How to solve the following for #x#? : 1) #x/(a-b) + 2x/(a+b) =1/(a^2 - b^2)# 2) #1/(x+a) + 1/(x+2a) = 2/(x+3a)#

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Answer 1 · 2018-01-22T04:56:43

1) #x/(a-b)+(2x)/(a+b)=1/(a^2-b^2)#

Multiplying both sides by #(a^2-b^2)# we get

#x/(a-b)(a^2-b^2)+(2x)/(a+b)(a^2-b^2)=1/(a^2-b^2)(a^2-b^2)#

#=>x(a+b)+2x(a-b)=1#

#=>x(a+b+2a-2b)=1#

#=>x(3a-b)=1#

#=>x=1/(3a-b)#

2) #1/(x+a)+1/(x+2a)=2/(x+3a)#

#=>1/(x+a)-1/(x+3a)=1/(x+3a)-1/(x+2a)#

#=>(x+3a-x-a)/((x+a)(x+3a))=(x+2a-x-3a)/((x+3a)(x+2a))#

#=>(2a)/((x+a)(x+3a))=(-a)/((x+3a)(x+2a))#
As #x+3a!=0#
#=>2/(x+a)=-1/(x+2a)#

#=>2x+4a=-x-a#

#=>2x+x=-a-4a#

#=>3x=-5a#

#=>x=-(5a)/3#