two squares:
#(a+b)(a-b) = a^2 - b^2#

#(y+5)(y-5) = y^2 - 5^2 = y^2-25#

common denominator: #y^2-25#
#y^2-25=(y^2-25)*1#
#y^2-25=(y-5)*(y+5)#

#(9y)/(y^2-25) = (9y)/(y^2-25)#

#8/(y+5) = (8(y-5))/(y+5)(y-5)=(8(y-5))/(y^2-25)#

#(9y)/(y^2-25)-8/(y+5) = (9y)/(y^2-25)-(8(y-5))/(y^2-25)#

#(9y)/(y^2-25)-(8(y-5))/(y^2-25) = (9y - 8(y-5))/(y^2-25)#

#=(9y-8y+40)/(y^2-25)#

#=(y+40)/(y^2-25)#