How do you solve #a/(a-5)+2=(3a)/(a+5)#?

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Answer 1 · 2015-05-24T13:16:27

Given:

#a/(a-5)+2=(3a)/(a+5)#

Multiply through by #(a^2-25) = (a-5)(a+5)# to get:

#a(a+5)+2(a^2-25) = 3a(a-5) = 3a^2-15a#

Subtract #3a^2-15a# from both sides to get:

#0 = a(a+5)+2(a^2-25) - 3a^2+15a#

#=a^2+5a+2a^2-50-3a^2+15a#

#=(a^2+2a^2-3a^2)+(5a+15a)-50#

#=20a-50#

Add #50# to both sides to get

#20a=50#

Divide both sides by #50# to get:

#a=5/2#

Check by substituting back in the original equation:

LHS #= a/(a-5)+2 = (5/2)/(5/2-5)+2#

#=5/(5-10)+2 = 5/-5+2 = -1 + 2 = 1#

RHS #= (3a)/(a+5) = (3(5/2))/(5/2+5) = (3(5/2))/(3(5/2)) = 1#