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)

1 Answer
Jan 22, 2018

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